A curve $C$ has parametric equations $x=2{{e}^{t}}$, $y={{t}^{3}}-t$, where $-1\le t\le \frac{3}{2}$

Question

A curve $C$ has parametric equations $x=2{{e}^{t}}$, $y={{t}^{3}}-t$, where $-1\le t\le \frac{3}{2}$. Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $t$. Hence find the exact equations of the normals to the curve which are parallel to the $y$-axis. [3] Sketch the curve $C$. [1] Find the equation of the tangent to $C$ at the point$\left( 2{{e}^{a}},{{a}^{3}}-a \right)$. Find the possible exact values of $a$ such that the tangent cuts the $y$-axis at $y=1$. [4]

Accepted Answer

(i) $x = 2e^{\frac{1}{\sqrt{3}}}$ \text{ and } $x = 2e^{-\frac{1}{\sqrt{3}}}$ (iii) $y = \left(\frac{3a^2 - 1}{2e^a}\right)x + a^3 - 3a^2 - a + 1$

2018 TJC P2 Q1

2018 TJC P2 Q1

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