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प्रश्न
If \[x = a\sin2t\left( 1 + \cos2t \right) \text { and y } = b\cos2t\left( 1 - \cos2t \right)\] , show that at \[t = \frac{\pi}{4}, \frac{dy}{dx} = \frac{b}{a}\] ?
उत्तर
\[x = a\sin2t\left( 1 + \cos2t \right) \text { and y} = b\cos2t\left( 1 - \cos2t \right)\]
\[ \Rightarrow \frac{dx}{dt} = 2a\cos2t\left( 1 + \cos2t \right) + 2a\sin2t\left( 1 - \cos2t \right) \text { and } \frac{dy}{dt} = - 2b\sin2t\left( 1 - \cos2t \right) + 2b\cos2t\left( 1 + \cos2t \right) \]
\[ \Rightarrow \frac{dx}{dt} = 2a\left( \cos2t + \cos^2 2t + \sin2t - \sin2t\cos2t \right) \text {and } \frac{dy}{dt} = 2b\left( - \sin2t + \sin2t\cos2t + \cos2t + \cos^2 2t \right)\]
\[ \therefore \frac{dy}{dt} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{- 2b\left( - \sin2t + \sin2t\cos2t + \cos2t + \cos^2 2t \right)}{2a\left( \cos2t + \cos^2 2t + \sin2t - \sin2t\cos2t \right)}\]
\[ \Rightarrow \left( \frac{dy}{dt} \right)_{t = \frac{\pi}{4}} = \frac{- 2b\left( - \sin\frac{\pi}{2} + \sin\frac{\pi}{2}\cos\frac{\pi}{2} + \cos\frac{\pi}{2} + \cos^2 \frac{\pi}{2} \right)}{2a\left( \cos\frac{\pi}{2} + \cos^2 \frac{\pi}{2} + \sin\frac{\pi}{2} - \sin\frac{\pi}{2}\cos\frac{\pi}{2} \right)} = \frac{b}{a}\]
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