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Question
Find \[\frac{dy}{dx}\] , when \[x = b \sin^2 \theta \text{ and } y = a \cos^2 \theta\] ?
Solution
\[\text{ We have, x } = b \sin^2 \theta \text{ and } y = a \cos^2 \theta\]
\[ \therefore \frac{dx}{d\theta} = \frac{d}{d\theta}\left( b \sin^2 \theta \right) = 2b \sin\theta\cos\theta\]
\[\text{ and }, \]
\[ \frac{dy}{d\theta} = \frac{d}{d\theta}\left( a \cos^2 \theta \right) = - 2a \cos\theta\sin\theta \]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{- 2a \cos\theta\sin\theta}{2b \sin\theta\cos\theta} = - \frac{a}{b}\]
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