Question

If $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 t}{1+t^{2}}$, 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 $y=\frac{2 t}{1+t^{2}}$
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, $\frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y}$