The parametric equations of a curve are $x=\frac12(\mathrm{e}^{3t}+2\mathrm{e}^{-3t})$ and \(y=\fr

Question

The parametric equations of a curve are $x=\frac12(\mathrm{e}^{3t}+2\mathrm{e}^{-3t})$ and $y=\frac12(\mathrm{e}^{3t}-2\mathrm{e}^{-3t})$. Using calculus, find the gradient of the normal to the curve at the point where $t=\frac13\ln 2$.[3] By considering $x^2$ and $y^2$ or otherwise, find the cartesian equation of the curve, stating any restriction on the values of $x$.[3]

Accepted Answer

(a) $-\frac13$ (b) $x^2-y^2=2,\ x\geq\sqrt{2}$

N2022 P1 Q3

N2022 P1 Q3

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