Question:
Find $\frac{d y}{d x}$, when
$x=b \sin ^{2} \theta$ and $y=a \cos ^{2} \theta$
Solution:
We have, $x=b \sin ^{2} \theta$ and $y=a \cos ^{2} \theta \therefore \frac{d x}{d \theta}=\frac{d}{d \theta}\left(b \sin ^{2} \theta\right)$
$=2 b \sin \theta \cos \theta$ and,$\frac{d y}{d \theta}=\frac{d}{d \theta}\left(a \cos ^{2} \theta\right)$
$=-2 a \cos \theta \sin \theta \quad \therefore \quad \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-2 a \cos \theta \sin \theta}{2 b \sin \theta \cos \theta}$
$=-\frac{a}{b}$