Question

If $x=\frac{2at}{1+t^2}$ and $y=\frac{b(1-t^2)}{1+t^2}$ where $a,b$ are non-zero constants then $\frac{dy}{dx}$ is equal to

• A. $-\frac{x}{y}$
• B. $\frac{y}{x}$
• C. $\frac{b^{2}x}{a^{2}y}$
• D. $-\frac{b^{2}x}{a^{2}y}$

Answer 1

The correct option is D $-\frac{b^{2}x}{a^{2}y}$

Put $t=\tan \theta$
$x=\frac{2a\tan \theta }{1+{\tan }^2\theta }=a\sin 2\theta \Rightarrow \frac{dx}{d\theta }=2a\cos 2\theta y=\frac{b(1-{\tan }^2\theta )}{1+{\tan }^2\theta }=b\cos 2\theta \Rightarrow \frac{dy}{d\theta }=-2b\sin 2\theta \frac{dy}{dx}=\frac{\left(\frac{dy}{d\theta }\right)}{\left(\frac{dx}{d\theta }\right)}=-\frac{2b\sin 2\theta }{2a\cos 2\theta }=-\frac{b\left(\frac{x}{a}\right)}{a\left(\frac{y}{b}\right)}\frac{dy}{dx}=-\frac{b^{2}x}{a^{2}y}$