A curve $C$ has parametric equations $x=\frac{1}{\sqrt{1-4{{t}^{2}}}}$ and $y={{\sin }^{-1}}2t$, for

Question

A curve $C$ has parametric equations $x=\frac{1}{\sqrt{1-4{{t}^{2}}}}$ and $y={{\sin }^{-1}}2t$, for $-\frac{1}{2} Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=2\left( 1-4{{t}^{2}} ight)$ and explain why the gradient is always positive for all points on the curve. [4] Describe the behaviour of $C$ as $x	o \pm \infty $. Sketch $C$, stating clearly the equations of any asymptotes and the coordinates of any points of intersection with the axes. [3] Given that $A$ is the point on $C$ with parameter $\frac{1}{2\sqrt{2}}$, find the equation of $l$, the tangent to $C$ at the point $A$, leaving your answer in exact form. [3] Show that $l$ meets the curve $C$ again at another point $B$ and find the coordinates of $B$. [2]

Accepted Answer

$$\frac{\mathrm{d}y}{\mathrm{d}x} = 2(1-4t^2)$$; $$y = x - \frac{1}{2} + \frac{\pi}{4}$$; $$B(-1.54, -1.26)$$

2020 ACJC PROMO Q11

2020 ACJC PROMO Q11

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