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Question
If x = a cos θ, y = b sin θ, show that \[\frac{d^2 y}{d x^2} = - \frac{b^4}{a^2 y^3}\] ?
Solution
Here,
\[x = a \cos\theta \text { and y } = b \sin\theta\]
\[\text { Differentiating w . r . t . } \theta, \text { we get}\]
\[\frac{d x}{d \theta} = - a \sin\theta \text { and } \frac{d y}{d \theta} = b \cos\theta\]
\[ \therefore \frac{d y}{d x} = \frac{b \cos\theta}{- a \sin\theta} = \frac{- b}{a}\cot\theta\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = - \frac{b}{a} \times \left( - {cosec}^2 \theta \right) \frac{d \theta}{d x}\]
\[ = \frac{b}{a} \times {cosec}^2 \theta \times \frac{1}{- a \sin\theta}\]
\[ = - \frac{b}{a^2} \times \frac{1}{\sin^3 \theta}\]
\[ = - \frac{b}{a^2} \times \frac{b^3}{y^3} \left[ \because y = b \sin\theta \right]\]
\[ = \frac{- b^4}{a^2 y^3}\]
Hence proved.
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