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bg-background","children":["$","$L3a",null,{"backHref":"/a-level/maths/edexcel/questions","basePath":"/a-level/maths/edexcel/a-level-pure-mathematics-8-3-curve-sketching/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/a-level-maths/edexcel/formula-book.pdf","initialIndex":2,"initialQuestionWindow":[{"id":"100c6223-1680-4a86-990e-d2ad9bf0528d","revisionId":"e0f47d68-4ddb-4bb2-843a-fe7b9eb8b33a","questionNumber":null,"questionType":"LA","specification":"A curve $K$ has parametric equations\n\n$$x = 3 \\sin t + 2, \\quad y = 2 \\cos^2 t - 5 \\quad 0 \\leq t \\leq 2\\pi$$","difficulty":1,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"d4ca8f1f-5461-4650-9ac2-fa339b43d9ee","label":"a","content":"Find an equation for $K$ in the form $y = f(x)$ and state the domain for which the curve is defined.","markscheme":"- $\\sin t = \\frac{x-2}{3} \\Rightarrow \\sin^2 t = \\frac{(x-2)^2}{9}$ M1\n- $y = 2(1 - \\sin^2 t) - 5 = 2(1 - \\frac{(x-2)^2}{9}) - 5 \\Rightarrow y = -3 - \\frac{2}{9}(x-2)^2$ A1\n- Domain: $-1 \\leq \\sin t \\leq 1 \\Rightarrow -3 \\leq 3\\sin t \\leq 3 \\Rightarrow -1 \\leq x \\leq 5$ B1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Attempt to express $\\sin^2 t$ in terms of $x$ and substitute into the equation for $y$ using the identity $\\cos^2 t + \\sin^2 t = 1$.","code":"M"},{"id":"p2","marks":1,"description":"Correct Cartesian equation in the form $y = f(x)$.","code":"A"},{"id":"p3","marks":1,"description":"Correct domain $-1 \\leq x \\leq 5$ identified from the range of the $x$ parametric equation.","code":"B"}],"finalAnswers":["$$y = -3 - \\frac{2}{9}(x-2)^2, \\quad -1 \\leq x \\leq 5$"],"generatedAt":"2026-05-06T14:05:21.991Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Rearrange for sin(t)","content":"To express $y$ in terms of $x$, we begin by isolating the trigonometric term in our equation for $x$.\n\nThe parametric equation for $x$ is given by:\n$$x = 3 \\sin t + 2$$\n\nSubtract $2$ from both sides and divide by $3$ to find an expression for $\\sin t$:\n$$\\sin t = \\frac{x - 2}{3}$$\n\nSquaring both sides gives us our expression for $\\sin^2 t$:\n$$\\sin^2 t = \\frac{(x - 2)^2}{9}$$"},{"type":"text","order":1,"title":"Rewrite y using identity","content":"Now we look at our parametric equation for $y$:\n$$y = 2 \\cos^2 t - 5$$\n\nRecall the fundamental trigonometric identity relating sine and cosine:\n$$\\cos^2 t + \\sin^2 t = 1 \\implies \\cos^2 t = 1 - \\sin^2 t$$\n\nSubstituting this identity into our equation for $y$ yields:\n$$y = 2(1 - \\sin^2 t) - 5$$\n\nSimplify this expression:\n$$y = 2 - 2\\sin^2 t - 5$$\n$$y = -3 - 2\\sin^2 t$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin^2 t = \\htmlData{anchor=sin-val}{\\color{@sky}{\\frac{(x - 2)^2}{9}}}$"},{"latex":"$$y = -3 - 2\\htmlData{anchor=sin-sub}{\\color{@sky}{\\sin^2 t}}$"},{"latex":"$$y = -3 - \\frac{2}{9}(x - 2)^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-val"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"sin-sub","from":"sin-val","color":"sky","label":"substitute","style":"solid"}]},"order":2,"title":"Substitute and simplify","description":"Now, we substitute our expression for $\\sin^2 t$ in terms of $x$ into our rewritten equation for $y$."},{"type":"text","order":3,"title":"Determine the domain","content":"To complete the solution, we must define the domain for $x$. The domain of the curve is determined by the range of the parametric variable $t$, where $0 \\leq t \\leq 2\\pi$.\n\nOver this interval, the value of the sine function is bounded as follows:\n$$-1 \\leq \\sin t \\leq 1$$\n\nTo find the boundaries of $x$, we multiply the inequality by $3$:\n$$-3 \\leq 3\\sin t \\leq 3$$\n\nThen, we add $2$ to all parts of the inequality:\n$$-3 + 2 \\leq 3\\sin t + 2 \\leq 3 + 2$$\n$$-1 \\leq x \\leq 5$$\n\nTherefore, the domain for the curve is $[-1, 5]$."}]},{"id":"pythagorean_identity","label":"Pythagorean Identity Method","steps":[{"type":"text","order":0,"title":"Rearrange the x equation","content":"To eliminate the parameter $t$, we first rearrange the parametric equation for $x$ to express $\\sin t$ and then $\\sin^2 t$ in terms of $x$.\n\nStarting with:\n$$x = 3 \\sin t + 2$$\n\nSubtract $2$ from both sides:\n$$x - 2 = 3 \\sin t$$\n\nDivide by $3$:\n$$\\sin t = \\frac{x - 2}{3}$$\n\nSquaring both sides gives:\n$$\\sin^2 t = \\frac{(x - 2)^2}{9}$$"},{"type":"text","order":1,"title":"Rearrange the y equation","content":"Next, we rearrange the parametric equation for $y$ to express $\\cos^2 t$ in terms of $y$.\n\nStarting with:\n$$y = 2 \\cos^2 t - 5$$\n\nAdd $5$ to both sides:\n$$y + 5 = 2 \\cos^2 t$$\n\nDivide by $2$:\n$$\\cos^2 t = \\frac{y + 5}{2}$$\n\nThis gives us our expression for $\\cos^2 t$ directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=identity}{\\color{@purple}{\\sin^2 t + \\cos^2 t = 1}}$"},{"latex":"$$\\htmlData{anchor=sub}{\\color{@sky}{\\frac{(x - 2)^2}{9}}} + \\htmlData{anchor=suby}{\\color{@amber}{\\frac{y + 5}{2}}} = 1$"}],"highlights":[{"color":"purple","style":"underline","anchor":"identity"},{"color":"sky","style":"box","anchor":"sub"},{"color":"amber","style":"box","anchor":"suby"}],"connections":[{"to":"sub","from":"identity","color":"purple","label":"substitute","style":"dashed"}]},"order":2,"title":"Apply the Pythagorean identity","description":"We now use the fundamental Pythagorean identity $\\sin^2 t + \\cos^2 t = 1$ to link our two expressions and eliminate $t$."},{"type":"text","order":3,"title":"Solve for y","content":"We can now rearrange this equation to express $y$ in the form $f(x)$:\n\n$$\\frac{y + 5}{2} = 1 - \\frac{(x - 2)^2}{9}$$\n\nMultiply both sides by $2$:\n$$y + 5 = 2\\left(1 - \\frac{(x - 2)^2}{9}\\right)$$\n\n$$y + 5 = 2 - \\frac{2}{9}(x - 2)^2$$\n\nSubtract $5$ from both sides:\n$$y = -3 - \\frac{2}{9}(x - 2)^2$$"},{"type":"text","order":4,"title":"Determine the domain","content":"To find the domain of $f(x)$, we use the restriction on the sine function. Since $\\sin t$ is defined for all real $t$, its range is:\n$$-1 \\le \\sin t \\le 1$$\n\nSubstitute $\\sin t = \\frac{x - 2}{3}$ into the inequality:\n$$-1 \\le \\frac{x - 2}{3} \\le 1$$\n\nMultiply the entire inequality by $3$:\n$$-3 \\le x - 2 \\le 3$$\n\nAdd $2$ to all parts:\n$$-1 \\le x \\le 5$$\n\nThus, the domain for which the curve is defined is $[-1, 5]$."}]}],"generatedAt":"2026-06-09T03:19:56.386Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"add0cbb1-5b08-467e-aad9-ac56e8a1b982","label":"b","content":"Sketch the curve","markscheme":"- Negative (inverted) parabola shape B1\n- Vertex at $(2, -3)$ B1\n- Curve starts at $(-1, -5)$ and ends at $(5, -5)$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct shape for a negative parabola within the defined domain.","code":"B"},{"id":"p2","marks":1,"description":"Correct coordinates for the vertex $(2, -3)$ indicated on the sketch.","code":"B"},{"id":"p3","marks":1,"description":"Correct endpoints at $(-1, -5)$ and $(5, -5)$ shown clearly.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:05:21.991Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"cartesian_conversion","label":"Cartesian Equation Method","steps":[{"type":"text","order":0,"title":"Express sin(t) in terms of x","content":"To find the Cartesian equation of the curve, we first isolate the trigonometric term $\\sin t$ from the parametric equation for $x$:\n\n$$x = 3 \\sin t + 2$$\n\nSubtracting $2$ and then dividing by $3$ gives:\n\n$$\\sin t = \\frac{x - 2}{3}$$"},{"type":"text","order":1,"title":"Apply trigonometric identity to y","content":"Next, we use the fundamental trigonometric identity:\n\n$$\\cos^2 t = 1 - \\sin^2 t$$\n\nSubstituting this identity into our parametric equation for $y$:\n\n$$y = 2 \\cos^2 t - 5$$\n\n$$y = 2(1 - \\sin^2 t) - 5$$\n\n$$y = -2\\sin^2 t - 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin t = \\htmlData{anchor=sin-expr}{\\color{@sky}{\\frac{x - 2}{3}}}$"},{"latex":"$$y = -2 (\\htmlData{anchor=sin-sub}{\\color{@sky}{\\sin t}})^2 - 3$"},{"latex":"$$y = -\\frac{2}{9}(x - 2)^2 - 3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-expr"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"sin-sub","from":"sin-expr","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Substitute to get Cartesian equation","description":"Now, we substitute our expression for $\\sin t$ into the simplified equation for $y$ to get the final Cartesian equation."},{"type":"text","order":3,"title":"Determine domain and endpoints","content":"Since $0 \\le t \\le 2\\pi$, the range of $\\sin t$ is restricted:\n\n$$-1 \\le \\sin t \\le 1$$\n\nWe can find the domain restriction for $x$ by substituting these limits into $x = 3 \\sin t + 2$:\n\n- Minimum value: when $\\sin t = -1$, $x = 3(-1) + 2 = -1$\n- Maximum value: when $\\sin t = 1$, $x = 3(1) + 2 = 5$\n\nSo, the domain restriction is $-1 \\le x \\le 5$.\n\nAt both endpoints $x = -1$ and $x = 5$, the corresponding $y$-value is:\n\n$$y = -\\frac{2}{9}(\\pm 3)^2 - 3 = -2 - 3 = -5$$\n\nThis gives the end coordinates of our curve: $(-1, -5)$ and $(5, -5)$."},{"type":"text","order":4,"title":"Identify the vertex","content":"The Cartesian equation is in the vertex form of a quadratic, $y = a(x-h)^2 + k$:\n\n$$y = -\\frac{2}{9}(x - 2)^2 - 3$$\n\n- Since the coefficient $-\\frac{2}{9}$ is negative, the parabola is **inverted** (opening downwards).\n- The vertex $(h, k)$ is located at $(2, -3)$, which is the maximum point of this inverted parabola."},{"type":"text","order":5,"title":"Sketch the curve","content":"Now we sketch the curve using our key features:\n\n1. Plot the vertex at $(2, -3)$ as a maximum point.\n2. Plot the endpoints at $(-1, -5)$ and $(5, -5)$.\n3. Draw a smooth, symmetric, inverted parabola shape connecting these points. Ensure your curve does not extend beyond $x = -1$ or $x = 5$.","calculator":[{"gdc":{"instructions":"1. Change your graphing mode to Parametric.\n2. Enter the equations: x(t) = 3*sin(t) + 2 and y(t) = 2*cos(t)^2 - 5.\n3. Set the domain variable t from 0 to 6.28.\n4. Set an appropriate window size to view the interval [-1, 5] on the x-axis and [-6, -2] on the y-axis."},"ti84":{"instructions":"1. Press [MODE], arrow down to 'FUNC', change it to 'PARAMETRIC' (PAR), and press [ENTER].\n2. Press [Y=] and enter:\n X1T = 3 * sin(T) + 2\n Y1T = 2 * (cos(T))^2 - 5\n3. Press [WINDOW] and set:\n Tmin = 0, Tmax = 6.28, Tstep = 0.1\n Xmin = -3, Xmax = 7, Xscl = 1\n Ymin = -7, Ymax = -1, Yscl = 1\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"(3*sin(t) + 2, 2*(cos(t))^2 - 5)","instructions":"1. Type the coordinate pair representing the parametric equations in a single line:\n (3*sin(t) + 2, 2*(cos(t))^2 - 5)\n2. Set the parameter interval beneath it to: 0 <= t <= 2pi."},"description":"Visualize the parametric curve to verify its shape, vertex, and domain endpoints."}]}]},{"id":"point_plotting","label":"Point-by-Point Plotting","steps":[{"type":"text","order":0,"title":"Set up a coordinate table","content":"To sketch the parametric curve using a point-by-point plotting approach, we calculate the coordinates $(x, y)$ for key values of $t$ in the interval $0 \\le t \\le 2\\pi$.\n\nLet's choose the standard angles:\n$$t = 0, \\quad \\frac{\\pi}{2}, \\quad \\pi, \\quad \\frac{3\\pi}{2}, \\quad 2\\pi$$\n\nOur parametric equations are:\n$$x = 3 \\sin t + 2$$\n$$y = 2 \\cos^2 t - 5$$"},{"type":"text","order":1,"title":"Evaluate points for t","content":"Let's calculate the coordinates $(x, y)$ for each chosen value of $t$:\n\n1. **For $t = 0$:**\n $$x = 3 \\sin(0) + 2 = 2$$\n $$y = 2 \\cos^2(0) - 5 = 2(1)^2 - 5 = -3$$\n Point: $(2, -3)$\n\n2. **For $t = \\frac{\\pi}{2}$:**\n $$x = 3 \\sin\\left(\\frac{\\pi}{2}\\right) + 2 = 3(1) + 2 = 5$$\n $$y = 2 \\cos^2\\left(\\frac{\\pi}{2}\\right) - 5 = 2(0)^2 - 5 = -5$$\n Point: $(5, -5)$\n\n3. **For $t = \\pi$:**\n $$x = 3 \\sin(\\pi) + 2 = 2$$\n $$y = 2 \\cos^2(\\pi) - 5 = 2(-1)^2 - 5 = -3$$\n Point: $(2, -3)$\n\n4. **For $t = \\frac{3\\pi}{2}$:**\n $$x = 3 \\sin\\left(\\frac{3\\pi}{2}\\right) + 2 = 3(-1) + 2 = -1$$\n $$y = 2 \\cos^2\\left(\\frac{3\\pi}{2}\\right) - 5 = 2(0)^2 - 5 = -5$$\n Point: $(-1, -5)$\n\n5. **For $t = 2\\pi$:**\n $$x = 3 \\sin(2\\pi) + 2 = 2$$\n $$y = 2 \\cos^2(2\\pi) - 5 = 2(1)^2 - 5 = -3$$\n Point: $(2, -3)$"},{"type":"text","order":2,"title":"Summarise coordinate table","content":"Let's compile these evaluated points into a clear summary table:\n\n$$\\begin{array}{|c|c|c|c|} \\hline t & x = 3\\sin t + 2 & y = 2\\cos^2 t - 5 & (x, y) \\\\ \\hline 0 & 2 & -3 & (2, -3) \\\\ \\hline \\frac{\\pi}{2} & 5 & -5 & (5, -5) \\\\ \\hline \\pi & 2 & -3 & (2, -3) \\\\ \\hline \\frac{3\\pi}{2} & -1 & -5 & (-1, -5) \\\\ \\hline 2\\pi & 2 & -3 & (2, -3) \\\\ \\hline \\end{array}$$\n\nNotice that as $t$ goes from $0$ to $2\\pi$:\n- The curve starts at the vertex $(2, -3)$.\n- It moves to the right endpoint $(5, -5)$ at $t = \\frac{\\pi}{2}$.\n- It returns to the vertex $(2, -3)$ at $t = \\pi$.\n- It moves to the left endpoint $(-1, -5)$ at $t = \\frac{3\\pi}{2}$.\n- It returns to the vertex $(2, -3)$ at $t = 2\\pi$."},{"type":"text","order":3,"title":"Analyze the shape","content":"Let's understand the underlying algebraic shape to draw the sketch accurately.\n\nFrom our table, we can see:\n- The maximum value of $y$ is $-3$, which occurs when $x = 2$. This indicates a peak or vertex at $(2, -3)$.\n- The minimum value of $y$ is $-5$, which occurs at the boundary points $x = -1$ and $x = 5$.\n- The coordinates are symmetric about the vertical line $x = 2$.\n\nThis behavior matches an inverted (negative) parabola opening downwards, with its vertex at $(2, -3)$ and endpoints at $(-1, -5)$ and $(5, -5)$."},{"type":"text","order":4,"title":"Verify with GDC","content":"We can easily verify the plotted path and the endpoints of the curve using a graphing calculator.","calculator":[{"gdc":{"instructions":"1. Set your graphing calculator's mode to Parametric equations.\n2. Define X1(t) = 3*sin(t) + 2 and Y1(t) = 2*cos(t)^2 - 5.\n3. Adjust the interval for t to [0, 6.28].\n4. Set a suitable window view (e.g., x from -2 to 6, y from -6 to -2).\n5. Plot the graph to verify the shape."},"ti84":{"instructions":"1. Press [MODE] and change the graphing mode to PAR (parametric).\n2. Press [Y=].\n3. Set X1T = 3 * sin(T) + 2 and Y1T = 2 * (cos(T))^2 - 5.\n4. Press [WINDOW] and set Tmin = 0, Tmax = 6.28, and Tstep = 0.1.\n5. Set Xmin = -2, Xmax = 6, Ymin = -6, Ymax = -2.\n6. Press [GRAPH] to view the curve."},"desmos":{"expression":"(3*sin(t) + 2, 2*cos(t)^2 - 5)","instructions":"1. Enter the parametric coordinate pair directly into an input row: (3*sin(t) + 2, 2*cos(t)^2 - 5).\n2. Set the parameter domain bounds next to the expression as 0 <= t <= 2*pi.\n3. Adjust the graph viewport to focus on the region around x in [-2, 6] and y in [-6, -2]."},"description":"Graph the parametric equations to visualize the negative parabola shape."}]},{"type":"text","order":5,"title":"Sketch the curve","content":"Now, sketch the curve on your axes:\n\n1. Plot the vertex at $(2, -3)$ as the highest point.\n2. Plot the left endpoint at $(-1, -5)$ and the right endpoint at $(5, -5)$.\n3. Draw a smooth, symmetric, downward-opening parabolic arc connecting $(-1, -5)$ to $(2, -3)$ and then to $(5, -5)$.\n\n**Marks awarded for:**\n- **B1**: Drawing an inverted (negative) parabola shape.\n- **B1**: Showing the vertex clearly located at $(2, -3)$.\n- **B1**: Showing the curve starting/ending at $(-1, -5)$ and $(5, -5)$ with no parts of the curve drawn outside this domain."}]}],"generatedAt":"2026-06-09T03:23:08.136Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"d4ca8f1f-5461-4650-9ac2-fa339b43d9ee:substitution","label":"(a) Substitution Method","steps":[{"type":"text","order":0,"title":"Rearrange for sin(t)","content":"To express $y$ in terms of $x$, we begin by isolating the trigonometric term in our equation for $x$.\n\nThe parametric equation for $x$ is given by:\n$$x = 3 \\sin t + 2$$\n\nSubtract $2$ from both sides and divide by $3$ to find an expression for $\\sin t$:\n$$\\sin t = \\frac{x - 2}{3}$$\n\nSquaring both sides gives us our expression for $\\sin^2 t$:\n$$\\sin^2 t = \\frac{(x - 2)^2}{9}$$"},{"type":"text","order":1,"title":"Rewrite y using identity","content":"Now we look at our parametric equation for $y$:\n$$y = 2 \\cos^2 t - 5$$\n\nRecall the fundamental trigonometric identity relating sine and cosine:\n$$\\cos^2 t + \\sin^2 t = 1 \\implies \\cos^2 t = 1 - \\sin^2 t$$\n\nSubstituting this identity into our equation for $y$ yields:\n$$y = 2(1 - \\sin^2 t) - 5$$\n\nSimplify this expression:\n$$y = 2 - 2\\sin^2 t - 5$$\n$$y = -3 - 2\\sin^2 t$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin^2 t = \\htmlData{anchor=sin-val}{\\color{@sky}{\\frac{(x - 2)^2}{9}}}$"},{"latex":"$$y = -3 - 2\\htmlData{anchor=sin-sub}{\\color{@sky}{\\sin^2 t}}$"},{"latex":"$$y = -3 - \\frac{2}{9}(x - 2)^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-val"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"sin-sub","from":"sin-val","color":"sky","label":"substitute","style":"solid"}]},"order":2,"title":"Substitute and simplify","description":"Now, we substitute our expression for $\\sin^2 t$ in terms of $x$ into our rewritten equation for $y$."},{"type":"text","order":3,"title":"Determine the domain","content":"To complete the solution, we must define the domain for $x$. The domain of the curve is determined by the range of the parametric variable $t$, where $0 \\leq t \\leq 2\\pi$.\n\nOver this interval, the value of the sine function is bounded as follows:\n$$-1 \\leq \\sin t \\leq 1$$\n\nTo find the boundaries of $x$, we multiply the inequality by $3$:\n$$-3 \\leq 3\\sin t \\leq 3$$\n\nThen, we add $2$ to all parts of the inequality:\n$$-3 + 2 \\leq 3\\sin t + 2 \\leq 3 + 2$$\n$$-1 \\leq x \\leq 5$$\n\nTherefore, the domain for the curve is $[-1, 5]$."}]},{"id":"d4ca8f1f-5461-4650-9ac2-fa339b43d9ee:pythagorean_identity","label":"(a) Pythagorean Identity Method","steps":[{"type":"text","order":0,"title":"Rearrange the x equation","content":"To eliminate the parameter $t$, we first rearrange the parametric equation for $x$ to express $\\sin t$ and then $\\sin^2 t$ in terms of $x$.\n\nStarting with:\n$$x = 3 \\sin t + 2$$\n\nSubtract $2$ from both sides:\n$$x - 2 = 3 \\sin t$$\n\nDivide by $3$:\n$$\\sin t = \\frac{x - 2}{3}$$\n\nSquaring both sides gives:\n$$\\sin^2 t = \\frac{(x - 2)^2}{9}$$"},{"type":"text","order":1,"title":"Rearrange the y equation","content":"Next, we rearrange the parametric equation for $y$ to express $\\cos^2 t$ in terms of $y$.\n\nStarting with:\n$$y = 2 \\cos^2 t - 5$$\n\nAdd $5$ to both sides:\n$$y + 5 = 2 \\cos^2 t$$\n\nDivide by $2$:\n$$\\cos^2 t = \\frac{y + 5}{2}$$\n\nThis gives us our expression for $\\cos^2 t$ directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=identity}{\\color{@purple}{\\sin^2 t + \\cos^2 t = 1}}$"},{"latex":"$$\\htmlData{anchor=sub}{\\color{@sky}{\\frac{(x - 2)^2}{9}}} + \\htmlData{anchor=suby}{\\color{@amber}{\\frac{y + 5}{2}}} = 1$"}],"highlights":[{"color":"purple","style":"underline","anchor":"identity"},{"color":"sky","style":"box","anchor":"sub"},{"color":"amber","style":"box","anchor":"suby"}],"connections":[{"to":"sub","from":"identity","color":"purple","label":"substitute","style":"dashed"}]},"order":2,"title":"Apply the Pythagorean identity","description":"We now use the fundamental Pythagorean identity $\\sin^2 t + \\cos^2 t = 1$ to link our two expressions and eliminate $t$."},{"type":"text","order":3,"title":"Solve for y","content":"We can now rearrange this equation to express $y$ in the form $f(x)$:\n\n$$\\frac{y + 5}{2} = 1 - \\frac{(x - 2)^2}{9}$$\n\nMultiply both sides by $2$:\n$$y + 5 = 2\\left(1 - \\frac{(x - 2)^2}{9}\\right)$$\n\n$$y + 5 = 2 - \\frac{2}{9}(x - 2)^2$$\n\nSubtract $5$ from both sides:\n$$y = -3 - \\frac{2}{9}(x - 2)^2$$"},{"type":"text","order":4,"title":"Determine the domain","content":"To find the domain of $f(x)$, we use the restriction on the sine function. Since $\\sin t$ is defined for all real $t$, its range is:\n$$-1 \\le \\sin t \\le 1$$\n\nSubstitute $\\sin t = \\frac{x - 2}{3}$ into the inequality:\n$$-1 \\le \\frac{x - 2}{3} \\le 1$$\n\nMultiply the entire inequality by $3$:\n$$-3 \\le x - 2 \\le 3$$\n\nAdd $2$ to all parts:\n$$-1 \\le x \\le 5$$\n\nThus, the domain for which the curve is defined is $[-1, 5]$."}]},{"id":"add0cbb1-5b08-467e-aad9-ac56e8a1b982:cartesian_conversion","label":"(b) Cartesian Equation Method","steps":[{"type":"text","order":0,"title":"Express sin(t) in terms of x","content":"To find the Cartesian equation of the curve, we first isolate the trigonometric term $\\sin t$ from the parametric equation for $x$:\n\n$$x = 3 \\sin t + 2$$\n\nSubtracting $2$ and then dividing by $3$ gives:\n\n$$\\sin t = \\frac{x - 2}{3}$$"},{"type":"text","order":1,"title":"Apply trigonometric identity to y","content":"Next, we use the fundamental trigonometric identity:\n\n$$\\cos^2 t = 1 - \\sin^2 t$$\n\nSubstituting this identity into our parametric equation for $y$:\n\n$$y = 2 \\cos^2 t - 5$$\n\n$$y = 2(1 - \\sin^2 t) - 5$$\n\n$$y = -2\\sin^2 t - 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\sin t = \\htmlData{anchor=sin-expr}{\\color{@sky}{\\frac{x - 2}{3}}}$"},{"latex":"$$y = -2 (\\htmlData{anchor=sin-sub}{\\color{@sky}{\\sin t}})^2 - 3$"},{"latex":"$$y = -\\frac{2}{9}(x - 2)^2 - 3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-expr"},{"color":"sky","style":"text-color","anchor":"sin-sub"}],"connections":[{"to":"sin-sub","from":"sin-expr","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Substitute to get Cartesian equation","description":"Now, we substitute our expression for $\\sin t$ into the simplified equation for $y$ to get the final Cartesian equation."},{"type":"text","order":3,"title":"Determine domain and endpoints","content":"Since $0 \\le t \\le 2\\pi$, the range of $\\sin t$ is restricted:\n\n$$-1 \\le \\sin t \\le 1$$\n\nWe can find the domain restriction for $x$ by substituting these limits into $x = 3 \\sin t + 2$:\n\n- Minimum value: when $\\sin t = -1$, $x = 3(-1) + 2 = -1$\n- Maximum value: when $\\sin t = 1$, $x = 3(1) + 2 = 5$\n\nSo, the domain restriction is $-1 \\le x \\le 5$.\n\nAt both endpoints $x = -1$ and $x = 5$, the corresponding $y$-value is:\n\n$$y = -\\frac{2}{9}(\\pm 3)^2 - 3 = -2 - 3 = -5$$\n\nThis gives the end coordinates of our curve: $(-1, -5)$ and $(5, -5)$."},{"type":"text","order":4,"title":"Identify the vertex","content":"The Cartesian equation is in the vertex form of a quadratic, $y = a(x-h)^2 + k$:\n\n$$y = -\\frac{2}{9}(x - 2)^2 - 3$$\n\n- Since the coefficient $-\\frac{2}{9}$ is negative, the parabola is **inverted** (opening downwards).\n- The vertex $(h, k)$ is located at $(2, -3)$, which is the maximum point of this inverted parabola."},{"type":"text","order":5,"title":"Sketch the curve","content":"Now we sketch the curve using our key features:\n\n1. Plot the vertex at $(2, -3)$ as a maximum point.\n2. Plot the endpoints at $(-1, -5)$ and $(5, -5)$.\n3. Draw a smooth, symmetric, inverted parabola shape connecting these points. Ensure your curve does not extend beyond $x = -1$ or $x = 5$.","calculator":[{"gdc":{"instructions":"1. Change your graphing mode to Parametric.\n2. Enter the equations: x(t) = 3*sin(t) + 2 and y(t) = 2*cos(t)^2 - 5.\n3. Set the domain variable t from 0 to 6.28.\n4. Set an appropriate window size to view the interval [-1, 5] on the x-axis and [-6, -2] on the y-axis."},"ti84":{"instructions":"1. Press [MODE], arrow down to 'FUNC', change it to 'PARAMETRIC' (PAR), and press [ENTER].\n2. Press [Y=] and enter:\n X1T = 3 * sin(T) + 2\n Y1T = 2 * (cos(T))^2 - 5\n3. Press [WINDOW] and set:\n Tmin = 0, Tmax = 6.28, Tstep = 0.1\n Xmin = -3, Xmax = 7, Xscl = 1\n Ymin = -7, Ymax = -1, Yscl = 1\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"(3*sin(t) + 2, 2*(cos(t))^2 - 5)","instructions":"1. Type the coordinate pair representing the parametric equations in a single line:\n (3*sin(t) + 2, 2*(cos(t))^2 - 5)\n2. Set the parameter interval beneath it to: 0 <= t <= 2pi."},"description":"Visualize the parametric curve to verify its shape, vertex, and domain endpoints."}]}]},{"id":"add0cbb1-5b08-467e-aad9-ac56e8a1b982:point_plotting","label":"(b) Point-by-Point Plotting","steps":[{"type":"text","order":0,"title":"Set up a coordinate table","content":"To sketch the parametric curve using a point-by-point plotting approach, we calculate the coordinates $(x, y)$ for key values of $t$ in the interval $0 \\le t \\le 2\\pi$.\n\nLet's choose the standard angles:\n$$t = 0, \\quad \\frac{\\pi}{2}, \\quad \\pi, \\quad \\frac{3\\pi}{2}, \\quad 2\\pi$$\n\nOur parametric equations are:\n$$x = 3 \\sin t + 2$$\n$$y = 2 \\cos^2 t - 5$$"},{"type":"text","order":1,"title":"Evaluate points for t","content":"Let's calculate the coordinates $(x, y)$ for each chosen value of $t$:\n\n1. **For $t = 0$:**\n $$x = 3 \\sin(0) + 2 = 2$$\n $$y = 2 \\cos^2(0) - 5 = 2(1)^2 - 5 = -3$$\n Point: $(2, -3)$\n\n2. **For $t = \\frac{\\pi}{2}$:**\n $$x = 3 \\sin\\left(\\frac{\\pi}{2}\\right) + 2 = 3(1) + 2 = 5$$\n $$y = 2 \\cos^2\\left(\\frac{\\pi}{2}\\right) - 5 = 2(0)^2 - 5 = -5$$\n Point: $(5, -5)$\n\n3. **For $t = \\pi$:**\n $$x = 3 \\sin(\\pi) + 2 = 2$$\n $$y = 2 \\cos^2(\\pi) - 5 = 2(-1)^2 - 5 = -3$$\n Point: $(2, -3)$\n\n4. **For $t = \\frac{3\\pi}{2}$:**\n $$x = 3 \\sin\\left(\\frac{3\\pi}{2}\\right) + 2 = 3(-1) + 2 = -1$$\n $$y = 2 \\cos^2\\left(\\frac{3\\pi}{2}\\right) - 5 = 2(0)^2 - 5 = -5$$\n Point: $(-1, -5)$\n\n5. **For $t = 2\\pi$:**\n $$x = 3 \\sin(2\\pi) + 2 = 2$$\n $$y = 2 \\cos^2(2\\pi) - 5 = 2(1)^2 - 5 = -3$$\n Point: $(2, -3)$"},{"type":"text","order":2,"title":"Summarise coordinate table","content":"Let's compile these evaluated points into a clear summary table:\n\n$$\\begin{array}{|c|c|c|c|} \\hline t & x = 3\\sin t + 2 & y = 2\\cos^2 t - 5 & (x, y) \\\\ \\hline 0 & 2 & -3 & (2, -3) \\\\ \\hline \\frac{\\pi}{2} & 5 & -5 & (5, -5) \\\\ \\hline \\pi & 2 & -3 & (2, -3) \\\\ \\hline \\frac{3\\pi}{2} & -1 & -5 & (-1, -5) \\\\ \\hline 2\\pi & 2 & -3 & (2, -3) \\\\ \\hline \\end{array}$$\n\nNotice that as $t$ goes from $0$ to $2\\pi$:\n- The curve starts at the vertex $(2, -3)$.\n- It moves to the right endpoint $(5, -5)$ at $t = \\frac{\\pi}{2}$.\n- It returns to the vertex $(2, -3)$ at $t = \\pi$.\n- It moves to the left endpoint $(-1, -5)$ at $t = \\frac{3\\pi}{2}$.\n- It returns to the vertex $(2, -3)$ at $t = 2\\pi$."},{"type":"text","order":3,"title":"Analyze the shape","content":"Let's understand the underlying algebraic shape to draw the sketch accurately.\n\nFrom our table, we can see:\n- The maximum value of $y$ is $-3$, which occurs when $x = 2$. This indicates a peak or vertex at $(2, -3)$.\n- The minimum value of $y$ is $-5$, which occurs at the boundary points $x = -1$ and $x = 5$.\n- The coordinates are symmetric about the vertical line $x = 2$.\n\nThis behavior matches an inverted (negative) parabola opening downwards, with its vertex at $(2, -3)$ and endpoints at $(-1, -5)$ and $(5, -5)$."},{"type":"text","order":4,"title":"Verify with GDC","content":"We can easily verify the plotted path and the endpoints of the curve using a graphing calculator.","calculator":[{"gdc":{"instructions":"1. Set your graphing calculator's mode to Parametric equations.\n2. Define X1(t) = 3*sin(t) + 2 and Y1(t) = 2*cos(t)^2 - 5.\n3. Adjust the interval for t to [0, 6.28].\n4. Set a suitable window view (e.g., x from -2 to 6, y from -6 to -2).\n5. Plot the graph to verify the shape."},"ti84":{"instructions":"1. Press [MODE] and change the graphing mode to PAR (parametric).\n2. Press [Y=].\n3. Set X1T = 3 * sin(T) + 2 and Y1T = 2 * (cos(T))^2 - 5.\n4. Press [WINDOW] and set Tmin = 0, Tmax = 6.28, and Tstep = 0.1.\n5. Set Xmin = -2, Xmax = 6, Ymin = -6, Ymax = -2.\n6. Press [GRAPH] to view the curve."},"desmos":{"expression":"(3*sin(t) + 2, 2*cos(t)^2 - 5)","instructions":"1. Enter the parametric coordinate pair directly into an input row: (3*sin(t) + 2, 2*cos(t)^2 - 5).\n2. Set the parameter domain bounds next to the expression as 0 <= t <= 2*pi.\n3. Adjust the graph viewport to focus on the region around x in [-2, 6] and y in [-6, -2]."},"description":"Graph the parametric equations to visualize the negative parabola shape."}]},{"type":"text","order":5,"title":"Sketch the curve","content":"Now, sketch the curve on your axes:\n\n1. Plot the vertex at $(2, -3)$ as the highest point.\n2. Plot the left endpoint at $(-1, -5)$ and the right endpoint at $(5, -5)$.\n3. Draw a smooth, symmetric, downward-opening parabolic arc connecting $(-1, -5)$ to $(2, -3)$ and then to $(5, -5)$.\n\n**Marks awarded for:**\n- **B1**: Drawing an inverted (negative) parabola shape.\n- **B1**: Showing the vertex clearly located at $(2, -3)$.\n- **B1**: Showing the curve starting/ending at $(-1, -5)$ and $(5, -5)$ with no parts of the curve drawn outside this domain."}]}],"generatedAt":"2026-06-09T03:23:21.981Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"543fca01-78c6-49af-ac8c-db2b576575df","revisionId":"e34ab4bf-b8c3-446e-94b5-d907def9ec3d","questionNumber":null,"questionType":"LA","specification":"A curve has the parametric equations\n\n$$x = t - 4, \\quad y = \\frac{8}{t}$$","difficulty":1,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"db8f8961-586a-48a0-a1e4-4ef7b0a462a8","label":"a","content":"Find a cartesian equation for the curve.","markscheme":"- $t = x + 4$ M1\n- $y = \\frac{8}{x+4}$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Attempt to rearrange the parametric equation for x to make t the subject.","code":"M"},{"id":"p2","marks":1,"description":"Substitute the expression for t into the equation for y to obtain a correct cartesian equation.","code":"A"}],"finalAnswers":["y = \\frac{8}{x+4}"],"generatedAt":"2026-05-06T14:05:12.893Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitute_t_from_x","label":"Substitution using x","steps":[{"type":"text","order":0,"title":"Make t the subject","content":"To find the Cartesian equation using this approach, we first rearrange the parametric equation for $x$ to make $t$ the subject.\n\nStart with the equation for $x$:\n$$x = t - 4$$\n\nAdd $4$ to both sides to isolate $t$:\n$$t = x + 4$$\n\nRearranging to find this expression for $t$ earns the first method mark (**M1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$t = \\htmlData{anchor=t-val}{\\color{@sky}{x + 4}}$"},{"latex":"$$y = \\frac{8}{\\htmlData{anchor=t-sub}{\\color{@sky}{t}}}$"},{"latex":"$$y = \\frac{8}{\\color{@sky}{x + 4}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-val"},{"color":"sky","style":"text-color","anchor":"t-sub"}],"connections":[{"to":"t-sub","from":"t-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into y","description":"Next, we substitute this expression for $t$ directly into the parametric equation for $y$ to obtain the Cartesian equation. Writing down this final Cartesian equation earns the accuracy mark (**A1**)."}]},{"id":"substitute_t_from_y","label":"Substitution using y","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"To find the Cartesian equation, we need to eliminate the parameter $t$ from our two parametric equations:\n\n1. $x = t - 4$\n2. $y = \\frac{8}{t}$\n\nWe will use the **substitution using $y$** approach. This means we make $t$ the subject of the equation for $y$, substitute it into the equation for $x$, and then rearrange to make $y$ the subject."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{8}{\\htmlData{anchor=t1}{t}}$"},{"latex":"$$\\htmlData{anchor=t2}{t} = \\frac{8}{y}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t1"},{"color":"sky","style":"text-color","anchor":"t2"}],"connections":[{"to":"t2","from":"t1","color":"sky","label":"rearrange","style":"solid"}]},"order":1,"title":"Make t the subject","description":"We start with the parametric equation for $y$ and rearrange it to isolate $t$. Multiplying both sides by $t$ and dividing by $y$ gives us our expression for $t$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x = \\htmlData{anchor=t-sub}{t} - 4$"},{"latex":"$$x = \\htmlData{anchor=sub-val}{\\left(\\frac{8}{y}\\right)} - 4$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"t-sub"},{"color":"purple","style":"text-color","anchor":"sub-val"}],"connections":[{"to":"t-sub","from":"sub-val","color":"purple","label":"substitute","style":"dashed"}]},"order":2,"title":"Substitute into x","description":"Next, we substitute our expression for $t$, which is $\\frac{8}{y}$, into the first parametric equation, $x = t - 4$."},{"type":"text","order":3,"title":"Solve for y","content":"Now we rearrange this equation to express $y$ in terms of $x$:\n\nFirst, add $4$ to both sides:\n$$x + 4 = \\frac{8}{y}$$\n\nNext, multiply both sides by $y$:\n$$y(x + 4) = 8$$\n\nFinally, divide by $(x + 4)$ to isolate $y$:\n$$y = \\frac{8}{x + 4}$$\n\nThis Cartesian equation matches the expected answer in the markscheme, earning you both marks (the method mark **M1** for showing a correct substitution step and the accuracy mark **A1** for the final equation)."}]}],"generatedAt":"2026-06-09T01:58:42.623Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"a00b6b50-0f4a-4ef8-ba01-0c054a99f596","label":"b","content":"Sketch the curve","markscheme":"- Correct hyperbola shape with asymptotes $x = -4$ and $y = 0$ B1\n- Correct branches in 1st and 3rd quadrants relative to asymptotes B1\n- $y$-intercept at $(0, 2)$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct hyperbola shape with vertical asymptote $x = -4$ and horizontal asymptote $y = 0$ clearly indicated.","code":"B"},{"id":"p2","marks":1,"description":"Branches of the hyperbola are correctly positioned in the 1st and 3rd quadrants relative to the asymptotes.","code":"B"},{"id":"p3","marks":1,"description":"The y-intercept is correctly calculated as $(0, 2)$ and marked on the sketch.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:05:12.893Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"cartesian_conversion","label":"Conversion to Cartesian Equation","steps":[{"type":"text","order":0,"title":"Rearrange for t","content":"To convert the parametric equations into a Cartesian equation, we first eliminate the parameter $t$.\n\nStart by rearranging the equation for $x$ to express $t$ in terms of $x$:\n\n$$x = t - 4$$\n\nAdding $4$ to both sides gives:\n\n$$t = x + 4$$","highlights":[{"text":"x = t - 4","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$t = \\htmlData{anchor=t-val}{\\color{@sky}{x + 4}}$"},{"latex":"$$y = \\frac{8}{\\htmlData{anchor=t-var}{\\color{@sky}{t}}}$"},{"latex":"$$y = \\frac{8}{\\color{@sky}{x + 4}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-val"},{"color":"sky","style":"text-color","anchor":"t-var"}],"connections":[{"to":"t-var","from":"t-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into y","description":"Now, substitute this expression for $t$ into the equation for $y$ to get the Cartesian equation of the curve."},{"type":"text","order":2,"title":"Identify Key Features","content":"The Cartesian equation $$y = \\frac{8}{x+4}$$ represents a translated standard hyperbola. Let's find its key features to draw an accurate sketch:\n\n1. **Asymptotes**:\n - **Vertical asymptote**: Set the denominator to zero:\n $$x + 4 = 0 \\implies x = -4$$\n - **Horizontal asymptote**: Find the behavior as $x \\to \\pm\\infty$:\n $$y \\to 0$$\n\n2. **$y$-intercept**:\n - Set $x = 0$:\n $$y = \\frac{8}{0 + 4} = 2$$\n So, the $y$-intercept is at $(0, 2)$.\n\n3. **Quadrants of the branches**:\n - For $x > -4$, $y > 0$ (the upper-right branch relative to the asymptotes).\n - For $x < -4$, $y < 0$ (the lower-left branch relative to the asymptotes)."},{"type":"text","order":3,"title":"Visualize with Calculator","content":"Using a graphing utility is a great way to verify your sketch of the hyperbola.","calculator":[{"gdc":{"expression":"8 / (x + 4)","instructions":"1. Go to the Graph menu.\n2. Enter the function f1(x) = 8 / (x + 4).\n3. Adjust the view window to see the vertical asymptote at x = -4 clearly.\n4. Use the G-Solv or Trace tool to find the y-intercept."},"ti84":{"expression":"8 / (X + 4)","instructions":"1. Press [Y=].\n2. Enter Y1 = 8 / (X + 4).\n3. Press [WINDOW] and set Xmin = -10, Xmax = 10, Ymin = -10, Ymax = 10.\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"y = 8 / (x + 4)","instructions":"1. Type y = 8 / (x + 4) in the first expression box.\n2. Type x = -4 in the second expression box to display the vertical asymptote as a dashed line.\n3. Click on the y-axis intersection to confirm the coordinate (0, 2)."},"description":"Graph the Cartesian equation to verify the asymptotes and intercepts."}]},{"type":"text","order":4,"title":"Draw the Sketch","content":"When drawing your sketch to secure all 3 marks, ensure you include:\n\n- **The Asymptotes**: Draw the vertical asymptote $x = -4$ and the horizontal asymptote $y = 0$ (the $x$-axis) as dashed lines. B1\n- **The Branches**: Sketch the two hyperbola branches in the first and third quadrants relative to the intersection point $(-4, 0)$ of your asymptotes. B1\n- **The Intercept**: Clearly label the $y$-intercept at $(0, 2)$. B1"}]},{"id":"table_of_values","label":"Direct Point Plotting","steps":[{"type":"text","order":0,"title":"Set up a table","content":"To sketch the curve using the **Direct Point Plotting** method, we will choose several positive and negative values for the parameter $t$, ensuring we avoid $t = 0$ because $y = \\frac{8}{t}$ is undefined there.\n\nLet's calculate the coordinates $(x, y)$ for each selected value of $t$ using the parametric equations:\n\n$$x = t - 4$$ \n\n$$y = \\frac{8}{t}$$"},{"type":"text","order":1,"title":"Calculate positive $t$ values","content":"Let's first choose some positive values for $t$:\n\n* **For $t = 1$:**\n $$x = 1 - 4 = -3, \\quad y = \\frac{8}{1} = 8 \\implies (-3, 8)$$\n* **For $t = 2$:**\n $$x = 2 - 4 = -2, \\quad y = \\frac{8}{2} = 4 \\implies (-2, 4)$$\n* **For $t = 4$:**\n $$x = 4 - 4 = 0, \\quad y = \\frac{8}{4} = 2 \\implies (0, 2)$$\n* **For $t = 8$:**\n $$x = 8 - 4 = 4, \\quad y = \\frac{8}{8} = 1 \\implies (4, 1)$$\n\nNotice that at $t = 4$, the curve crosses the $y$-axis at **$(0, 2)$**.","highlights":[{"text":"y-intercept at (0, 2)","location":"specification"}]},{"type":"text","order":2,"title":"Calculate negative $t$ values","content":"Now let's choose some negative values for $t$:\n\n* **For $t = -1$:**\n $$x = -1 - 4 = -5, \\quad y = \\frac{8}{-1} = -8 \\implies (-5, -8)$$\n* **For $t = -2$:**\n $$x = -2 - 4 = -6, \\quad y = \\frac{8}{-2} = -4 \\implies (-6, -4)$$\n* **For $t = -4$:**\n $$x = -4 - 4 = -8, \\quad y = \\frac{8}{-4} = -2 \\implies (-8, -2)$$\n* **For $t = -8$:**\n $$x = -8 - 4 = -12, \\quad y = \\frac{8}{-8} = -1 \\implies (-12, -1)$$ \n\nWe now have a strong set of points to plot on both sides of $t = 0$."},{"type":"text","order":3,"title":"Identify the asymptotes","content":"Before we plot, let's look at the behaviour of $x$ and $y$ as $t$ approaches key values:\n\n* **As $t \\to 0$:** \n $x \\to -4$ and $y \\to \\pm\\infty$. This means there is a **vertical asymptote** at:\n $$x = -4$$\n* **As $t \\to \\pm\\infty$:** \n $x \\to \\pm\\infty$ and $y \\to 0$. This means there is a **horizontal asymptote** at:\n $$y = 0$$ (the $x$-axis)","highlights":[{"text":"Correct hyperbola shape with asymptotes x = -4 and y = 0","location":"specification"}]},{"type":"text","order":4,"title":"Verify with a calculator","content":"You can use a graphing tool to verify the shape of your plotted points. Enter the parametric equations directly into your calculator.","calculator":[{"gdc":{"instructions":"1. Change your graphing mode to 'Parametric'.\n2. Enter the equations as x(t) = t - 4 and y(t) = 8/t.\n3. Adjust the viewing window to see the asymptotes clearly at x = -4 and y = 0."},"ti84":{"instructions":"1. Press [MODE], go to the 4th line, and select 'PAR' (Parametric mode).\n2. Press [Y=] and enter:\n X1T = T - 4\n Y1T = 8 / T\n3. Press [WINDOW] and set Tmin = -10, Tmax = 10, Tstep = 0.1, Xmin = -15, Xmax = 10, Ymin = -10, Ymax = 10.\n4. Press [GRAPH] to view."},"desmos":{"expression":"(t - 4, 8/t)","instructions":"1. Type the coordinate pair (t - 4, 8/t) into an expression line.\n2. Set the parameter domain for t (for example, -10 <= t <= 10).\n3. Plot the asymptotes by typing x = -4 and y = 0 to verify the shape."},"description":"Plot the parametric equations to view the hyperbola."}]},{"type":"text","order":5,"title":"Sketch the final curve","content":"To complete your sketch:\n\n1. Draw dashed lines representing the asymptotes at $$x = -4 \\quad \\text{and} \\quad y = 0$$\n2. Plot the points computed in the previous steps.\n3. Connect the points with two smooth branches. Relative to the translated axes formed by the asymptotes, the branches lie in the upper-right (1st quadrant relative to the asymptotes) and lower-left (3rd quadrant relative to the asymptotes) regions.\n4. Ensure you mark the $y$-intercept clearly at $$(0, 2)$$","highlights":[{"text":"Correct branches in 1st and 3rd quadrants relative to asymptotes","location":"specification"}]}]}],"generatedAt":"2026-06-09T02:00:27.112Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"db8f8961-586a-48a0-a1e4-4ef7b0a462a8:substitute_t_from_x","label":"(a) Substitution using x","steps":[{"type":"text","order":0,"title":"Make t the subject","content":"To find the Cartesian equation using this approach, we first rearrange the parametric equation for $x$ to make $t$ the subject.\n\nStart with the equation for $x$:\n$$x = t - 4$$\n\nAdd $4$ to both sides to isolate $t$:\n$$t = x + 4$$\n\nRearranging to find this expression for $t$ earns the first method mark (**M1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$t = \\htmlData{anchor=t-val}{\\color{@sky}{x + 4}}$"},{"latex":"$$y = \\frac{8}{\\htmlData{anchor=t-sub}{\\color{@sky}{t}}}$"},{"latex":"$$y = \\frac{8}{\\color{@sky}{x + 4}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-val"},{"color":"sky","style":"text-color","anchor":"t-sub"}],"connections":[{"to":"t-sub","from":"t-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into y","description":"Next, we substitute this expression for $t$ directly into the parametric equation for $y$ to obtain the Cartesian equation. Writing down this final Cartesian equation earns the accuracy mark (**A1**)."}]},{"id":"db8f8961-586a-48a0-a1e4-4ef7b0a462a8:substitute_t_from_y","label":"(a) Substitution using y","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"To find the Cartesian equation, we need to eliminate the parameter $t$ from our two parametric equations:\n\n1. $x = t - 4$\n2. $y = \\frac{8}{t}$\n\nWe will use the **substitution using $y$** approach. This means we make $t$ the subject of the equation for $y$, substitute it into the equation for $x$, and then rearrange to make $y$ the subject."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{8}{\\htmlData{anchor=t1}{t}}$"},{"latex":"$$\\htmlData{anchor=t2}{t} = \\frac{8}{y}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t1"},{"color":"sky","style":"text-color","anchor":"t2"}],"connections":[{"to":"t2","from":"t1","color":"sky","label":"rearrange","style":"solid"}]},"order":1,"title":"Make t the subject","description":"We start with the parametric equation for $y$ and rearrange it to isolate $t$. Multiplying both sides by $t$ and dividing by $y$ gives us our expression for $t$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x = \\htmlData{anchor=t-sub}{t} - 4$"},{"latex":"$$x = \\htmlData{anchor=sub-val}{\\left(\\frac{8}{y}\\right)} - 4$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"t-sub"},{"color":"purple","style":"text-color","anchor":"sub-val"}],"connections":[{"to":"t-sub","from":"sub-val","color":"purple","label":"substitute","style":"dashed"}]},"order":2,"title":"Substitute into x","description":"Next, we substitute our expression for $t$, which is $\\frac{8}{y}$, into the first parametric equation, $x = t - 4$."},{"type":"text","order":3,"title":"Solve for y","content":"Now we rearrange this equation to express $y$ in terms of $x$:\n\nFirst, add $4$ to both sides:\n$$x + 4 = \\frac{8}{y}$$\n\nNext, multiply both sides by $y$:\n$$y(x + 4) = 8$$\n\nFinally, divide by $(x + 4)$ to isolate $y$:\n$$y = \\frac{8}{x + 4}$$\n\nThis Cartesian equation matches the expected answer in the markscheme, earning you both marks (the method mark **M1** for showing a correct substitution step and the accuracy mark **A1** for the final equation)."}]},{"id":"a00b6b50-0f4a-4ef8-ba01-0c054a99f596:cartesian_conversion","label":"(b) Conversion to Cartesian Equation","steps":[{"type":"text","order":0,"title":"Rearrange for t","content":"To convert the parametric equations into a Cartesian equation, we first eliminate the parameter $t$.\n\nStart by rearranging the equation for $x$ to express $t$ in terms of $x$:\n\n$$x = t - 4$$\n\nAdding $4$ to both sides gives:\n\n$$t = x + 4$$","highlights":[{"text":"x = t - 4","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$t = \\htmlData{anchor=t-val}{\\color{@sky}{x + 4}}$"},{"latex":"$$y = \\frac{8}{\\htmlData{anchor=t-var}{\\color{@sky}{t}}}$"},{"latex":"$$y = \\frac{8}{\\color{@sky}{x + 4}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-val"},{"color":"sky","style":"text-color","anchor":"t-var"}],"connections":[{"to":"t-var","from":"t-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into y","description":"Now, substitute this expression for $t$ into the equation for $y$ to get the Cartesian equation of the curve."},{"type":"text","order":2,"title":"Identify Key Features","content":"The Cartesian equation $$y = \\frac{8}{x+4}$$ represents a translated standard hyperbola. Let's find its key features to draw an accurate sketch:\n\n1. **Asymptotes**:\n - **Vertical asymptote**: Set the denominator to zero:\n $$x + 4 = 0 \\implies x = -4$$\n - **Horizontal asymptote**: Find the behavior as $x \\to \\pm\\infty$:\n $$y \\to 0$$\n\n2. **$y$-intercept**:\n - Set $x = 0$:\n $$y = \\frac{8}{0 + 4} = 2$$\n So, the $y$-intercept is at $(0, 2)$.\n\n3. **Quadrants of the branches**:\n - For $x > -4$, $y > 0$ (the upper-right branch relative to the asymptotes).\n - For $x < -4$, $y < 0$ (the lower-left branch relative to the asymptotes)."},{"type":"text","order":3,"title":"Visualize with Calculator","content":"Using a graphing utility is a great way to verify your sketch of the hyperbola.","calculator":[{"gdc":{"expression":"8 / (x + 4)","instructions":"1. Go to the Graph menu.\n2. Enter the function f1(x) = 8 / (x + 4).\n3. Adjust the view window to see the vertical asymptote at x = -4 clearly.\n4. Use the G-Solv or Trace tool to find the y-intercept."},"ti84":{"expression":"8 / (X + 4)","instructions":"1. Press [Y=].\n2. Enter Y1 = 8 / (X + 4).\n3. Press [WINDOW] and set Xmin = -10, Xmax = 10, Ymin = -10, Ymax = 10.\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"y = 8 / (x + 4)","instructions":"1. Type y = 8 / (x + 4) in the first expression box.\n2. Type x = -4 in the second expression box to display the vertical asymptote as a dashed line.\n3. Click on the y-axis intersection to confirm the coordinate (0, 2)."},"description":"Graph the Cartesian equation to verify the asymptotes and intercepts."}]},{"type":"text","order":4,"title":"Draw the Sketch","content":"When drawing your sketch to secure all 3 marks, ensure you include:\n\n- **The Asymptotes**: Draw the vertical asymptote $x = -4$ and the horizontal asymptote $y = 0$ (the $x$-axis) as dashed lines. B1\n- **The Branches**: Sketch the two hyperbola branches in the first and third quadrants relative to the intersection point $(-4, 0)$ of your asymptotes. B1\n- **The Intercept**: Clearly label the $y$-intercept at $(0, 2)$. B1"}]},{"id":"a00b6b50-0f4a-4ef8-ba01-0c054a99f596:table_of_values","label":"(b) Direct Point Plotting","steps":[{"type":"text","order":0,"title":"Set up a table","content":"To sketch the curve using the **Direct Point Plotting** method, we will choose several positive and negative values for the parameter $t$, ensuring we avoid $t = 0$ because $y = \\frac{8}{t}$ is undefined there.\n\nLet's calculate the coordinates $(x, y)$ for each selected value of $t$ using the parametric equations:\n\n$$x = t - 4$$ \n\n$$y = \\frac{8}{t}$$"},{"type":"text","order":1,"title":"Calculate positive $t$ values","content":"Let's first choose some positive values for $t$:\n\n* **For $t = 1$:**\n $$x = 1 - 4 = -3, \\quad y = \\frac{8}{1} = 8 \\implies (-3, 8)$$\n* **For $t = 2$:**\n $$x = 2 - 4 = -2, \\quad y = \\frac{8}{2} = 4 \\implies (-2, 4)$$\n* **For $t = 4$:**\n $$x = 4 - 4 = 0, \\quad y = \\frac{8}{4} = 2 \\implies (0, 2)$$\n* **For $t = 8$:**\n $$x = 8 - 4 = 4, \\quad y = \\frac{8}{8} = 1 \\implies (4, 1)$$\n\nNotice that at $t = 4$, the curve crosses the $y$-axis at **$(0, 2)$**.","highlights":[{"text":"y-intercept at (0, 2)","location":"specification"}]},{"type":"text","order":2,"title":"Calculate negative $t$ values","content":"Now let's choose some negative values for $t$:\n\n* **For $t = -1$:**\n $$x = -1 - 4 = -5, \\quad y = \\frac{8}{-1} = -8 \\implies (-5, -8)$$\n* **For $t = -2$:**\n $$x = -2 - 4 = -6, \\quad y = \\frac{8}{-2} = -4 \\implies (-6, -4)$$\n* **For $t = -4$:**\n $$x = -4 - 4 = -8, \\quad y = \\frac{8}{-4} = -2 \\implies (-8, -2)$$\n* **For $t = -8$:**\n $$x = -8 - 4 = -12, \\quad y = \\frac{8}{-8} = -1 \\implies (-12, -1)$$ \n\nWe now have a strong set of points to plot on both sides of $t = 0$."},{"type":"text","order":3,"title":"Identify the asymptotes","content":"Before we plot, let's look at the behaviour of $x$ and $y$ as $t$ approaches key values:\n\n* **As $t \\to 0$:** \n $x \\to -4$ and $y \\to \\pm\\infty$. This means there is a **vertical asymptote** at:\n $$x = -4$$\n* **As $t \\to \\pm\\infty$:** \n $x \\to \\pm\\infty$ and $y \\to 0$. This means there is a **horizontal asymptote** at:\n $$y = 0$$ (the $x$-axis)","highlights":[{"text":"Correct hyperbola shape with asymptotes x = -4 and y = 0","location":"specification"}]},{"type":"text","order":4,"title":"Verify with a calculator","content":"You can use a graphing tool to verify the shape of your plotted points. Enter the parametric equations directly into your calculator.","calculator":[{"gdc":{"instructions":"1. Change your graphing mode to 'Parametric'.\n2. Enter the equations as x(t) = t - 4 and y(t) = 8/t.\n3. Adjust the viewing window to see the asymptotes clearly at x = -4 and y = 0."},"ti84":{"instructions":"1. Press [MODE], go to the 4th line, and select 'PAR' (Parametric mode).\n2. Press [Y=] and enter:\n X1T = T - 4\n Y1T = 8 / T\n3. Press [WINDOW] and set Tmin = -10, Tmax = 10, Tstep = 0.1, Xmin = -15, Xmax = 10, Ymin = -10, Ymax = 10.\n4. Press [GRAPH] to view."},"desmos":{"expression":"(t - 4, 8/t)","instructions":"1. Type the coordinate pair (t - 4, 8/t) into an expression line.\n2. Set the parameter domain for t (for example, -10 <= t <= 10).\n3. Plot the asymptotes by typing x = -4 and y = 0 to verify the shape."},"description":"Plot the parametric equations to view the hyperbola."}]},{"type":"text","order":5,"title":"Sketch the final curve","content":"To complete your sketch:\n\n1. Draw dashed lines representing the asymptotes at $$x = -4 \\quad \\text{and} \\quad y = 0$$\n2. Plot the points computed in the previous steps.\n3. Connect the points with two smooth branches. Relative to the translated axes formed by the asymptotes, the branches lie in the upper-right (1st quadrant relative to the asymptotes) and lower-left (3rd quadrant relative to the asymptotes) regions.\n4. Ensure you mark the $y$-intercept clearly at $$(0, 2)$$","highlights":[{"text":"Correct branches in 1st and 3rd quadrants relative to asymptotes","location":"specification"}]}]}],"generatedAt":"2026-06-09T02:00:45.917Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"5adb3bf5-a974-44f3-b829-cd9f56fa4014","revisionId":"951cf01f-06d1-4cd0-b1fb-9e48c0b9555d","questionNumber":null,"questionType":"LA","specification":"A curve has the parametric equations\n\n$$x = t - 4, \\quad y = t^2 + 2$$","difficulty":1,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"c0401f07-1f3a-412c-94de-614c31b624a9","label":"a","content":"Find a cartesian equation for the curve.","markscheme":"- Rearrange for $t$: $t = x + 4$ M1\n- Substitute into $y$: $y = (x + 4)^2 + 2$ M1\n- Expand and simplify: $y = x^2 + 8x + 18$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Rearranging the equation for $x$ to make $t$ the subject.","code":"M"},{"id":"p2","marks":1,"description":"Substituting the expression for $t$ into the equation for $y$.","code":"M"},{"id":"p3","marks":1,"description":"Expanding and simplifying to obtain the final quadratic equation.","code":"A"}],"finalAnswers":["y = x^2 + 8x + 18"],"generatedAt":"2026-05-06T14:04:57.260Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution_t_from_x","label":"Substitution of t in terms of x","steps":[{"type":"text","order":0,"title":"Rearrange x for t","content":"To find a cartesian equation, we need to eliminate the parameter $t$. We start with the linear parametric equation for $x$:\n\n$$x = t - 4$$\n\nTo express $t$ in terms of $x$, we add $4$ to both sides:\n\n$$t = x + 4$$","highlights":[{"text":"x = t - 4","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$t = \\htmlData{anchor=t-val}{\\color{@sky}{x + 4}}$"},{"latex":"$$y = \\htmlData{anchor=t-var}{\\color{@sky}{t}}^2 + 2$"},{"latex":"$$y = \\htmlData{anchor=substituted}{\\color{@sky}{(x + 4)}}^2 + 2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-val"},{"color":"sky","style":"text-color","anchor":"substituted"}],"connections":[{"to":"t-var","from":"t-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute t into y","description":"Now, we substitute this expression for $t$ into the parametric equation for $y$:"},{"type":"text","order":2,"title":"Expand and simplify","content":"Now we expand the squared term and simplify the expression to get the final cartesian equation:\n\n$$\\begin{aligned} y &= (x + 4)^2 + 2 \\\\ y &= (x^2 + 8x + 16) + 2 \\\\ y &= x^2 + 8x + 18 \\end{aligned}$$\n\nThis gives us a quadratic curve (a parabola) representing the cartesian equation of the curve."}]}],"generatedAt":"2026-06-09T05:26:46.586Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"a8e4be13-e471-4a11-8007-5f182789200b","label":"b","content":"Sketch the curve","markscheme":"- Correct shape (parabola opening upwards) B1\n- Minimum point at $(-4, 2)$ B1\n- $y$-intercept at 18 B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct parabolic shape opening upwards.","code":"B"},{"id":"p2","marks":1,"description":"Identifying and labeling the minimum point at $(-4, 2)$.","code":"B"},{"id":"p3","marks":1,"description":"Identifying and labeling the $y$-intercept at 18.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:04:57.260Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"cartesian_conversion","label":"Conversion to Cartesian Form","steps":[{"type":"text","order":0,"title":"Rearrange for t","content":"To convert the parametric equations into Cartesian form, we start by rearranging the equation for $x$ to express $t$ in terms of $x$.\n\nGiven:\n$$x = t - 4$$\n\nAdding $4$ to both sides gives:\n$$t = x + 4$$","highlights":[{"text":"x = t - 4","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=t-var}{\\color{@sky}{t}}^2 + 2$"},{"latex":"$$y = \\htmlData{anchor=t-sub}{\\color{@sky}{(x + 4)}}^2 + 2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-var"},{"color":"sky","style":"text-color","anchor":"t-sub"}],"connections":[{"to":"t-sub","from":"t-var","color":"sky","label":"substitute t","style":"solid"}]},"order":1,"title":"Substitute into y","description":"Now, we substitute this expression for $t$ into the parametric equation for $y$ to get the Cartesian equation of the curve."},{"type":"text","order":2,"title":"Identify key features","content":"From the Cartesian equation $y = (x + 4)^2 + 2$, we can determine the key features required to sketch the curve:\n\n1. **Shape**: Since the coefficient of $x^2$ is positive, the curve is a parabola opening upwards B1.\n\n2. **Vertex (Minimum Point)**: The equation is already in completed-square form $y = a(x - h)^2 + k$, where the vertex is $(h, k)$. \n Comparing this, we get the minimum point at:\n $$(-4, 2)$$\n B1\n\n3. **$y$-intercept**: Set $x = 0$ to find where the curve crosses the $y$-axis:\n $$\\begin{aligned} y &= (0 + 4)^2 + 2 \\\\ y &= 16 + 2 = 18 \\end{aligned}$$\n B1"},{"type":"text","order":3,"title":"Sketch the curve","content":"Using these key properties, we can now sketch the parabola:\n- Draw a set of $x$ and $y$ axes.\n- Plot the minimum point at $(-4, 2)$ in the second quadrant.\n- Mark the $y$-intercept at $(0, 18)$.\n- Draw a smooth, symmetric, upward-opening U-shaped curve passing through these points.","calculator":[{"gdc":{"expression":"(x + 4)^2 + 2","instructions":"1. Go to the Graph application.\n2. Enter the function \$ f1(x) = (x + 4)^2 + 2 \$.\n3. Adjust the window settings to capture the vertex at \$(-4, 2)\$ and \$y\$-intercept at \$18\$.\n4. View the graph."},"ti84":{"expression":"(X + 4)^2 + 2","instructions":"1. Press [Y=].\n2. Enter \$ Y_1 = (X + 4)^2 + 2 \$.\n3. Press [WINDOW] and set Xmin = -8, Xmax = 2, Ymin = 0, Ymax = 25.\n4. Press [GRAPH] to view the parabola."},"desmos":{"expression":"y = (x+4)^2 + 2","instructions":"1. In the expression bar, type the equation: \$ y = (x + 4)^2 + 2 \$.\n2. Click on the gray intercept dots on the graph to confirm the vertex is at \$(-4, 2)\$ and the \$y\$-intercept is at \$(0, 18)\$."},"description":"You can verify your sketch by graphing the Cartesian equation on a calculator."}]}]},{"id":"table_of_values","label":"Table of Values Approach","steps":[{"type":"text","order":0,"title":"Select values of $t$","content":"To sketch the parametric curve using a **Table of Values**, we will choose several values for the parameter $t$ and calculate the corresponding $x$ and $y$ coordinates.\n\nLet's choose a range of integers for $t$ around $0$, for example from $t = -2$ to $t = 6$, to get a clear picture of how the curve behaves."},{"type":"text","order":1,"title":"Calculate coordinates","content":"Using the parametric equations:\n\n$$x = t - 4$$\n$$y = t^2 + 2$$\n\nLet's calculate the coordinates $(x, y)$ for each chosen $t$ value:\n\n* **For $t = -2$:**\n $$x = -2 - 4 = -6$$\n $$y = (-2)^2 + 2 = 6 \\implies (-6, 6)$$\n* **For $t = -1$:**\n $$x = -1 - 4 = -5$$\n $$y = (-1)^2 + 2 = 3 \\implies (-5, 3)$$\n* **For $t = 0$:**\n $$x = 0 - 4 = -4$$\n $$y = 0^2 + 2 = 2 \\implies (-4, 2)$$\n* **For $t = 1$:**\n $$x = 1 - 4 = -3$$\n $$y = 1^2 + 2 = 3 \\implies (-3, 3)$$\n* **For $t = 2$:**\n $$x = 2 - 4 = -2$$\n $$y = 2^2 + 2 = 6 \\implies (-2, 6)$$\n* **For $t = 4$:**\n $$x = 4 - 4 = 0$$\n $$y = 4^2 + 2 = 18 \\implies (0, 18)$$\n\nLet's compile these values into a table."},{"type":"text","order":2,"title":"Create the table","content":"Here is our completed table of values:\n\n$$\\begin{array}{|c|c|c|c|} \\hline t & x = t - 4 & y = t^2 + 2 & (x, y) \\\\ \\hline -2 & -6 & 6 & (-6, 6) \\\\ \\hline -1 & -5 & 3 & (-5, 3) \\\\ \\hline 0 & -4 & 2 & (-4, 2) \\\\ \\hline 1 & -3 & 3 & (-3, 3) \\\\ \\hline 2 & -2 & 6 & (-2, 6) \\\\ \\hline 4 & 0 & 18 & (0, 18) \\\\ \\hline \\end{array}$$"},{"type":"text","order":3,"title":"Identify key features","content":"Before sketching, we can identify key features of the curve from our calculations:\n\n1. **Shape:** As $t$ increases or decreases from $0$, $y$ increases symmetrically around $t = 0$. Since $y$ is quadratic in $t$ and $x$ is linear, this curve is a parabola that opens upwards.\n2. **Minimum Point:** The minimum value of $y$ occurs when $t = 0$. At this point, the coordinates are $\\mathbf{(-4, 2)}$.\n3. **$y$-intercept:** This occurs where $x = 0$, which corresponds to $t = 4$. The coordinates of this intercept are $\\mathbf{(0, 18)}$."},{"type":"text","order":4,"title":"Sketch the curve","content":"Now, plot the coordinate points on a Cartesian plane and connect them with a smooth, continuous curve:\n\n1. Plot the vertex (minimum point) at $(-4, 2)$.\n2. Plot the nearby symmetrical points like $(-5, 3)$ and $(-3, 3)$, as well as $(-6, 6)$ and $(-2, 6)$.\n3. Plot the $y$-intercept at $(0, 18)$.\n4. Draw a smooth, upward-opening parabola passing through these plotted points.\n\nBe sure to clearly label the minimum point $(-4, 2)$ and the $y$-intercept $(0, 18)$ to secure all **3 marks** from the markscheme.","calculator":[{"gdc":{"expression":"x(t) = t - 4, y(t) = t^2 + 2","instructions":"1. Open the Graphing app and change the graph type to Parametric.\n2. Enter x(t) = t - 4 and y(t) = t^2 + 2.\n3. Adjust the window settings to capture the vertex at x = -4 and the y-intercept at y = 18.\n4. Draw the graph."},"ti84":{"expression":"X1T = T - 4, Y1T = T^2 + 2","instructions":"1. Press [MODE], arrow down to 'FUNCTION' and change it to 'PARAMETRIC' (PAR).\n2. Press [Y=].\n3. Enter X1T = T - 4 and Y1T = T^2 + 2.\n4. Press [WINDOW] and set Tmin = -5, Tmax = 5, Xmin = -10, Xmax = 5, Ymin = -2, Ymax = 25.\n5. Press [GRAPH] to view the curve."},"desmos":{"expression":"(t - 4, t^2 + 2)","instructions":"1. In the input bar, type the parametric coordinates as an ordered pair: (t - 4, t^2 + 2).\n2. Adjust the parameter range for t, for example: -5 <= t <= 5.\n3. Zoom out to clearly see the minimum point at (-4, 2) and the y-intercept at (0, 18)."},"description":"Visualize the parametric curve directly using a graphing utility."}]}]}],"generatedAt":"2026-06-09T05:30:06.718Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c0401f07-1f3a-412c-94de-614c31b624a9:substitution_t_from_x","label":"(a) Substitution of t in terms of x","steps":[{"type":"text","order":0,"title":"Rearrange x for t","content":"To find a cartesian equation, we need to eliminate the parameter $t$. We start with the linear parametric equation for $x$:\n\n$$x = t - 4$$\n\nTo express $t$ in terms of $x$, we add $4$ to both sides:\n\n$$t = x + 4$$","highlights":[{"text":"x = t - 4","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$t = \\htmlData{anchor=t-val}{\\color{@sky}{x + 4}}$"},{"latex":"$$y = \\htmlData{anchor=t-var}{\\color{@sky}{t}}^2 + 2$"},{"latex":"$$y = \\htmlData{anchor=substituted}{\\color{@sky}{(x + 4)}}^2 + 2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-val"},{"color":"sky","style":"text-color","anchor":"substituted"}],"connections":[{"to":"t-var","from":"t-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute t into y","description":"Now, we substitute this expression for $t$ into the parametric equation for $y$:"},{"type":"text","order":2,"title":"Expand and simplify","content":"Now we expand the squared term and simplify the expression to get the final cartesian equation:\n\n$$\\begin{aligned} y &= (x + 4)^2 + 2 \\\\ y &= (x^2 + 8x + 16) + 2 \\\\ y &= x^2 + 8x + 18 \\end{aligned}$$\n\nThis gives us a quadratic curve (a parabola) representing the cartesian equation of the curve."}]},{"id":"a8e4be13-e471-4a11-8007-5f182789200b:cartesian_conversion","label":"(b) Conversion to Cartesian Form","steps":[{"type":"text","order":0,"title":"Rearrange for t","content":"To convert the parametric equations into Cartesian form, we start by rearranging the equation for $x$ to express $t$ in terms of $x$.\n\nGiven:\n$$x = t - 4$$\n\nAdding $4$ to both sides gives:\n$$t = x + 4$$","highlights":[{"text":"x = t - 4","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=t-var}{\\color{@sky}{t}}^2 + 2$"},{"latex":"$$y = \\htmlData{anchor=t-sub}{\\color{@sky}{(x + 4)}}^2 + 2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-var"},{"color":"sky","style":"text-color","anchor":"t-sub"}],"connections":[{"to":"t-sub","from":"t-var","color":"sky","label":"substitute t","style":"solid"}]},"order":1,"title":"Substitute into y","description":"Now, we substitute this expression for $t$ into the parametric equation for $y$ to get the Cartesian equation of the curve."},{"type":"text","order":2,"title":"Identify key features","content":"From the Cartesian equation $y = (x + 4)^2 + 2$, we can determine the key features required to sketch the curve:\n\n1. **Shape**: Since the coefficient of $x^2$ is positive, the curve is a parabola opening upwards B1.\n\n2. **Vertex (Minimum Point)**: The equation is already in completed-square form $y = a(x - h)^2 + k$, where the vertex is $(h, k)$. \n Comparing this, we get the minimum point at:\n $$(-4, 2)$$\n B1\n\n3. **$y$-intercept**: Set $x = 0$ to find where the curve crosses the $y$-axis:\n $$\\begin{aligned} y &= (0 + 4)^2 + 2 \\\\ y &= 16 + 2 = 18 \\end{aligned}$$\n B1"},{"type":"text","order":3,"title":"Sketch the curve","content":"Using these key properties, we can now sketch the parabola:\n- Draw a set of $x$ and $y$ axes.\n- Plot the minimum point at $(-4, 2)$ in the second quadrant.\n- Mark the $y$-intercept at $(0, 18)$.\n- Draw a smooth, symmetric, upward-opening U-shaped curve passing through these points.","calculator":[{"gdc":{"expression":"(x + 4)^2 + 2","instructions":"1. Go to the Graph application.\n2. Enter the function \$ f1(x) = (x + 4)^2 + 2 \$.\n3. Adjust the window settings to capture the vertex at \$(-4, 2)\$ and \$y\$-intercept at \$18\$.\n4. View the graph."},"ti84":{"expression":"(X + 4)^2 + 2","instructions":"1. Press [Y=].\n2. Enter \$ Y_1 = (X + 4)^2 + 2 \$.\n3. Press [WINDOW] and set Xmin = -8, Xmax = 2, Ymin = 0, Ymax = 25.\n4. Press [GRAPH] to view the parabola."},"desmos":{"expression":"y = (x+4)^2 + 2","instructions":"1. In the expression bar, type the equation: \$ y = (x + 4)^2 + 2 \$.\n2. Click on the gray intercept dots on the graph to confirm the vertex is at \$(-4, 2)\$ and the \$y\$-intercept is at \$(0, 18)\$."},"description":"You can verify your sketch by graphing the Cartesian equation on a calculator."}]}]},{"id":"a8e4be13-e471-4a11-8007-5f182789200b:table_of_values","label":"(b) Table of Values Approach","steps":[{"type":"text","order":0,"title":"Select values of $t$","content":"To sketch the parametric curve using a **Table of Values**, we will choose several values for the parameter $t$ and calculate the corresponding $x$ and $y$ coordinates.\n\nLet's choose a range of integers for $t$ around $0$, for example from $t = -2$ to $t = 6$, to get a clear picture of how the curve behaves."},{"type":"text","order":1,"title":"Calculate coordinates","content":"Using the parametric equations:\n\n$$x = t - 4$$\n$$y = t^2 + 2$$\n\nLet's calculate the coordinates $(x, y)$ for each chosen $t$ value:\n\n* **For $t = -2$:**\n $$x = -2 - 4 = -6$$\n $$y = (-2)^2 + 2 = 6 \\implies (-6, 6)$$\n* **For $t = -1$:**\n $$x = -1 - 4 = -5$$\n $$y = (-1)^2 + 2 = 3 \\implies (-5, 3)$$\n* **For $t = 0$:**\n $$x = 0 - 4 = -4$$\n $$y = 0^2 + 2 = 2 \\implies (-4, 2)$$\n* **For $t = 1$:**\n $$x = 1 - 4 = -3$$\n $$y = 1^2 + 2 = 3 \\implies (-3, 3)$$\n* **For $t = 2$:**\n $$x = 2 - 4 = -2$$\n $$y = 2^2 + 2 = 6 \\implies (-2, 6)$$\n* **For $t = 4$:**\n $$x = 4 - 4 = 0$$\n $$y = 4^2 + 2 = 18 \\implies (0, 18)$$\n\nLet's compile these values into a table."},{"type":"text","order":2,"title":"Create the table","content":"Here is our completed table of values:\n\n$$\\begin{array}{|c|c|c|c|} \\hline t & x = t - 4 & y = t^2 + 2 & (x, y) \\\\ \\hline -2 & -6 & 6 & (-6, 6) \\\\ \\hline -1 & -5 & 3 & (-5, 3) \\\\ \\hline 0 & -4 & 2 & (-4, 2) \\\\ \\hline 1 & -3 & 3 & (-3, 3) \\\\ \\hline 2 & -2 & 6 & (-2, 6) \\\\ \\hline 4 & 0 & 18 & (0, 18) \\\\ \\hline \\end{array}$$"},{"type":"text","order":3,"title":"Identify key features","content":"Before sketching, we can identify key features of the curve from our calculations:\n\n1. **Shape:** As $t$ increases or decreases from $0$, $y$ increases symmetrically around $t = 0$. Since $y$ is quadratic in $t$ and $x$ is linear, this curve is a parabola that opens upwards.\n2. **Minimum Point:** The minimum value of $y$ occurs when $t = 0$. At this point, the coordinates are $\\mathbf{(-4, 2)}$.\n3. **$y$-intercept:** This occurs where $x = 0$, which corresponds to $t = 4$. The coordinates of this intercept are $\\mathbf{(0, 18)}$."},{"type":"text","order":4,"title":"Sketch the curve","content":"Now, plot the coordinate points on a Cartesian plane and connect them with a smooth, continuous curve:\n\n1. Plot the vertex (minimum point) at $(-4, 2)$.\n2. Plot the nearby symmetrical points like $(-5, 3)$ and $(-3, 3)$, as well as $(-6, 6)$ and $(-2, 6)$.\n3. Plot the $y$-intercept at $(0, 18)$.\n4. Draw a smooth, upward-opening parabola passing through these plotted points.\n\nBe sure to clearly label the minimum point $(-4, 2)$ and the $y$-intercept $(0, 18)$ to secure all **3 marks** from the markscheme.","calculator":[{"gdc":{"expression":"x(t) = t - 4, y(t) = t^2 + 2","instructions":"1. Open the Graphing app and change the graph type to Parametric.\n2. Enter x(t) = t - 4 and y(t) = t^2 + 2.\n3. Adjust the window settings to capture the vertex at x = -4 and the y-intercept at y = 18.\n4. Draw the graph."},"ti84":{"expression":"X1T = T - 4, Y1T = T^2 + 2","instructions":"1. Press [MODE], arrow down to 'FUNCTION' and change it to 'PARAMETRIC' (PAR).\n2. Press [Y=].\n3. Enter X1T = T - 4 and Y1T = T^2 + 2.\n4. Press [WINDOW] and set Tmin = -5, Tmax = 5, Xmin = -10, Xmax = 5, Ymin = -2, Ymax = 25.\n5. Press [GRAPH] to view the curve."},"desmos":{"expression":"(t - 4, t^2 + 2)","instructions":"1. In the input bar, type the parametric coordinates as an ordered pair: (t - 4, t^2 + 2).\n2. Adjust the parameter range for t, for example: -5 <= t <= 5.\n3. Zoom out to clearly see the minimum point at (-4, 2) and the y-intercept at (0, 18)."},"description":"Visualize the parametric curve directly using a graphing utility."}]}]}],"generatedAt":"2026-06-09T05:30:19.029Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"0cf3cec0-7d3e-4255-a08a-cc01a4ed3f3b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"100c6223-1680-4a86-990e-d2ad9bf0528d","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"543fca01-78c6-49af-ac8c-db2b576575df","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"5adb3bf5-a974-44f3-b829-cd9f56fa4014","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"afd9c275-74ca-442b-82d8-681c50290b45","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":false},{"id":"cb0ab090-0a4d-4b4a-ad34-c6028469b06f","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"d6b8dd59-ebfe-49f9-b11e-a5a5e346e2b8","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"1f3b9e4e-4c6e-4a3b-a095-ff140a7d50d0","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"7b3f3471-2160-4d33-8e91-64f2aeec3242","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"7c254a8f-4c38-4870-99b5-0105397c86cf","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"96d41ac0-231d-4cf5-83f2-5ea8f306db12","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"e2db1103-cf17-4e61-8786-dbdbbf13a927","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":false},{"id":"e9d7a7f9-40a0-46fd-8404-cce418aa8297","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"1b974b05-47bd-4b1b-9b1f-3b22927a212f","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"50d7e278-f5cc-4135-ac25-412e2b9abfaf","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":false},{"id":"5a32a335-0d40-411d-87e1-2723a51994e4","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"6b9dcbff-22e8-4551-b86a-2e2781f1df7b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"cce8998b-de31-4c16-af74-d67bb0fbe522","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"d1ae58c3-5694-4151-9072-a8c360ff032a","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false}],"title":"8.3 Curve Sketching","topicId":"13f22ac1-2516-4b0f-93f4-a700d39f70f3","topicSidebar":"$L3b","topicSlug":"a-level-pure-mathematics-8-3-curve-sketching","topicTitle":"8.3 Curve Sketching"}]}],"$L3c","$L3d"]}]