A curve $C$ is defined parametrically by \[x = a \tan t, \quad y = a \sec^2 t \sin t, \quad -\frac

Question

A curve $C$ is defined parametrically by $$x = a \tan t, \quad y = a \sec^2 t \sin t, \quad -\frac{\pi}{4} < t < \frac{\pi}{4},$$ where $a$ is a positive constant. Show that $\frac{\tan t}{\sqrt{1 + \tan^2 t}} = \sin t$.[1] By using part (a) or otherwise, find the cartesian equation of $C$ in the form $y = \mathrm{f}(x)$, simplifying your answer.[2] Show that $\mathrm{f}(-x) = -\mathrm{f}(x)$. Hence sketch $C$.[2] Find the exact area of the region bounded by $C$, the $x$-axis and the lines $x = -A$ and $x = A$, where $0 < A < a$, leaving your answer in terms of $A$ and $a$.[3]

Accepted Answer

(b) $y=\dfrac{x\sqrt{x^2+a^2}}a$, $-a<x<a$ (c) $f$ is odd; sketch shown (d) $\dfrac{2}{3a}\left[(A^2+a^2)^{3/2}-a^3\right]$

2025 DHS P1 Q6

2025 DHS P1 Q6

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