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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":7,"initialQuestionWindow":[{"id":"d6b8dd59-ebfe-49f9-b11e-a5a5e346e2b8","revisionId":"b84cc5d2-9af3-42d2-9d87-a0876d2d5d92","questionNumber":null,"questionType":"LA","specification":"A curve $C$ has parametric equations\n\n$$x = \\tan t + 1, \\quad y = \\sec^2 t + 4, \\quad 0 \\leq t \\leq 2\\pi, \\, t \\neq \\frac{\\pi}{2}, \\frac{3\\pi}{2}$$","difficulty":1,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"25fd70a3-6433-406e-92b2-a975e70aa0fc","label":"a","content":"Find an equation for $C$ in the form $y = f(x)$ and state the domain for which the curve is defined.","markscheme":"- Attempt to use the trigonometric identity $1 + \\tan^2 t = \\sec^2 t$ M1\n- Substitute $\\tan t = x - 1$ and $\\sec^2 t = y - 4$ into the identity: $y - 4 = 1 + (x - 1)^2$ M1\n- $y = (x - 1)^2 + 5$ with domain $x \\in \\mathbb{R}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Evidence of using the identity $\\sec^2 t = 1 + \\tan^2 t$.","code":"M"},{"id":"p2","marks":1,"description":"Correct substitution of $x$ and $y$ expressions into the identity.","code":"M"},{"id":"p3","marks":1,"description":"Correct Cartesian equation $y = (x - 1)^2 + 5$ (or $y = x^2 - 2x + 6$) and states domain $x \\in \\mathbb{R}$.","code":"A"}],"finalAnswers":["$$y = (x - 1)^2 + 5, x \\in \\mathbb{R}$","$$y = x^2 - 2x + 6, x \\in \\mathbb{R}$"],"generatedAt":"2026-05-06T14:04:59.675Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"trig_identity","label":"Using Pythagorean Identity","steps":[{"type":"text","order":0,"title":"Rearrange for trigonometric terms","content":"To eliminate the parameter $t$, we first express the trigonometric functions $\\tan t$ and $\\sec^2 t$ in terms of $x$ and $y$.\n\nFrom the first parametric equation:\n$$x = \\tan t + 1 \\implies \\tan t = x - 1$$\n\nFrom the second parametric equation:\n$$y = \\sec^2 t + 4 \\implies \\sec^2 t = y - 4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=identity}{\\color{@purple}{\\sec^2 t}} = 1 + \\htmlData{anchor=tan-sq}{\\color{@sky}{\\tan^2 t}}$"},{"latex":"$$\\htmlData{anchor=sec-sub}{\\color{@purple}{(y - 4)}} = 1 + \\htmlData{anchor=tan-sub}{\\color{@sky}{(x - 1)^2}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"identity"},{"color":"sky","style":"text-color","anchor":"tan-sq"}],"connections":[{"to":"sec-sub","from":"identity","color":"purple","label":"substitute sec² t = y - 4","style":"solid"},{"to":"tan-sub","from":"tan-sq","color":"sky","label":"substitute tan t = x - 1","style":"solid"}]},"order":1,"title":"Use Pythagorean identity","description":"We recall the fundamental trigonometric identity:\n\n$$\\sec^2 t = 1 + \\tan^2 t$$\n\nWe then substitute our expressions for $\\tan t$ and $\\sec^2 t$ into this identity."},{"type":"text","order":2,"title":"Solve for y","content":"Rearrange the equation into the form $y = f(x)$ by adding $4$ to both sides of the equation:\n\n$$y - 4 = 1 + (x - 1)^2$$\n\n$$y = (x - 1)^2 + 5$$"},{"type":"text","order":3,"title":"Determine the domain","content":"The domain of a parametric curve is the set of all possible values that $x$ can take.\n\nSince $x = \\tan t + 1$ and $t$ is defined for $0 \\leq t \\leq 2\\pi$, $t \\neq \\frac{\\pi}{2}, \\frac{3\\pi}{2}$, we look at the range of the tangent function. Over this interval, the function $\\tan t$ takes all real values.\n\nTherefore, $x$ can take any real value, meaning the domain is all real numbers:\n\n$$x \\in \\mathbb{R}$$"}]},{"id":"inverse_trig","label":"Using Inverse Trigonometric Functions","steps":[{"type":"text","order":0,"title":"Rearrange for the parameter","content":"First, let's rearrange the parametric equation for $x$ to express the parameter $t$ in terms of $x$.\n\nWe start with:\n$$x = \\tan t + 1$$\n\nSubtract $1$ from both sides:\n$$x - 1 = \\tan t$$\n\nNow, we can take the inverse tangent (arctangent) of both sides to isolate $t$:\n$$t = \\arctan(x - 1)$$"},{"type":"text","order":1,"title":"Substitute into the equation","content":"Next, we substitute our expression for $t$ into the parametric equation for $y$:\n$$y = \\sec^2 t + 4$$\n\nReplacing $t$ with $\\arctan(x - 1)$, we get:\n$$y = \\sec^2(\\arctan(x - 1)) + 4$$"},{"type":"text","order":2,"title":"Simplify using identities","content":"To simplify this expression, recall the fundamental trigonometric identity:\n\n$$\\sec^2 \\theta = 1 + \\tan^2 \\theta$$\n\nIf we let $\\theta = \\arctan(x - 1)$, then:\n\n$$\\tan(\\arctan(x - 1)) = x - 1$$\n\nSubstituting this into the identity gives:\n\n$$\\sec^2(\\arctan(x - 1)) = 1 + (x - 1)^2$$\n\nNow we substitute this back into our equation for $y$:\n\n$$y = [1 + (x - 1)^2] + 4$$\n\n$$y = (x - 1)^2 + 5$$"},{"type":"text","order":3,"title":"Determine the domain","content":"Now let's find the domain of the Cartesian equation.\n\nThe parameter $t$ is defined for $0 \\le t \\le 2\\pi$ excluding $t = \\frac{\\pi}{2}$ and $t = \\frac{3\\pi}{2}$.\n\nSince the function $x = \\tan t + 1$ has vertical asymptotes at $t = \\frac{\\pi}{2}$ and $t = \\frac{3\\pi}{2}$, the range of $\\tan t$ over the given interval is all real numbers, $(-\\infty, \\infty)$.\n\nTherefore, $x$ can take any real value:\n\n$$x \\in \\mathbb{R}$$"}]}],"generatedAt":"2026-06-09T08:19:56.846Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"40ee5759-6761-4c71-8909-826de6f1539c","label":"b","content":"Sketch the curve $C$.","markscheme":"- Correct shape of a parabola opening upwards B1\n- Vertex correctly identified at $(1, 5)$ B1\n- $y$-intercept correctly identified at $(0, 6)$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"A sketch showing the characteristic shape of a positive quadratic (parabola).","code":"B"},{"id":"p2","marks":1,"description":"The vertex of the parabola is correctly located at $(1, 5)$.","code":"B"},{"id":"p3","marks":1,"description":"The $y$-intercept is correctly located or labeled at $(0, 6)$.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:04:59.675Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"cartesian","label":"Cartesian Equation","steps":[{"type":"text","order":0,"title":"Find the Cartesian equation","content":"To sketch the curve, we first convert the parametric equations into a Cartesian equation by eliminating the parameter $t$.\n\nWe are given:\n$$x = \\tan t + 1 \\implies \\tan t = x - 1$$\n$$y = \\sec^2 t + 4$$\n\nNow we use the fundamental trigonometric identity from the formula booklet:\n$$\\sec^2 t = 1 + \\tan^2 t$$\n\nSubstituting this identity into our equation for $y$:\n$$\\begin{aligned} y &= (1 + \\tan^2 t) + 4 \\\\ y &= \\tan^2 t + 5 \\end{aligned}$$\n\nNow substitute $\\tan t = x - 1$ into this equation:\n$$y = (x - 1)^2 + 5$$"},{"type":"text","order":1,"title":"Analyze the curve's features","content":"The Cartesian equation is in the vertex form of a quadratic equation:\n$$y = a(x - h)^2 + k$$\n\nFrom our equation $$y = (x - 1)^2 + 5$$:\n- Since $a = 1 > 0$, the curve is a **parabola opening upwards**.\n- The **vertex** $(h, k)$ is at $(1, 5)$.\n- To find the **$y$-intercept**, substitute $x = 0$:\n $$y = (0 - 1)^2 + 5 = 1 + 5 = 6$$\n So, the $y$-intercept is at $(0, 6)$.\n\n*Note on domain:* Since $0 \\leq t \\leq 2\\pi$ and $t \\neq \\frac{\\pi}{2}, \\frac{3\\pi}{2}$, the term $\\tan t$ takes all real values, which means $x$ can be any real number ($-\\infty < x < \\infty$). Thus, the entire parabola is sketched."},{"type":"text","order":2,"title":"Sketch the parabola","content":"Now we sketch the curve using these key features:\n\n1. Draw a set of coordinate axes.\n2. Plot the minimum point (vertex) at $(1, 5)$.\n3. Plot the $y$-intercept at $(0, 6)$.\n4. Draw a smooth, symmetric U-shaped parabola opening upwards that passes through $(0, 6)$, reaches its lowest point at $(1, 5)$, and curves back upwards.\n\n### Markscheme Distribution:\n- **B1**: Correct shape of a parabola opening upwards.\n- **B1**: Vertex correctly identified and plotted at $(1, 5)$.\n- **B1**: $y$-intercept correctly identified and plotted at $(0, 6)$.","calculator":[{"gdc":{"expression":"(x - 1)^2 + 5","instructions":"Go to the Graph app. Enter f1(x) = (x - 1)^2 + 5. View the graph and use the 'Analyze Graph' tool to find the minimum point (vertex) and the y-intercept."},"ti84":{"expression":"(X - 1)^2 + 5","instructions":"Press [Y=], enter Y1 = (X - 1)^2 + 5. Press [WINDOW] and set Xmin = -4, Xmax = 6, Ymin = 0, Ymax = 15. Press [GRAPH]. Press [2nd] [CALC] and select 'value' with X = 0 to verify the y-intercept (0, 6), and select 'minimum' to verify the vertex (1, 5)."},"desmos":{"expression":"y = (x - 1)^2 + 5","instructions":"Type y = (x - 1)^2 + 5 into the expression box. Click on the gray dots at the vertex and y-intercept to display their coordinates."},"description":"Verify your sketch by graphing the Cartesian equation."}]}]},{"id":"plotting","label":"Point Plotting","steps":[{"type":"text","order":0,"title":"Set up a table of values","content":"To sketch the curve $C$ using point plotting, we calculate $(x, y)$ coordinates for various values of $t$. Since the functions are periodic and undefined at $t = \\frac{\\pi}{2}$ and $t = \\frac{3\\pi}{2}$, we can choose values of $t$ in the interval $\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)$ to understand the shape of the main branch of the curve.\n\nLet's recall the parametric equations:\n\n$$x = \\tan t + 1$$\n$$y = \\sec^2 t + 4$$"},{"type":"text","order":1,"title":"Calculate coordinates for key values","content":"Let's calculate $(x, y)$ for specific values of $t$:\n\n* **For $t = 0$:**\n $$x = \\tan(0) + 1 = 0 + 1 = 1$$\n $$y = \\sec^2(0) + 4 = 1^2 + 4 = 5$$\n This gives the coordinate **$(1, 5)$**.\n\n* **For $t = -\\frac{\\pi}{4}$:**\n $$x = \\tan\\left(-\\frac{\\pi}{4}\\right) + 1 = -1 + 1 = 0$$\n $$y = \\sec^2\\left(-\\frac{\\pi}{4}\\right) + 4 = (\\sqrt{2})^2 + 4 = 2 + 4 = 6$$\n This gives the coordinate **$(0, 6)$**.\n\n* **For $t = \\frac{\\pi}{4}$:**\n $$x = \\tan\\left(\\frac{\\pi}{4}\\right) + 1 = 1 + 1 = 2$$\n $$y = \\sec^2\\left(\\frac{\\pi}{4}\\right) + 4 = (\\sqrt{2})^2 + 4 = 2 + 4 = 6$$\n This gives the coordinate **$(2, 6)$**."},{"type":"text","order":2,"title":"Analyze the points and shape","content":"Let's summarize our plotted points:\n\n$$\\begin{array}{c|c|c} t & x & y \\\\ \\hline -\\frac{\\pi}{4} & 0 & 6 \\\\ 0 & 1 & 5 \\\\ \\frac{\\pi}{4} & 2 & 6 \\\\ \\end{array}$$\n\n- The lowest point on this curve is at $(1, 5)$, which represents the vertex.\n- The point $(0, 6)$ is the $y$-intercept.\n- The points are symmetric about the vertical line $x = 1$.\n- Since $y = \\sec^2 t + 4 = (\\tan^2 t + 1) + 4 = \\tan^2 t + 5$, and $x - 1 = \\tan t$, we have:\n $$y = (x - 1)^2 + 5$$\n This is a parabola opening upwards with its vertex at $(1, 5)$.."},{"type":"text","order":3,"title":"Sketch the curve","content":"To sketch the curve $C$:\n\n1. Draw a set of coordinate axes.\n2. Plot the vertex at **$(1, 5)$**.\n3. Plot the $y$-intercept at **$(0, 6)$** and its symmetric counterpart at **$(2, 6)$**.\n4. Draw a smooth, upward-opening parabolic curve passing through these points.\n\nSince $t$ can take any value in $[0, 2\\pi]$ except $\\frac{\\pi}{2}$ and $\\frac{3\\pi}{2}$, $x = \\tan t + 1$ can take any real value ($-\\infty < x < \\infty$). Thus, the curve is a complete parabola opening upwards.\n\n**Markscheme criteria check:**\n- Correct shape of an upward-opening parabola B1\n- Vertex correctly shown at $(1, 5)$ B1\n- $y$-intercept correctly shown at $(0, 6)$ B1"}]}],"generatedAt":"2026-06-09T08:24:27.802Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"25fd70a3-6433-406e-92b2-a975e70aa0fc:trig_identity","label":"(a) Using Pythagorean Identity","steps":[{"type":"text","order":0,"title":"Rearrange for trigonometric terms","content":"To eliminate the parameter $t$, we first express the trigonometric functions $\\tan t$ and $\\sec^2 t$ in terms of $x$ and $y$.\n\nFrom the first parametric equation:\n$$x = \\tan t + 1 \\implies \\tan t = x - 1$$\n\nFrom the second parametric equation:\n$$y = \\sec^2 t + 4 \\implies \\sec^2 t = y - 4$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=identity}{\\color{@purple}{\\sec^2 t}} = 1 + \\htmlData{anchor=tan-sq}{\\color{@sky}{\\tan^2 t}}$"},{"latex":"$$\\htmlData{anchor=sec-sub}{\\color{@purple}{(y - 4)}} = 1 + \\htmlData{anchor=tan-sub}{\\color{@sky}{(x - 1)^2}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"identity"},{"color":"sky","style":"text-color","anchor":"tan-sq"}],"connections":[{"to":"sec-sub","from":"identity","color":"purple","label":"substitute sec² t = y - 4","style":"solid"},{"to":"tan-sub","from":"tan-sq","color":"sky","label":"substitute tan t = x - 1","style":"solid"}]},"order":1,"title":"Use Pythagorean identity","description":"We recall the fundamental trigonometric identity:\n\n$$\\sec^2 t = 1 + \\tan^2 t$$\n\nWe then substitute our expressions for $\\tan t$ and $\\sec^2 t$ into this identity."},{"type":"text","order":2,"title":"Solve for y","content":"Rearrange the equation into the form $y = f(x)$ by adding $4$ to both sides of the equation:\n\n$$y - 4 = 1 + (x - 1)^2$$\n\n$$y = (x - 1)^2 + 5$$"},{"type":"text","order":3,"title":"Determine the domain","content":"The domain of a parametric curve is the set of all possible values that $x$ can take.\n\nSince $x = \\tan t + 1$ and $t$ is defined for $0 \\leq t \\leq 2\\pi$, $t \\neq \\frac{\\pi}{2}, \\frac{3\\pi}{2}$, we look at the range of the tangent function. Over this interval, the function $\\tan t$ takes all real values.\n\nTherefore, $x$ can take any real value, meaning the domain is all real numbers:\n\n$$x \\in \\mathbb{R}$$"}]},{"id":"25fd70a3-6433-406e-92b2-a975e70aa0fc:inverse_trig","label":"(a) Using Inverse Trigonometric Functions","steps":[{"type":"text","order":0,"title":"Rearrange for the parameter","content":"First, let's rearrange the parametric equation for $x$ to express the parameter $t$ in terms of $x$.\n\nWe start with:\n$$x = \\tan t + 1$$\n\nSubtract $1$ from both sides:\n$$x - 1 = \\tan t$$\n\nNow, we can take the inverse tangent (arctangent) of both sides to isolate $t$:\n$$t = \\arctan(x - 1)$$"},{"type":"text","order":1,"title":"Substitute into the equation","content":"Next, we substitute our expression for $t$ into the parametric equation for $y$:\n$$y = \\sec^2 t + 4$$\n\nReplacing $t$ with $\\arctan(x - 1)$, we get:\n$$y = \\sec^2(\\arctan(x - 1)) + 4$$"},{"type":"text","order":2,"title":"Simplify using identities","content":"To simplify this expression, recall the fundamental trigonometric identity:\n\n$$\\sec^2 \\theta = 1 + \\tan^2 \\theta$$\n\nIf we let $\\theta = \\arctan(x - 1)$, then:\n\n$$\\tan(\\arctan(x - 1)) = x - 1$$\n\nSubstituting this into the identity gives:\n\n$$\\sec^2(\\arctan(x - 1)) = 1 + (x - 1)^2$$\n\nNow we substitute this back into our equation for $y$:\n\n$$y = [1 + (x - 1)^2] + 4$$\n\n$$y = (x - 1)^2 + 5$$"},{"type":"text","order":3,"title":"Determine the domain","content":"Now let's find the domain of the Cartesian equation.\n\nThe parameter $t$ is defined for $0 \\le t \\le 2\\pi$ excluding $t = \\frac{\\pi}{2}$ and $t = \\frac{3\\pi}{2}$.\n\nSince the function $x = \\tan t + 1$ has vertical asymptotes at $t = \\frac{\\pi}{2}$ and $t = \\frac{3\\pi}{2}$, the range of $\\tan t$ over the given interval is all real numbers, $(-\\infty, \\infty)$.\n\nTherefore, $x$ can take any real value:\n\n$$x \\in \\mathbb{R}$$"}]},{"id":"40ee5759-6761-4c71-8909-826de6f1539c:cartesian","label":"(b) Cartesian Equation","steps":[{"type":"text","order":0,"title":"Find the Cartesian equation","content":"To sketch the curve, we first convert the parametric equations into a Cartesian equation by eliminating the parameter $t$.\n\nWe are given:\n$$x = \\tan t + 1 \\implies \\tan t = x - 1$$\n$$y = \\sec^2 t + 4$$\n\nNow we use the fundamental trigonometric identity from the formula booklet:\n$$\\sec^2 t = 1 + \\tan^2 t$$\n\nSubstituting this identity into our equation for $y$:\n$$\\begin{aligned} y &= (1 + \\tan^2 t) + 4 \\\\ y &= \\tan^2 t + 5 \\end{aligned}$$\n\nNow substitute $\\tan t = x - 1$ into this equation:\n$$y = (x - 1)^2 + 5$$"},{"type":"text","order":1,"title":"Analyze the curve's features","content":"The Cartesian equation is in the vertex form of a quadratic equation:\n$$y = a(x - h)^2 + k$$\n\nFrom our equation $$y = (x - 1)^2 + 5$$:\n- Since $a = 1 > 0$, the curve is a **parabola opening upwards**.\n- The **vertex** $(h, k)$ is at $(1, 5)$.\n- To find the **$y$-intercept**, substitute $x = 0$:\n $$y = (0 - 1)^2 + 5 = 1 + 5 = 6$$\n So, the $y$-intercept is at $(0, 6)$.\n\n*Note on domain:* Since $0 \\leq t \\leq 2\\pi$ and $t \\neq \\frac{\\pi}{2}, \\frac{3\\pi}{2}$, the term $\\tan t$ takes all real values, which means $x$ can be any real number ($-\\infty < x < \\infty$). Thus, the entire parabola is sketched."},{"type":"text","order":2,"title":"Sketch the parabola","content":"Now we sketch the curve using these key features:\n\n1. Draw a set of coordinate axes.\n2. Plot the minimum point (vertex) at $(1, 5)$.\n3. Plot the $y$-intercept at $(0, 6)$.\n4. Draw a smooth, symmetric U-shaped parabola opening upwards that passes through $(0, 6)$, reaches its lowest point at $(1, 5)$, and curves back upwards.\n\n### Markscheme Distribution:\n- **B1**: Correct shape of a parabola opening upwards.\n- **B1**: Vertex correctly identified and plotted at $(1, 5)$.\n- **B1**: $y$-intercept correctly identified and plotted at $(0, 6)$.","calculator":[{"gdc":{"expression":"(x - 1)^2 + 5","instructions":"Go to the Graph app. Enter f1(x) = (x - 1)^2 + 5. View the graph and use the 'Analyze Graph' tool to find the minimum point (vertex) and the y-intercept."},"ti84":{"expression":"(X - 1)^2 + 5","instructions":"Press [Y=], enter Y1 = (X - 1)^2 + 5. Press [WINDOW] and set Xmin = -4, Xmax = 6, Ymin = 0, Ymax = 15. Press [GRAPH]. Press [2nd] [CALC] and select 'value' with X = 0 to verify the y-intercept (0, 6), and select 'minimum' to verify the vertex (1, 5)."},"desmos":{"expression":"y = (x - 1)^2 + 5","instructions":"Type y = (x - 1)^2 + 5 into the expression box. Click on the gray dots at the vertex and y-intercept to display their coordinates."},"description":"Verify your sketch by graphing the Cartesian equation."}]}]},{"id":"40ee5759-6761-4c71-8909-826de6f1539c:plotting","label":"(b) Point Plotting","steps":[{"type":"text","order":0,"title":"Set up a table of values","content":"To sketch the curve $C$ using point plotting, we calculate $(x, y)$ coordinates for various values of $t$. Since the functions are periodic and undefined at $t = \\frac{\\pi}{2}$ and $t = \\frac{3\\pi}{2}$, we can choose values of $t$ in the interval $\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)$ to understand the shape of the main branch of the curve.\n\nLet's recall the parametric equations:\n\n$$x = \\tan t + 1$$\n$$y = \\sec^2 t + 4$$"},{"type":"text","order":1,"title":"Calculate coordinates for key values","content":"Let's calculate $(x, y)$ for specific values of $t$:\n\n* **For $t = 0$:**\n $$x = \\tan(0) + 1 = 0 + 1 = 1$$\n $$y = \\sec^2(0) + 4 = 1^2 + 4 = 5$$\n This gives the coordinate **$(1, 5)$**.\n\n* **For $t = -\\frac{\\pi}{4}$:**\n $$x = \\tan\\left(-\\frac{\\pi}{4}\\right) + 1 = -1 + 1 = 0$$\n $$y = \\sec^2\\left(-\\frac{\\pi}{4}\\right) + 4 = (\\sqrt{2})^2 + 4 = 2 + 4 = 6$$\n This gives the coordinate **$(0, 6)$**.\n\n* **For $t = \\frac{\\pi}{4}$:**\n $$x = \\tan\\left(\\frac{\\pi}{4}\\right) + 1 = 1 + 1 = 2$$\n $$y = \\sec^2\\left(\\frac{\\pi}{4}\\right) + 4 = (\\sqrt{2})^2 + 4 = 2 + 4 = 6$$\n This gives the coordinate **$(2, 6)$**."},{"type":"text","order":2,"title":"Analyze the points and shape","content":"Let's summarize our plotted points:\n\n$$\\begin{array}{c|c|c} t & x & y \\\\ \\hline -\\frac{\\pi}{4} & 0 & 6 \\\\ 0 & 1 & 5 \\\\ \\frac{\\pi}{4} & 2 & 6 \\\\ \\end{array}$$\n\n- The lowest point on this curve is at $(1, 5)$, which represents the vertex.\n- The point $(0, 6)$ is the $y$-intercept.\n- The points are symmetric about the vertical line $x = 1$.\n- Since $y = \\sec^2 t + 4 = (\\tan^2 t + 1) + 4 = \\tan^2 t + 5$, and $x - 1 = \\tan t$, we have:\n $$y = (x - 1)^2 + 5$$\n This is a parabola opening upwards with its vertex at $(1, 5)$.."},{"type":"text","order":3,"title":"Sketch the curve","content":"To sketch the curve $C$:\n\n1. Draw a set of coordinate axes.\n2. Plot the vertex at **$(1, 5)$**.\n3. Plot the $y$-intercept at **$(0, 6)$** and its symmetric counterpart at **$(2, 6)$**.\n4. Draw a smooth, upward-opening parabolic curve passing through these points.\n\nSince $t$ can take any value in $[0, 2\\pi]$ except $\\frac{\\pi}{2}$ and $\\frac{3\\pi}{2}$, $x = \\tan t + 1$ can take any real value ($-\\infty < x < \\infty$). Thus, the curve is a complete parabola opening upwards.\n\n**Markscheme criteria check:**\n- Correct shape of an upward-opening parabola B1\n- Vertex correctly shown at $(1, 5)$ B1\n- $y$-intercept correctly shown at $(0, 6)$ B1"}]}],"generatedAt":"2026-06-09T08:24:46.119Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"1f3b9e4e-4c6e-4a3b-a095-ff140a7d50d0","revisionId":"2541d8f4-cb7c-4d0a-a3e7-43bd30bb088c","questionNumber":null,"questionType":"LA","specification":"A curve has the parametric equations\n\n$$x = t + 2, \\quad y = \\frac{10}{t}$$","difficulty":1,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"654493d1-d934-4e9a-85cc-85ff0172a300","label":"a","content":"Find a cartesian equation for the curve.","markscheme":"- $t = x - 2$ M1\n- $y = \\frac{10}{x-2}$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Rearrange the equation for x to express t as x - 2.","code":"M"},{"id":"p2","marks":1,"description":"Substitute t into the y equation to find the correct cartesian form.","code":"A"}],"finalAnswers":["y = \\frac{10}{x-2}"],"generatedAt":"2026-05-06T14:05:14.975Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitute_t_from_x","label":"Substitution from x-equation","steps":[{"type":"text","order":0,"title":"Identify the method","content":"To find the Cartesian equation, we need to eliminate the parameter $t$. We can do this by rearranging the parametric equation for $x$ to express $t$ in terms of $x$, and then substituting this expression into the equation for $y$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=orig-x}{x} = \\htmlData{anchor=orig-t}{t} + 2$"},{"latex":"$$\\htmlData{anchor=new-t}{t} = \\htmlData{anchor=new-x}{x} - 2$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"new-t"},{"color":"sky","style":"text-color","anchor":"new-x"}],"connections":[{"to":"new-x","from":"orig-x","color":"sky","label":null,"style":"solid"},{"to":"new-t","from":"orig-t","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Rearrange the x-equation","description":"We start with the equation for $x$ and isolate the variable $t$ by subtracting $2$ from both sides."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{10}{\\htmlData{anchor=y-t}{\\color{@amber}{t}}}$"},{"latex":"$$\\text{Since } \\htmlData{anchor=sub-t}{\\color{@amber}{t}} = \\htmlData{anchor=sub-expr}{\\color{@sky}{x - 2}}$"},{"latex":"$$y = \\frac{10}{\\htmlData{anchor=final-expr}{\\color{@sky}{x - 2}}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"y-t"},{"color":"sky","style":"text-color","anchor":"sub-expr"},{"color":"sky","style":"text-color","anchor":"final-expr"}],"connections":[{"to":"final-expr","from":"sub-expr","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Substitute into y-equation","description":"Next, substitute this expression for $t$ into our parametric equation for $y$."},{"type":"text","order":3,"title":"Write the final equation","content":"The final Cartesian equation for the curve is $y = \\frac{10}{x - 2}$. In an exam, you would earn one mark for successfully rearranging to find $t = x - 2$, and a second mark for substituting this correctly to obtain the final equation."}]},{"id":"substitute_t_from_y","label":"Substitution from y-equation","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=y-eq}{\\color{@sky}{y}} = \\frac{10}{\\htmlData{anchor=t-var}{\\color{@purple}{t}}}$"},{"text":"Rearranging to make $t$ the subject:"},{"latex":"$$\\htmlData{anchor=t-eq}{\\color{@purple}{t}} = \\frac{10}{\\htmlData{anchor=y-var}{\\color{@sky}{y}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-eq"},{"color":"purple","style":"text-color","anchor":"t-var"}],"connections":[{"to":"t-eq","from":"t-var","color":"purple","label":"rearrange","style":"solid"}]},"order":0,"title":"Rearrange the y equation","description":"We start with the parametric equation for $y$ and rearrange it to make $t$ the subject. Multiplying both sides by $t$ and dividing by $y$ gives us $t$ in terms of $y$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=t-expr}{\\color{@purple}{t}} = \\frac{10}{\\htmlData{anchor=y-expr}{\\color{@sky}{y}}}$"},{"latex":"$$x = \\htmlData{anchor=t-sub}{\\color{@purple}{t}} + 2$"},{"latex":"$$x = \\frac{10}{\\htmlData{anchor=y-sub}{\\color{@sky}{y}}} + 2$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"t-expr"},{"color":"sky","style":"text-color","anchor":"y-expr"}],"connections":[{"to":"t-sub","from":"t-expr","color":"purple","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into x equation","description":"Next, we substitute our expression for $t$ into the parametric equation for $x$."},{"type":"text","order":2,"title":"Solve for y","content":"Now, we rearrange the equation to express $y$ in terms of $x$:\n\n$$x = \\frac{10}{y} + 2$$\n\n$$x - 2 = \\frac{10}{y}$$\n\n$$y(x - 2) = 10$$\n\n$$y = \\frac{10}{x - 2}$$\n\nThis gives us the Cartesian equation of the curve."},{"type":"text","order":3,"title":"Understand the markscheme","content":"The markscheme awards marks as follows:\n\n- **M1**: For a correct attempt to find $t$ in terms of $x$ or $y$ and substitute, which is represented by our step of obtaining $x - 2 = \\frac{10}{y}$ or equivalent.\n- **A1**: For finding the correct final Cartesian equation:\n\n$$y = \\frac{10}{x - 2}$$\n"}]}],"generatedAt":"2026-06-09T03:42:24.589Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"230e5fe0-76c8-4af3-9d11-0d6718e28739","label":"b","content":"Sketch the curve","markscheme":"- Correct hyperbola shape with asymptotes $x = 2$ and $y = 0$ B1\n- Correct branches in 1st and 3rd quadrants relative to asymptotes B1\n- $y$-intercept at $(0, -5)$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Shape of a hyperbola shown with vertical asymptote at $x = 2$ and horizontal asymptote $y = 0$.","code":"B"},{"id":"p2","marks":1,"description":"Correct placement of branches in the upper-right and lower-left regions relative to the intersection of the asymptotes.","code":"B"},{"id":"p3","marks":1,"description":"The y-intercept is correctly identified and labeled as $(0, -5)$.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:05:14.975Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"cartesian_transformation","label":"Conversion to Cartesian Form","steps":[{"type":"text","order":0,"title":"Convert to Cartesian form","content":"To sketch the curve, we first eliminate the parameter $t$ to find the Cartesian equation.\n\nFrom the parametric equations:\n$$x = t + 2 \\implies t = x - 2$$\n\nNow, substitute this expression for $t$ into the equation for $y$:\n$$y = \\frac{10}{t} \\implies y = \\frac{10}{x - 2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Standard hyperbola: } y = \\frac{10}{\\htmlData{anchor=orig-x}{x}}$"},{"latex":"$$\\text{Translated hyperbola: } y = \\frac{10}{\\htmlData{anchor=trans-x}{x - 2}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"trans-x"}],"connections":[{"to":"trans-x","from":"orig-x","color":"sky","label":"Shift right by 2","style":"solid"}]},"order":1,"title":"Identify graph transformations","description":"We recognize $y = \\frac{10}{x-2}$ as a horizontal translation of the standard hyperbola $y = \\frac{10}{x}$ by $+2$ units in the $x$-direction (to the right)."},{"type":"text","order":2,"title":"Find the asymptotes","content":"The standard hyperbola $y = \\frac{10}{x}$ has asymptotes at $x = 0$ (vertical) and $y = 0$ (horizontal).\n\nApplying the translation of $2$ units to the right:\n- **Vertical asymptote:** The denominator cannot be zero, so:\n $$x - 2 = 0 \\implies x = 2$$\n- **Horizontal asymptote:** As $x \\to \\pm\\infty$, $y \\to 0$, so:\n $$y = 0$$\n\nThese asymptotes will guide our sketch and earn us the first mark B1."},{"type":"text","order":3,"title":"Determine the intercepts","content":"Let's find where the curve crosses the axes:\n- **$x$-intercept:** Since $y = \\frac{10}{x-2}$ can never equal $0$, there are no $x$-intercepts (which makes sense as $y = 0$ is a horizontal asymptote).\n- **$y$-intercept:** Set $x = 0$ to find the $y$-intercept:\n $$y = \\frac{10}{0 - 2} = -5$$\n\nSo, the curve crosses the $y$-axis at $(0, -5)$. Correctly identifying this intercept earns the third mark B1."},{"type":"text","order":4,"title":"Sketch the curve","content":"Now draw the axes and add the features we found:\n1. Draw the vertical asymptote as a dashed line at $x = 2$ and label the horizontal asymptote along the $x$-axis ($y = 0$).\n2. Plot the $y$-intercept at $(0, -5)$.\n3. Draw the two branches of the hyperbola relative to the asymptotes in the first and third quadrants relative to the intersection point $(2, 0)$:\n - One branch is in the region $x > 2, y > 0$.\n - The other branch is in the region $x < 2, y < 0$, passing smoothly through $(0, -5)$.\n\nEnsuring the correct shape and position of the branches earns the final mark B1."}]},{"id":"parametric_plotting","label":"Direct Parametric Plotting","steps":[{"type":"text","order":0,"title":"Set up a table of values","content":"To sketch the curve using the **Direct Parametric Plotting** approach, we begin by calculating $(x, y)$ coordinate pairs for selected values of the parameter $t$.\n\nSince $y = \\frac{10}{t}$, the parameter $t$ cannot be equal to $0$. Let's choose some positive and negative values of $t$ to see how the coordinates behave:\n\n* **For negative values of $t$:**\n * When $t = -10$: $x = -10 + 2 = -8$, $y = \\frac{10}{-10} = -1 \\implies (-8, -1)$\n * When $t = -5$: $x = -5 + 2 = -3$, $y = \\frac{10}{-5} = -2 \\implies (-3, -2)$\n * When $t = -2$: $x = -2 + 2 = 0$, $y = \\frac{10}{-2} = -5 \\implies (0, -5)$\n * When $t = -1$: $x = -1 + 2 = 1$, $y = \\frac{10}{-1} = -10 \\implies (1, -10)$\n\n* **For positive values of $t$:**\n * When $t = 1$: $x = 1 + 2 = 3$, $y = \\frac{10}{1} = 10 \\implies (3, 10)$\n * When $t = 2$: $x = 2 + 2 = 4$, $y = \\frac{10}{2} = 5 \\implies (4, 5)$\n * When $t = 5$: $x = 5 + 2 = 7$, $y = \\frac{10}{5} = 2 \\implies (7, 2)$\n * When $t = 10$: $x = 10 + 2 = 12$, $y = \\frac{10}{10} = 1 \\implies (12, 1)$"},{"type":"text","order":1,"title":"Identify key intercepts","content":"From our table of values, we can immediately identify the $y$-intercept where the curve crosses the $y$-axis ($x = 0$).\n\nSetting $x = 0$ in the first equation:\n\n$$0 = t + 2 \\implies t = -2$$\n\nSubstituting $t = -2$ into the equation for $y$ gives:\n\n$$y = \\frac{10}{-2} = -5$$\n\nThus, the $y$-intercept is located at $(0, -5)$. \n\nThere is no $x$-intercept because setting $y = 0$ would require $\\frac{10}{t} = 0$, which has no real solution."},{"type":"text","order":2,"title":"Analyze behavior at boundaries","content":"Now we determine the behavior of the curve as $t$ approaches critical boundaries to locate any asymptotes:\n\n* **As $t \\to 0$:**\n * The $x$-values approach $0 + 2 = 2$.\n * If $t$ approaches $0$ from the positive side ($t \\to 0^+$), then $y \\to +\\infty$.\n * If $t$ approaches $0$ from the negative side ($t \\to 0^-$), then $y \\to -\\infty$.\n * This indicates a **vertical asymptote** at $x = 2$.\n\n* **As $t \\to \\pm\\infty$:**\n * The $x$-values approach $\\pm\\infty$.\n * The $y$-values approach $\\frac{10}{\\pm\\infty} = 0$.\n * This indicates a **horizontal asymptote** at $y = 0$ (the $x$-axis)."},{"type":"text","order":3,"title":"Sketch the final curve","content":"Using all the gathered information, we can now sketch the curve:\n\n1. **Draw the asymptotes:** Draw the vertical asymptote $x = 2$ as a dashed line and recognize the $x$-axis ($y = 0$) as the horizontal asymptote.\n2. **Plot key points:** Mark the $y$-intercept at $(0, -5)$ and plot a few representative points like $(4, 5)$ and $(-3, -2)$.\n3. **Draw the branches:** \n * For $t > 0$, the curve lies entirely in the region $x > 2, y > 0$ (the first quadrant relative to the asymptotes), approaching $y = 0$ as $x$ increases and $x = 2$ as $y$ increases.\n * For $t < 0$, the curve lies in the region $x < 2, y < 0$ (the third quadrant relative to the asymptotes), passing through the $y$-intercept at $(0, -5)$ and approaching the asymptotes $x = 2$ and $y = 0$.\n\nThis yields a rectangular hyperbola shifted $2$ units to the right.","calculator":[{"gdc":{"instructions":"1. Change your graphing type to **Parametric**.\n2. Enter \$x(t) = t + 2\$ and \$y(t) = 10/t\$.\n3. Set your active domain for \$t\$ to exclude 0 (e.g., plot from \$-10\$ to \$10\$).\n4. Graph to verify the shape and the asymptotes at \$x = 2\$ and \$y = 0\$."},"ti84":{"instructions":"1. Press **MODE** and change the graphing mode from **FUNC** to **PAR** (Parametric).\n2. Press **Y=** and enter the parametric equations:\n * \$X_{1T} = T + 2\$\n * \$Y_{1T} = 10/T\$\n3. Press **WINDOW** and set \$Tmin = -10\$, \$Tmax = 10\$, \$Tstep = 0.1\$.\n4. Adjust the window to \$Xmin = -10\$, \$Xmax = 10\$, \$Ymin = -10\$, \$Ymax = 10\$.\n5. Press **GRAPH** to view the shifted hyperbola."},"desmos":{"expression":"(t + 2, 10/t)","instructions":"1. In the expression bar, enter the parametric coordinates as an ordered pair: `(t + 2, 10/t)`.\n2. Desmos will prompt a range for \$t\$. Set it to `[-10, 10]` to view both branches of the hyperbola."},"description":"Visualize the parametric curve directly on your graphing utility."}]}]}],"generatedAt":"2026-06-09T03:43:32.229Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"654493d1-d934-4e9a-85cc-85ff0172a300:substitute_t_from_x","label":"(a) Substitution from x-equation","steps":[{"type":"text","order":0,"title":"Identify the method","content":"To find the Cartesian equation, we need to eliminate the parameter $t$. We can do this by rearranging the parametric equation for $x$ to express $t$ in terms of $x$, and then substituting this expression into the equation for $y$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=orig-x}{x} = \\htmlData{anchor=orig-t}{t} + 2$"},{"latex":"$$\\htmlData{anchor=new-t}{t} = \\htmlData{anchor=new-x}{x} - 2$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"new-t"},{"color":"sky","style":"text-color","anchor":"new-x"}],"connections":[{"to":"new-x","from":"orig-x","color":"sky","label":null,"style":"solid"},{"to":"new-t","from":"orig-t","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Rearrange the x-equation","description":"We start with the equation for $x$ and isolate the variable $t$ by subtracting $2$ from both sides."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{10}{\\htmlData{anchor=y-t}{\\color{@amber}{t}}}$"},{"latex":"$$\\text{Since } \\htmlData{anchor=sub-t}{\\color{@amber}{t}} = \\htmlData{anchor=sub-expr}{\\color{@sky}{x - 2}}$"},{"latex":"$$y = \\frac{10}{\\htmlData{anchor=final-expr}{\\color{@sky}{x - 2}}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"y-t"},{"color":"sky","style":"text-color","anchor":"sub-expr"},{"color":"sky","style":"text-color","anchor":"final-expr"}],"connections":[{"to":"final-expr","from":"sub-expr","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Substitute into y-equation","description":"Next, substitute this expression for $t$ into our parametric equation for $y$."},{"type":"text","order":3,"title":"Write the final equation","content":"The final Cartesian equation for the curve is $y = \\frac{10}{x - 2}$. In an exam, you would earn one mark for successfully rearranging to find $t = x - 2$, and a second mark for substituting this correctly to obtain the final equation."}]},{"id":"654493d1-d934-4e9a-85cc-85ff0172a300:substitute_t_from_y","label":"(a) Substitution from y-equation","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=y-eq}{\\color{@sky}{y}} = \\frac{10}{\\htmlData{anchor=t-var}{\\color{@purple}{t}}}$"},{"text":"Rearranging to make $t$ the subject:"},{"latex":"$$\\htmlData{anchor=t-eq}{\\color{@purple}{t}} = \\frac{10}{\\htmlData{anchor=y-var}{\\color{@sky}{y}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-eq"},{"color":"purple","style":"text-color","anchor":"t-var"}],"connections":[{"to":"t-eq","from":"t-var","color":"purple","label":"rearrange","style":"solid"}]},"order":0,"title":"Rearrange the y equation","description":"We start with the parametric equation for $y$ and rearrange it to make $t$ the subject. Multiplying both sides by $t$ and dividing by $y$ gives us $t$ in terms of $y$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=t-expr}{\\color{@purple}{t}} = \\frac{10}{\\htmlData{anchor=y-expr}{\\color{@sky}{y}}}$"},{"latex":"$$x = \\htmlData{anchor=t-sub}{\\color{@purple}{t}} + 2$"},{"latex":"$$x = \\frac{10}{\\htmlData{anchor=y-sub}{\\color{@sky}{y}}} + 2$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"t-expr"},{"color":"sky","style":"text-color","anchor":"y-expr"}],"connections":[{"to":"t-sub","from":"t-expr","color":"purple","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute into x equation","description":"Next, we substitute our expression for $t$ into the parametric equation for $x$."},{"type":"text","order":2,"title":"Solve for y","content":"Now, we rearrange the equation to express $y$ in terms of $x$:\n\n$$x = \\frac{10}{y} + 2$$\n\n$$x - 2 = \\frac{10}{y}$$\n\n$$y(x - 2) = 10$$\n\n$$y = \\frac{10}{x - 2}$$\n\nThis gives us the Cartesian equation of the curve."},{"type":"text","order":3,"title":"Understand the markscheme","content":"The markscheme awards marks as follows:\n\n- **M1**: For a correct attempt to find $t$ in terms of $x$ or $y$ and substitute, which is represented by our step of obtaining $x - 2 = \\frac{10}{y}$ or equivalent.\n- **A1**: For finding the correct final Cartesian equation:\n\n$$y = \\frac{10}{x - 2}$$\n"}]},{"id":"230e5fe0-76c8-4af3-9d11-0d6718e28739:cartesian_transformation","label":"(b) Conversion to Cartesian Form","steps":[{"type":"text","order":0,"title":"Convert to Cartesian form","content":"To sketch the curve, we first eliminate the parameter $t$ to find the Cartesian equation.\n\nFrom the parametric equations:\n$$x = t + 2 \\implies t = x - 2$$\n\nNow, substitute this expression for $t$ into the equation for $y$:\n$$y = \\frac{10}{t} \\implies y = \\frac{10}{x - 2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Standard hyperbola: } y = \\frac{10}{\\htmlData{anchor=orig-x}{x}}$"},{"latex":"$$\\text{Translated hyperbola: } y = \\frac{10}{\\htmlData{anchor=trans-x}{x - 2}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"trans-x"}],"connections":[{"to":"trans-x","from":"orig-x","color":"sky","label":"Shift right by 2","style":"solid"}]},"order":1,"title":"Identify graph transformations","description":"We recognize $y = \\frac{10}{x-2}$ as a horizontal translation of the standard hyperbola $y = \\frac{10}{x}$ by $+2$ units in the $x$-direction (to the right)."},{"type":"text","order":2,"title":"Find the asymptotes","content":"The standard hyperbola $y = \\frac{10}{x}$ has asymptotes at $x = 0$ (vertical) and $y = 0$ (horizontal).\n\nApplying the translation of $2$ units to the right:\n- **Vertical asymptote:** The denominator cannot be zero, so:\n $$x - 2 = 0 \\implies x = 2$$\n- **Horizontal asymptote:** As $x \\to \\pm\\infty$, $y \\to 0$, so:\n $$y = 0$$\n\nThese asymptotes will guide our sketch and earn us the first mark B1."},{"type":"text","order":3,"title":"Determine the intercepts","content":"Let's find where the curve crosses the axes:\n- **$x$-intercept:** Since $y = \\frac{10}{x-2}$ can never equal $0$, there are no $x$-intercepts (which makes sense as $y = 0$ is a horizontal asymptote).\n- **$y$-intercept:** Set $x = 0$ to find the $y$-intercept:\n $$y = \\frac{10}{0 - 2} = -5$$\n\nSo, the curve crosses the $y$-axis at $(0, -5)$. Correctly identifying this intercept earns the third mark B1."},{"type":"text","order":4,"title":"Sketch the curve","content":"Now draw the axes and add the features we found:\n1. Draw the vertical asymptote as a dashed line at $x = 2$ and label the horizontal asymptote along the $x$-axis ($y = 0$).\n2. Plot the $y$-intercept at $(0, -5)$.\n3. Draw the two branches of the hyperbola relative to the asymptotes in the first and third quadrants relative to the intersection point $(2, 0)$:\n - One branch is in the region $x > 2, y > 0$.\n - The other branch is in the region $x < 2, y < 0$, passing smoothly through $(0, -5)$.\n\nEnsuring the correct shape and position of the branches earns the final mark B1."}]},{"id":"230e5fe0-76c8-4af3-9d11-0d6718e28739:parametric_plotting","label":"(b) Direct Parametric Plotting","steps":[{"type":"text","order":0,"title":"Set up a table of values","content":"To sketch the curve using the **Direct Parametric Plotting** approach, we begin by calculating $(x, y)$ coordinate pairs for selected values of the parameter $t$.\n\nSince $y = \\frac{10}{t}$, the parameter $t$ cannot be equal to $0$. Let's choose some positive and negative values of $t$ to see how the coordinates behave:\n\n* **For negative values of $t$:**\n * When $t = -10$: $x = -10 + 2 = -8$, $y = \\frac{10}{-10} = -1 \\implies (-8, -1)$\n * When $t = -5$: $x = -5 + 2 = -3$, $y = \\frac{10}{-5} = -2 \\implies (-3, -2)$\n * When $t = -2$: $x = -2 + 2 = 0$, $y = \\frac{10}{-2} = -5 \\implies (0, -5)$\n * When $t = -1$: $x = -1 + 2 = 1$, $y = \\frac{10}{-1} = -10 \\implies (1, -10)$\n\n* **For positive values of $t$:**\n * When $t = 1$: $x = 1 + 2 = 3$, $y = \\frac{10}{1} = 10 \\implies (3, 10)$\n * When $t = 2$: $x = 2 + 2 = 4$, $y = \\frac{10}{2} = 5 \\implies (4, 5)$\n * When $t = 5$: $x = 5 + 2 = 7$, $y = \\frac{10}{5} = 2 \\implies (7, 2)$\n * When $t = 10$: $x = 10 + 2 = 12$, $y = \\frac{10}{10} = 1 \\implies (12, 1)$"},{"type":"text","order":1,"title":"Identify key intercepts","content":"From our table of values, we can immediately identify the $y$-intercept where the curve crosses the $y$-axis ($x = 0$).\n\nSetting $x = 0$ in the first equation:\n\n$$0 = t + 2 \\implies t = -2$$\n\nSubstituting $t = -2$ into the equation for $y$ gives:\n\n$$y = \\frac{10}{-2} = -5$$\n\nThus, the $y$-intercept is located at $(0, -5)$. \n\nThere is no $x$-intercept because setting $y = 0$ would require $\\frac{10}{t} = 0$, which has no real solution."},{"type":"text","order":2,"title":"Analyze behavior at boundaries","content":"Now we determine the behavior of the curve as $t$ approaches critical boundaries to locate any asymptotes:\n\n* **As $t \\to 0$:**\n * The $x$-values approach $0 + 2 = 2$.\n * If $t$ approaches $0$ from the positive side ($t \\to 0^+$), then $y \\to +\\infty$.\n * If $t$ approaches $0$ from the negative side ($t \\to 0^-$), then $y \\to -\\infty$.\n * This indicates a **vertical asymptote** at $x = 2$.\n\n* **As $t \\to \\pm\\infty$:**\n * The $x$-values approach $\\pm\\infty$.\n * The $y$-values approach $\\frac{10}{\\pm\\infty} = 0$.\n * This indicates a **horizontal asymptote** at $y = 0$ (the $x$-axis)."},{"type":"text","order":3,"title":"Sketch the final curve","content":"Using all the gathered information, we can now sketch the curve:\n\n1. **Draw the asymptotes:** Draw the vertical asymptote $x = 2$ as a dashed line and recognize the $x$-axis ($y = 0$) as the horizontal asymptote.\n2. **Plot key points:** Mark the $y$-intercept at $(0, -5)$ and plot a few representative points like $(4, 5)$ and $(-3, -2)$.\n3. **Draw the branches:** \n * For $t > 0$, the curve lies entirely in the region $x > 2, y > 0$ (the first quadrant relative to the asymptotes), approaching $y = 0$ as $x$ increases and $x = 2$ as $y$ increases.\n * For $t < 0$, the curve lies in the region $x < 2, y < 0$ (the third quadrant relative to the asymptotes), passing through the $y$-intercept at $(0, -5)$ and approaching the asymptotes $x = 2$ and $y = 0$.\n\nThis yields a rectangular hyperbola shifted $2$ units to the right.","calculator":[{"gdc":{"instructions":"1. Change your graphing type to **Parametric**.\n2. Enter \$x(t) = t + 2\$ and \$y(t) = 10/t\$.\n3. Set your active domain for \$t\$ to exclude 0 (e.g., plot from \$-10\$ to \$10\$).\n4. Graph to verify the shape and the asymptotes at \$x = 2\$ and \$y = 0\$."},"ti84":{"instructions":"1. Press **MODE** and change the graphing mode from **FUNC** to **PAR** (Parametric).\n2. Press **Y=** and enter the parametric equations:\n * \$X_{1T} = T + 2\$\n * \$Y_{1T} = 10/T\$\n3. Press **WINDOW** and set \$Tmin = -10\$, \$Tmax = 10\$, \$Tstep = 0.1\$.\n4. Adjust the window to \$Xmin = -10\$, \$Xmax = 10\$, \$Ymin = -10\$, \$Ymax = 10\$.\n5. Press **GRAPH** to view the shifted hyperbola."},"desmos":{"expression":"(t + 2, 10/t)","instructions":"1. In the expression bar, enter the parametric coordinates as an ordered pair: `(t + 2, 10/t)`.\n2. Desmos will prompt a range for \$t\$. Set it to `[-10, 10]` to view both branches of the hyperbola."},"description":"Visualize the parametric curve directly on your graphing utility."}]}]}],"generatedAt":"2026-06-09T03:43:52.526Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"7b3f3471-2160-4d33-8e91-64f2aeec3242","revisionId":"3a9a6ae1-ef8a-433c-a44d-26e9885d6689","questionNumber":null,"questionType":"LA","specification":"A curve has the parametric equations\n\n$$x = t + 3, \\quad y = t^2 - 5$$","difficulty":1,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"14cd238c-ad91-4be9-9822-776e884e74f8","label":"a","content":"Find a cartesian equation for the curve.","markscheme":"- Rearrange for $t$: $t = x - 3$ M1\n- Substitute into $y$: $y = (x - 3)^2 - 5$ M1\n- Expand and simplify: $y = x^2 - 6x + 4$ 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 - 6x + 4"],"generatedAt":"2026-05-06T14:04:59.249Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Rearrange for t","content":"To find a cartesian equation, we need to eliminate the parameter $t$. Let's start with the linear parametric equation for $x$:\n\n$$x = t + 3$$\n\nRearranging this to make $t$ the subject gives:\n\n$$t = x - 3$$\n\nThis step earns you the first mark (**M1**)."},{"type":"text","order":1,"title":"Substitute into y","content":"Now, we substitute this expression for $t$ into the parametric equation for $y$:\n\n$$y = t^2 - 5$$\n\nSubstituting $t = x - 3$ gives:\n\n$$y = (x - 3)^2 - 5$$\n\nWriting down this substitution earns you the second mark (**M1**)."},{"type":"text","order":2,"title":"Expand and simplify","content":"Finally, we expand the quadratic expression and simplify to get the cartesian equation in its standard form:\n\n$$y = (x^2 - 6x + 9) - 5$$\n\n$$y = x^2 - 6x + 4$$\n\nThis correct simplified equation earns you the final mark (**A1**)."}]}],"generatedAt":"2026-06-09T07:21:29.566Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"30a6f5b7-0c7e-4f1d-8fe2-3143930c1ea8","label":"b","content":"Sketch the curve","markscheme":"- Correct shape (parabola opening upwards) B1\n- Minimum point at $(3, -5)$ B1\n- $y$-intercept at 4 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 $(3, -5)$.","code":"B"},{"id":"p3","marks":1,"description":"Identifying and labeling the $y$-intercept at 4.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:04:59.249Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"cartesian_elimination","label":"Converting to Cartesian Form","steps":[{"type":"text","order":0,"title":"Rearrange for parameter t","content":"To eliminate the parameter $t$ from the parametric equations, we first rearrange the equation for $x$ to express $t$ in terms of $x$.\n\nGiven:\n$$x = t + 3$$\n\nSubtracting $3$ from both sides gives:\n$$t = x - 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=t-var}{\\color{@sky}{t}}^2 - 5$"},{"latex":"$$y = \\left(\\htmlData{anchor=t-sub}{\\color{@sky}{x - 3}}\\right)^2 - 5$"}],"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","style":"solid"}]},"order":1,"title":"Substitute into y","description":"Now, we substitute this expression for $t$ into the equation for $y$ to get the Cartesian equation of the curve."},{"type":"text","order":2,"title":"Analyze the Cartesian equation","content":"The Cartesian equation is in the standard vertex form of a quadratic:\n$$y = (x - 3)^2 - 5$$\n\nFrom this vertex form, we can directly identify two key features:\n1. **Shape**: Since the coefficient of the squared term is positive ($+1$), the curve is a **parabola opening upwards** (U-shape).\n2. **Minimum Point (Vertex)**: The minimum point of the curve occurs at $(h, k)$ from $y = a(x - h)^2 + k$, which gives us the minimum point at $(3, -5)$."},{"type":"text","order":3,"title":"Calculate the y-intercept","content":"To find where the curve crosses the $y$-axis, we set $x = 0$ in our Cartesian equation:\n\n$$y = (0 - 3)^2 - 5$$\n\n$$y = 9 - 5$$\n\n$$y = 4$$\n\nSo, the $y$-intercept is at $(0, 4)$."},{"type":"text","order":4,"title":"Sketch the curve","content":"Using the features we have derived, sketch the curve on your axes:\n- Draw a smooth, symmetrical **U-shaped parabola** opening upwards.\n- Clearly label the minimum point (vertex) at $(3, -5)$.\n- Clearly mark the $y$-intercept where the curve crosses the vertical axis at $y = 4$.\n\nAccording to the markscheme, the 3 marks are awarded as follows:\n- **B1**: Correct shape (parabola opening upwards).\n- **B1**: Minimum point labeled at $(3, -5)$.\n- **B1**: $y$-intercept labeled at $4$."}]},{"id":"table_of_values","label":"Plotting a Table of Values","steps":[{"type":"text","order":0,"title":"Create a table of values","content":"To sketch the curve using a table of values, we select several integer values for the parameter $t$ and calculate the corresponding values of $x = t + 3$ and $y = t^2 - 5$.\n\nLet's choose $t$ values from $-3$ to $3$:\n\n| $t$ | $x = t + 3$ | $y = t^2 - 5$ | $(x, y)$ |\n| :---: | :---: | :---: | :---: |\n| $-3$ | $0$ | $4$ | $(0, 4)$ |\n| $-2$ | $1$ | $-1$ | $(1, -1)$ |\n| $-1$ | $2$ | $-4$ | $(2, -4)$ |\n| $0$ | $3$ | $-5$ | $(3, -5)$ |\n| $1$ | $4$ | $-4$ | $(4, -4)$ |\n| $2$ | $5$ | $-1$ | $(5, -1)$ |\n| $3$ | $6$ | $4$ | $(6, 4)$ |","calculator":[{"gdc":{"expression":"x(t) = t + 3, y(t) = t^2 - 5","instructions":"1. Go to the Graph/Table menu.\n2. Set the graph type to Parametric.\n3. Enter $x(t) = t + 3$ and $y(t) = t^2 - 5$.\n4. Generate a table of values from $t = -3$ to $t = 3$ with a step size of 1."},"ti84":{"expression":"X1T = T + 3, Y1T = T^2 - 5","instructions":"1. Change the mode to parametric: Press [MODE], arrow down to the line with 'Func', select 'Parametric' (PAR), and press [ENTER].\n2. Enter the equations: Press [Y=] and input $X_{1T} = T + 3$ and $Y_{1T} = T^2 - 5$.\n3. Set up the table: Press [2nd] [WINDOW] (TblSet) and set TblStart = -3 and Delta Tbl = 1.\n4. View the table: Press [2nd] [GRAPH] (TABLE) to view the coordinate list."},"desmos":{"expression":"t_1 = [-3...3], x_1 = t_1 + 3, y_1 = t_1^2 - 5","instructions":"1. Click the '+' icon and select 'Table'.\n2. In the first column, type a list of values for $t$: `t = [-3, -2, -1, 0, 1, 2, 3]`.\n3. In the second column, define $x_1 = t_1 + 3$ and in the third column $y_1 = t_1^2 - 5$ to plot the points directly."},"description":"You can use the Table feature on your calculator to quickly generate these coordinates."}]},{"type":"text","order":1,"title":"Identify key features","content":"From our table, we can identify key structural points that will earn marks on the exam:\n\n- **Minimum point**: The lowest $y$-value occurs when $t = 0$, giving the vertex at $(3, -5)$.\n- **$y$-intercept**: This is where the curve crosses the $y$-axis ($x = 0$). This occurs at $t = -3$, yielding the point $(0, 4)$."},{"type":"text","order":2,"title":"Sketch the curve","content":"Now, plot the coordinate points $(0,4)$, $(1,-1)$, $(2,-4)$, $(3,-5)$, $(4,-4)$, $(5,-1)$, and $(6,4)$ on a Cartesian plane.\n\nConnect the points with a smooth curve:\n- Draw a smooth parabola opening upwards.\n- Clearly label the minimum point at $(3, -5)$ and the $y$-intercept at $(0, 4)$.","calculator":[{"gdc":{"instructions":"1. In Parametric graphing mode, set the window range for $t$ to $[-5, 5]$.\n2. Press Draw/Graph to display the curve."},"ti84":{"instructions":"1. Press [WINDOW] and set Tmin = -5, Tmax = 5, Xmin = -2, Xmax = 8, Ymin = -6, Ymax = 6.\n2. Press [GRAPH] to view the sketched parabola."},"desmos":{"expression":"(t + 3, t^2 - 5)","instructions":"1. In a blank expression line, type the parametric coordinates as an ordered pair: `(t + 3, t^2 - 5)`.\n2. Adjust the parameter bounds to `-5 <= t <= 5` to see the full shape."},"description":"Verify your sketch by graphing the parametric equations directly on your calculator."}]}]}],"generatedAt":"2026-06-09T07:24:25.733Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"14cd238c-ad91-4be9-9822-776e884e74f8:substitution","label":"(a) Substitution Method","steps":[{"type":"text","order":0,"title":"Rearrange for t","content":"To find a cartesian equation, we need to eliminate the parameter $t$. Let's start with the linear parametric equation for $x$:\n\n$$x = t + 3$$\n\nRearranging this to make $t$ the subject gives:\n\n$$t = x - 3$$\n\nThis step earns you the first mark (**M1**)."},{"type":"text","order":1,"title":"Substitute into y","content":"Now, we substitute this expression for $t$ into the parametric equation for $y$:\n\n$$y = t^2 - 5$$\n\nSubstituting $t = x - 3$ gives:\n\n$$y = (x - 3)^2 - 5$$\n\nWriting down this substitution earns you the second mark (**M1**)."},{"type":"text","order":2,"title":"Expand and simplify","content":"Finally, we expand the quadratic expression and simplify to get the cartesian equation in its standard form:\n\n$$y = (x^2 - 6x + 9) - 5$$\n\n$$y = x^2 - 6x + 4$$\n\nThis correct simplified equation earns you the final mark (**A1**)."}]},{"id":"30a6f5b7-0c7e-4f1d-8fe2-3143930c1ea8:cartesian_elimination","label":"(b) Converting to Cartesian Form","steps":[{"type":"text","order":0,"title":"Rearrange for parameter t","content":"To eliminate the parameter $t$ from the parametric equations, we first rearrange the equation for $x$ to express $t$ in terms of $x$.\n\nGiven:\n$$x = t + 3$$\n\nSubtracting $3$ from both sides gives:\n$$t = x - 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\htmlData{anchor=t-var}{\\color{@sky}{t}}^2 - 5$"},{"latex":"$$y = \\left(\\htmlData{anchor=t-sub}{\\color{@sky}{x - 3}}\\right)^2 - 5$"}],"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","style":"solid"}]},"order":1,"title":"Substitute into y","description":"Now, we substitute this expression for $t$ into the equation for $y$ to get the Cartesian equation of the curve."},{"type":"text","order":2,"title":"Analyze the Cartesian equation","content":"The Cartesian equation is in the standard vertex form of a quadratic:\n$$y = (x - 3)^2 - 5$$\n\nFrom this vertex form, we can directly identify two key features:\n1. **Shape**: Since the coefficient of the squared term is positive ($+1$), the curve is a **parabola opening upwards** (U-shape).\n2. **Minimum Point (Vertex)**: The minimum point of the curve occurs at $(h, k)$ from $y = a(x - h)^2 + k$, which gives us the minimum point at $(3, -5)$."},{"type":"text","order":3,"title":"Calculate the y-intercept","content":"To find where the curve crosses the $y$-axis, we set $x = 0$ in our Cartesian equation:\n\n$$y = (0 - 3)^2 - 5$$\n\n$$y = 9 - 5$$\n\n$$y = 4$$\n\nSo, the $y$-intercept is at $(0, 4)$."},{"type":"text","order":4,"title":"Sketch the curve","content":"Using the features we have derived, sketch the curve on your axes:\n- Draw a smooth, symmetrical **U-shaped parabola** opening upwards.\n- Clearly label the minimum point (vertex) at $(3, -5)$.\n- Clearly mark the $y$-intercept where the curve crosses the vertical axis at $y = 4$.\n\nAccording to the markscheme, the 3 marks are awarded as follows:\n- **B1**: Correct shape (parabola opening upwards).\n- **B1**: Minimum point labeled at $(3, -5)$.\n- **B1**: $y$-intercept labeled at $4$."}]},{"id":"30a6f5b7-0c7e-4f1d-8fe2-3143930c1ea8:table_of_values","label":"(b) Plotting a Table of Values","steps":[{"type":"text","order":0,"title":"Create a table of values","content":"To sketch the curve using a table of values, we select several integer values for the parameter $t$ and calculate the corresponding values of $x = t + 3$ and $y = t^2 - 5$.\n\nLet's choose $t$ values from $-3$ to $3$:\n\n| $t$ | $x = t + 3$ | $y = t^2 - 5$ | $(x, y)$ |\n| :---: | :---: | :---: | :---: |\n| $-3$ | $0$ | $4$ | $(0, 4)$ |\n| $-2$ | $1$ | $-1$ | $(1, -1)$ |\n| $-1$ | $2$ | $-4$ | $(2, -4)$ |\n| $0$ | $3$ | $-5$ | $(3, -5)$ |\n| $1$ | $4$ | $-4$ | $(4, -4)$ |\n| $2$ | $5$ | $-1$ | $(5, -1)$ |\n| $3$ | $6$ | $4$ | $(6, 4)$ |","calculator":[{"gdc":{"expression":"x(t) = t + 3, y(t) = t^2 - 5","instructions":"1. Go to the Graph/Table menu.\n2. Set the graph type to Parametric.\n3. Enter $x(t) = t + 3$ and $y(t) = t^2 - 5$.\n4. Generate a table of values from $t = -3$ to $t = 3$ with a step size of 1."},"ti84":{"expression":"X1T = T + 3, Y1T = T^2 - 5","instructions":"1. Change the mode to parametric: Press [MODE], arrow down to the line with 'Func', select 'Parametric' (PAR), and press [ENTER].\n2. Enter the equations: Press [Y=] and input $X_{1T} = T + 3$ and $Y_{1T} = T^2 - 5$.\n3. Set up the table: Press [2nd] [WINDOW] (TblSet) and set TblStart = -3 and Delta Tbl = 1.\n4. View the table: Press [2nd] [GRAPH] (TABLE) to view the coordinate list."},"desmos":{"expression":"t_1 = [-3...3], x_1 = t_1 + 3, y_1 = t_1^2 - 5","instructions":"1. Click the '+' icon and select 'Table'.\n2. In the first column, type a list of values for $t$: `t = [-3, -2, -1, 0, 1, 2, 3]`.\n3. In the second column, define $x_1 = t_1 + 3$ and in the third column $y_1 = t_1^2 - 5$ to plot the points directly."},"description":"You can use the Table feature on your calculator to quickly generate these coordinates."}]},{"type":"text","order":1,"title":"Identify key features","content":"From our table, we can identify key structural points that will earn marks on the exam:\n\n- **Minimum point**: The lowest $y$-value occurs when $t = 0$, giving the vertex at $(3, -5)$.\n- **$y$-intercept**: This is where the curve crosses the $y$-axis ($x = 0$). This occurs at $t = -3$, yielding the point $(0, 4)$."},{"type":"text","order":2,"title":"Sketch the curve","content":"Now, plot the coordinate points $(0,4)$, $(1,-1)$, $(2,-4)$, $(3,-5)$, $(4,-4)$, $(5,-1)$, and $(6,4)$ on a Cartesian plane.\n\nConnect the points with a smooth curve:\n- Draw a smooth parabola opening upwards.\n- Clearly label the minimum point at $(3, -5)$ and the $y$-intercept at $(0, 4)$.","calculator":[{"gdc":{"instructions":"1. In Parametric graphing mode, set the window range for $t$ to $[-5, 5]$.\n2. Press Draw/Graph to display the curve."},"ti84":{"instructions":"1. Press [WINDOW] and set Tmin = -5, Tmax = 5, Xmin = -2, Xmax = 8, Ymin = -6, Ymax = 6.\n2. Press [GRAPH] to view the sketched parabola."},"desmos":{"expression":"(t + 3, t^2 - 5)","instructions":"1. In a blank expression line, type the parametric coordinates as an ordered pair: `(t + 3, t^2 - 5)`.\n2. Adjust the parameter bounds to `-5 <= t <= 5` to see the full shape."},"description":"Verify your sketch by graphing the parametric equations directly on your calculator."}]}]}],"generatedAt":"2026-06-09T07:24:42.101Z","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"]}]