If $x=a \cos \theta, y=b \sin \theta$, then $\left[\frac{d^2 y}{d x^2}\right]_{\theta=\frac{\pi}{4}}=$

If $x=a \cos \theta, y=b \sin \theta$, then $\left[\frac{d^2 y}{d x^2}\right]_{\theta=\frac{\pi}{4}}=$
  1. $2\left(\frac{\mathrm{a}^2}{\mathrm{~b}}\right)$
  2. $\sqrt{2}\left(\frac{\mathrm{a}^2}{\mathrm{~b}}\right)$
  3. $-2 \sqrt{2}\left(\frac{b}{a^2}\right)$
  4. $2 \sqrt{2}\left(\frac{\mathrm{b}}{\mathrm{a}^2}\right)$

Solution

$\begin{aligned} & \mathrm{x}=\mathrm{a} \cos \theta, \mathrm{y}=\mathrm{b} \sin \theta \\ & \therefore \frac{\mathrm{dx}}{\mathrm{d} \theta}=-\mathrm{a} \sin \theta, \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{b} \cos \theta \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{b} \cos \theta}{-\mathrm{a} \sin \theta}=\left(\frac{-\mathrm{b}}{\mathrm{a}}\right) \cot \theta \\ & \therefore \frac{\mathrm{d}^2}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{d} \theta}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \cdot \frac{\mathrm{d} \theta}{\mathrm{dx}} \\ & =\frac{\mathrm{d}}{\mathrm{d} \theta}\left[\left(\frac{-\mathrm{b}}{2}\right) \cot \theta\right] \times \frac{1}{\left(\frac{\mathrm{dx}}{\mathrm{d} \theta}\right)} \\ & =\frac{\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)\left(-\operatorname{cosec}{ }^2 \theta\right)}{-\mathrm{a} \sin \theta}=\left(\frac{-\mathrm{b}}{\mathrm{a}}\right) \times \frac{1}{\mathrm{a}} \times \frac{1}{\sin ^3 \theta} \\ & \therefore\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{4}}=\left(\frac{-\mathrm{b}}{\mathrm{a}^2}\right)\left[\frac{1}{\left(\sin \frac{\pi}{4}\right)^3}\right]=\left(\frac{-\mathrm{b}}{\mathrm{a}^2}\right)(\sqrt{2})^3=-2 \sqrt{2}\left(\frac{\mathrm{b}}{\mathrm{a}^2}\right)\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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