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, $\frac{dy}{dx}=\frac{dy}{d\theta}\times\frac{d\theta}{dx} =- \frac{\cos 2\theta}{\sin 2\theta} = -\frac{x}{y}$