Question

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

• A. $\frac{-b^4}{a^2 y^3}$
• B. $\frac{b^2}{ay^2}$
• 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{d x}{d \theta}=-a \sin \theta, \frac{d y}{d \theta}=b \cos \theta $
$\frac{d y}{d x}=-\frac{b}{a} \cot \theta$
On differentiating w.r.t. $\theta$, we get
$\frac{d^{2} y}{d x^{2}}=-\frac{b}{a}\left(-\operatorname{cosec}^{2} \theta\right) \frac{d \theta}{d x}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{b \operatorname{cosec}^{2} \theta}{-a^{2} \sin \theta}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{b}{a^{2} \sin ^{3} \theta}$
$\Rightarrow \frac{d^{2} y}{d x}=\frac{-b^{4}}{a^{2} y^{3}} [\because \sin \theta=\frac{y}{b}]$