The parametric equations of a curve are $x=\frac{1}{2}{{\mathrm{e}}^{t}}-2{{\mathrm{e}}^{-t}}$ and

Question

The parametric equations of a curve are $x=\frac{1}{2}{{\mathrm{e}}^{t}}-2{{\mathrm{e}}^{-t}}$ and $y={{\mathrm{e}}^{t}}+2{{\mathrm{e}}^{-t}}$. Using calculus, find the exact gradient of the normal to the curve at the point where $t=\ln 4$.[3] Show that the equation of the normal is $5x+7y=39$.[2] Find the exact value of $t$ where the normal meets the curve again.[4]

Accepted Answer

(a) Gradient of normal $= -\dfrac{5}{7}$; (b) $7y + 5x = 39$ shown; (c) $t = \ln\left(\dfrac{2}{19}\right)$

2023 NYJC Promo Q7

2023 NYJC Promo Q7

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