Answer:
Option C
Explanation:
Given , $x=a\left(t-\frac{1}{t}\right),y=a\left(t+\frac{1}{t}\right)$
Now, $y^{2}-x^{2}=a^{2}\left[\left(t+\frac{1}{t}\right)^{2}-a^{2}\left(t-\frac{1}{t}\right)^{2}\right]$
= $a^{2}\left[t^{2}+\frac{1}{t^{2}}+2-t^{2}-\frac{1}{t^{2}}+2\right]$
$\Rightarrow $ $y^{2}-x^{2}=4a^{2}$
On differentiating both sides w.r.t 'x' , we get
2y $\frac{dy}{dx}$-2x=0
$\Rightarrow$ $ 2\left( y\frac{dy}{dx}-x\right)=0$
$\therefore$ $\frac{dy}{dx}=\frac{x}{y}$
