SOLUTION —
Put $t=\tan \theta$ in both the equations, we get
$x=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\cos 2 \theta \text { and } y=\frac{2 \tan \theta}{1+\tan ^2 \theta}=\sin 2 \theta$
$\Rightarrow \quad \frac{d x}{d \theta}=-2 \sin 2 \theta$
and, $\frac{d y}{d \theta}=2 \cos 2 \theta$
$\therefore \quad \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y}$
So, The correct option will be (C).