Question

If $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$, then $\frac{dy}{dx}$ is equal to

• A. $-\frac{y}{x}$
• B. $\frac{y}{x}$
• C. $-\frac{x}{y}$
• D. $\frac{x}{y}$

Answer 1

Answer: (C)

Let $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2t}{1+t^2}$
Put $t=\tan \theta$, we get
$x=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$ and $y=\frac{2\tan \theta }{1+{\tan }^2\theta }$
$\Rightarrow x=\cos 2\theta$ and $y=\sin 2\theta$
$\therefore \frac{dx}{d\theta }=-2\sin 2\theta$ and $\frac{dy}{d\theta }=2\cos 2\theta$
Now, $\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=-\frac{\cos 2\theta }{\sin 2\theta }=-\frac{x}{y}$