Q.If $x=a\cos\theta$, $y=b\sin\theta$, show that $a^2y\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^2+b^2=0$

$\frac{dx}{d\theta}=-a\sin\theta,\qquad \frac{dy}{d\theta}=b\cos\theta,\qquad \frac{dy}{dx}=\frac{b\cos\theta}{-a\sin\theta}=-\frac{b}{a}\cot\theta.$

$\frac{d}{d\theta}\left(\frac{dy}{dx}\right)=\frac{b}{a}\csc^2\theta,\qquad \frac{d^2y}{dx^2}=\frac{\dfrac{b}{a}\csc^2\theta}{-a\sin\theta}=-\frac{b}{a^2}\csc^3\theta=-\frac{b}{a^2\sin^3\theta}.$

$y\,y''=b\sin\theta\cdot\left(-\frac{b}{a^2\sin^3\theta}\right)=-\frac{b^2}{a^2\sin^2\theta}=-\frac{b^2}{a^2}\csc^2\theta.$

$y\,y''+(y')^2=\frac{b^2}{a^2}\left(\cot^2\theta-\csc^2\theta\right)=\frac{b^2}{a^2}(-1)=-\frac{b^2}{a^2}.$

$y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2+\frac{b^2}{a^2}=0.$