Question

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ then $\frac{d^{2}y}{d{x}^2}=$

• A. $\frac{-b^4}{a^2{y}^3}$
• B. $\frac{b^2}{a{y}^2}$
• C. $\frac{-b^3}{a^2{y}^3}$
• D. $\frac{b^3}{a^2{y}^2}$

Answer 1

Answer: (A)

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
$\frac{d^{2}y}{d{x}^2}=-\frac{b}{a}\left(-{\mathrm{cosec}}^2\theta \right)\frac{d\theta }{dx}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{b{\mathrm{cosec}}^2\theta }{-a^2\sin \theta }$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-\frac{b}{a^2{\sin }^3\theta }$
$\Rightarrow \frac{d^{2}y}{dx}=\frac{-b^4}{a^2{y}^3}[\because \sin \theta =\frac{y}{b}]$