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