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If \( x=\frac{2 b t}{1+t^{2}}, y=a\left(\frac{1-t^{2}}{1+t^{2}}\right) \) then show that \( \frac{d y}{d x}=\frac{-b^{2} y}{a^{2} x} \) |
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Answer» x \(=\frac{2bt}{1+t^2}\) ∴ \(\frac{dx}{dt}\) \(\frac{(1+t^2)\times2b-2bt\times2t}{(1+t^2)^2}\) \(\left(\because\frac{d}{dt}\frac{u}{v}=\cfrac{v\frac{du}{dt}-u\frac{dv}{dt}}{v^2}\right)\) \(=\frac{2b(1+t^2-2t^2)}{(1+t^2)^2}\) \(=\frac{2b(1-t^2)}{(1+t^2)^2}\) And y \(=\frac{a(1-t^2)}{1+t^2}\) ∴ \(\frac{dy}{dt}\) \(\frac{(1+t^2)\times-2at-a(1-t^2)\times2t}{(1+t^2)^2}\) (By u/v formula) \(=\frac{2at(-1-t^2-1+t^2)}{(1+t^2)^2}\) \(=\frac{-4at}{(1+t^2)^2}\) Now, \(\frac{dy}{dx}\) \(=\cfrac{\frac{dy}{dt}}{\frac{dx}{dt}}\) \(=\cfrac{\frac{-4at}{(1+t^2)^2}}{\frac{2b(1-t^2)}{(1+t^2)^2}}\) \(=\frac{-2at}{b(1-t^2)}\) = -a/b x \(\frac{2bt}{b(1+t^2)}\times\frac{a(1+t^2)}{a(1-t^2)}\) \(=-\frac{a^2}{b^2}\times\frac{2bt}{1+t^2}\times\frac{1+t^2}{a(1-t^2)}\) \(=\frac{-a^2}{b^2}\times x\times\frac1y=-\frac{a^2x}{b^2y}\) |
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