Answer:
Option A
Explanation:
We have,
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Let $x=a\cos \theta, y=b \sin \theta$
$\therefore$ $\frac{dx}{d\theta}=-a \sin \theta, \frac{dy}{d \theta}=b \cos \theta$
$\frac{dy}{dx}=-\frac{b}{a} \cot\theta$
On differentiating w.r.t $\theta$ , we get,
$\frac{d^{2}y}{dx^{2}}=-\frac{b}{a}(- cosec^{2} \theta) \frac{d \theta}{dx}$
$\Rightarrow$ $\frac{d^{2}y}{dx^{2}}=\frac{b cosec^{2} \theta}{-a^{2} \sin \theta}$
$\Rightarrow$ $\frac{d^{2}y}{dx^{2}}=-\frac{b}{a^{2}\sin^{3} \theta}$
$\Rightarrow$ $\frac{d^{2}y}{dx}=\frac{-b^{4}}{a^{2}y^{3}}$
$\left[\because \sin \theta=\frac{y}{b}\right]$
