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

The correct option is C $-\frac{x}{y}$

Given, $x=\frac{1-t^2}{1+t^2}$
and $y=\frac{2t}{1+t^2}$
Put $t=\tan \theta$ in both the equations, we get
$x=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\cos 2\theta$ ....$\left(i\right)$
and $y=\frac{2\tan \theta }{1+{\tan }^2\theta }=\sin 2\theta$ ......$\left(ii\right)$
On differentiating both the Eqs. $\left(i\right)$ and $\left(ii\right)$ w.r.t. $theta$, we get
$\frac{dx}{d\theta }=2\sin 2\theta$
and $\frac{dy}{d\theta }=2\cos 2\theta$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=-\frac{2\cos 2\theta }{2\sin 2\theta }$
$=-\frac{x}{y}$ [From Eqs. $\left(i\right)$ and $\left(ii\right)$]