If $x=3 \cos t$ and $y=4 \sin t$, then $\frac{d^2 y}{d x^2}$ at the point $\left(x_0, y_0\right)=\left(\frac{3}{2} \sqrt{2}, 2 \sqrt{2}\right)$, is

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Aicharya · Accepted Answer

We have, $ \begin{aligned} & x=3 \cos t 	ext { and } y=4 \sin t \ & \Rightarrow \quad \frac{x^2}{9}+\frac{y^2}{16}=\cos ^2 t+\sin ^2 t \ & \Rightarrow \quad \frac{x^2}{9}+\frac{y^2}{16}=1 \ & \Rightarrow \frac{2 x}{9}+\frac{2 y}{16} \frac{d y}{d x}=0 \ & \Rightarrow \quad \frac{d y}{d x}=\frac{-16}{9} \frac{x}{y} \ & \Rightarrow \quad \frac{d^2 y}{d x^2}=\frac{-16}{9}\left[\frac{1 \cdot y-x \frac{d y}{d x}}{y^2}ight] \ & \Rightarrow \quad \frac{d^2 y}{d x^2}=\frac{-16}{9}\left[\frac{y-x-\left(\frac{-16}{9} \frac{x}{y}ight)}{y^2}ight] \ & =\frac{-16}{9}\left[\frac{9 y^2+16 x^2}{9 y^3}ight]=\frac{-16}{9} 	imes \frac{144}{9 y^3}\left[\because \frac{x^2}{9}+\frac{y^2}{16}=1ight] \ & =\frac{-16 	imes 144}{81 	imes(2 \sqrt{2})^3} \quad\left[\because\left(x_0, y_0ight)=\left(\frac{3 \sqrt{2}}{2}, 2 \sqrt{2}ight)ight] \ & =\frac{-16 	imes 144}{81 	imes 16 \sqrt{2}}=\frac{-144}{81 	imes \sqrt{2}}=\frac{-16}{9 \sqrt{2}}=\frac{-8 \sqrt{2}}{9} \ & \end{aligned} $

If $x=3 \cos t$ and $y=4 \sin t$, then $\frac{d^2 y}{d x^2}$ at the point $\left(x_0,…

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