If \(x=\frac{1-t^2}{1+t^2}\) and \(y=\frac{2 a t}{1+t^2}\), then \(\frac{d y}{d x}=\)

We have \(\mathrm{x}=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\) and \(\mathrm{y}=\frac{2 \mathrm{at}}{1+\mathrm{t}^2}\)
Put \(\mathrm{t}=\tan \theta \Rightarrow \mathrm{x}=\cos 2 \theta\) and \(\mathrm{y}=\mathrm{a} \sin 2 \theta\)
\(
\frac{d x}{d \theta}=-2 \sin 2 \theta \text { and } \frac{d y}{d \theta}=2 a \cos 2 \theta
\)
\(
\therefore \frac{d y}{d x}=\frac{2 a \cos 2 \theta}{-2 \sin 2 \theta}=\frac{-a}{\tan 2 \theta}=\frac{-a}{\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)}
\)
\(
=\frac{-\mathrm{a}}{\left(\frac{2 \mathrm{t}}{1-\mathrm{t}^2}\right)}=\frac{-\mathrm{a}\left(1-\mathrm{t}^2\right)}{2 \mathrm{t}}=\frac{\mathrm{a}\left(\mathrm{t}^2-1\right)}{2 \mathrm{t}}
\)