\\begin{aligned} 2t &= e^x - 1 \\\\ t &= \\frac{e^x - 1}{2} \\end{aligned}$$"},{"type":"text","order":1,"title":"Substitute $t$ into $y$","content":"Now we substitute this expression for $t$ into our equation for $y$ to get the Cartesian equation of the curve:\n\n$$y = t^2 - 16$$\n\nSubstituting $t = \\frac{e^x - 1}{2}$ gives:\n$$y = \\left(\\frac{e^x - 1}{2}\\right)^2 - 16$$\n\nSimplifying this expression:\n$$y = \\frac{(e^x - 1)^2}{4} - 16$$"},{"type":"text","order":2,"title":"Find the $y$-intercept","content":"To find where the curve crosses the $y$-axis, we set $x = 0$ in our Cartesian equation:\n\n$$y = \\frac{(e^0 - 1)^2}{4} - 16$$\n\nSince $e^0 = 1$:\n$$\\begin{aligned} y &= \\frac{(1 - 1)^2}{4} - 16 \\\\ y &= 0 - 16 \\\\ y &= -16 \\end{aligned}$$\n\nSo, the $y$-intercept is at the point **$(0, -16)$**. This earns A1."},{"type":"text","order":3,"title":"Find the $x$-intercept","content":"To find where the curve crosses the $x$-axis, we set $y = 0$ in our Cartesian equation:\n\n$$0 = \\frac{(e^x - 1)^2}{4} - 16$$\n\nNow, solve for $x$:\n$$\\begin{aligned} \\frac{(e^x - 1)^2}{4} &= 16 \\\\ (e^x - 1)^2 &= 64 \\\\ e^x - 1 &= \\pm 8 \\end{aligned}$$\n\nThis gives us two possibilities:\n1. $e^x - 1 = 8 \\implies e^x = 9 \\implies x = \\ln 9$\n2. $e^x - 1 = -8 \\implies e^x = -7$ (which has no real solutions since $e^x$ must be positive)\n\nTherefore, we have $x = \\ln 9$, giving the $x$-intercept **$(\\ln 9, 0)$**. This earns A1."}]}],"generatedAt":"2026-06-11T03:31:24.307Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"def190c5-d557-4cd0-abdb-af12b45047d6","label":"b","content":"Find an equation for the tangent to the curve when $t = 1$.","markscheme":"- At $t = 1$, $x = \\ln 3$ and $y = 1^2 - 16 = -15$ B1\n- $\\frac{dx}{dt} = \\frac{2}{2t+1}$, $\\frac{dy}{dt} = 2t$ M1\n- $\\frac{dy}{dx} = \\frac{2t}{(2/(2t+1))} = t(2t + 1)$. At $t=1$, $m = 1(3) = 3$ M1\n- $y - (-15) = 3(x - \\ln 3)$ M1\n- $y = 3x - 3\\ln 3 - 15$ A1","marks":5,"order":1,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Finding coordinates $(\\ln 3, -15)$.","code":"B"},{"id":"p2","marks":1,"description":"Correct differentiation of $x$ and $y$ (requires chain rule for $x$).","code":"M"},{"id":"p3","marks":1,"description":"Combining derivatives to find $dy/dx$ and substituting $t=1$.","code":"M"},{"id":"p4","marks":1,"description":"Method to find tangent line equation.","code":"M"},{"id":"p5","marks":1,"description":"Correct final equation.","code":"A"}],"finalAnswers":["y = 3x - 3\\ln 3 - 15"],"generatedAt":"2026-05-06T14:04:48.974Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"parametric","label":"Parametric Differentiation","steps":[{"type":"text","order":0,"title":"Find the point of contact","content":"To find the equation of the tangent line, we first need the coordinates $(x, y)$ of the point on the curve where $t = 1$.\n\nSubstitute $t = 1$ into both parametric equations:\n\n$$\\begin{aligned} x &= \\ln(2(1) + 1) = \\ln 3 \\\\ y &= (1)^2 - 16 = -15 \\end{aligned}$$\n\nSo, the point of contact is $(\\ln 3, -15)$. This earns you the first mark **(B1)**."},{"type":"text","order":1,"title":"Find the individual derivatives","content":"Next, we differentiate both $x$ and $y$ with respect to $t$.\n\nFor $x = \\ln(2t + 1)$, we use the chain rule:\n$$\\frac{dx}{dt} = \\frac{1}{2t+1} \\cdot 2 = \\frac{2}{2t+1}$$\n\nFor $y = t^2 - 16$, we use the power rule:\n$$\\frac{dy}{dt} = 2t$$\n\nFinding these two derivatives correctly earns you the second mark **(M1)**."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$${\\frac{dy}{dx}} = \\frac{\\htmlData{anchor=dydt-val}{\\color{@sky}{\\frac{dy}{dt}}}}{\\htmlData{anchor=dxdt-val}{\\color{@amber}{\\frac{dx}{dt}}}}$"},{"latex":"$${\\frac{dy}{dx}} = \\frac{\\htmlData{anchor=dydt-sub}{\\color{@sky}{2t}}}{\\htmlData{anchor=dxdt-sub}{\\color{@amber}{\\frac{2}{2t+1}}}}$"},{"latex":"$${\\frac{dy}{dx}} = \\frac{2t(2t+1)}{2} = t(

1.5.9 Intersection of a line and a curve

Question 2

1.5.9 Intersection of a line and a curve Questions

1.5.9 Intersection of a line and a curve

Question 2

1.5.9 Intersection of a line and a curve Questions