Question

Find \[\frac{dy}{dx}\] , when \[x = b   \sin^2   \theta  \text{ and }  y = a   \cos^2   \theta\] ?

Answer 1

\[\text{ We have, x } = b \sin^2 \theta \text{ and } y = a \cos^2 \theta\]

\[ \therefore \frac{dx}{d\theta} = \frac{d}{d\theta}\left( b \sin^2 \theta \right) = 2b \sin\theta\cos\theta\]

\[\text{ and }, \]

\[ \frac{dy}{d\theta} = \frac{d}{d\theta}\left( a \cos^2 \theta \right) = - 2a \cos\theta\sin\theta \]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{- 2a \cos\theta\sin\theta}{2b \sin\theta\cos\theta} = - \frac{a}{b}\]