The length of the tangent drawn at the point | vedclass

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(B) Given the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2^{\frac{2}{3}}$.Let $x=2 \cos^3 	heta$ and $y=2 \sin^3 	heta$.At $	heta = \frac{\pi}{4}$,the

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(B) Given the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2^{\frac{2}{3}}$.Let $x=2 \cos^3 	heta$ and $y=2 \sin^3 	heta$.At $	heta = \frac{\pi}{4}$,the coordinates are $x = 2(\frac{1}{\sqrt{2}})^3 = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$ and $y = 2(\frac{1}{\sqrt{2}})^3 = \frac{1}{\sqrt{2}}$.The derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{6 \sin^2 	heta \cos 	heta}{-6 \cos^2 	heta \sin 	heta} = -	an 	heta$.At $	heta = \frac{\pi}{4}$,$\frac{dy}{dx} = -	an(\frac{\pi}{4}) = -1$.The equation of the tangent at $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ is $y - \frac{1}{\sqrt{2}} = -1(x - \frac{1}{\sqrt{2}})$,which simplifies to $x + y = \sqrt{2}$.The tangent intersects the $x$-axis at $y=0$,giving $x = \sqrt{2}$,so the point is $(\sqrt{2}, 0)$.The length of the tangent segment from the point of tangency $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ to the $x$-axis intersection $(\sqrt{2}, 0)$ is $\sqrt{(\sqrt{2} - \frac{1}{\sqrt{2}})^2 + (0 - \frac{1}{\sqrt{2}})^2} = \sqrt{(\frac{1}{\sqrt{2}})^2 + (-\frac{1}{\sqrt{2}})^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$.

The length of the tangent drawn at the point $P\left(\frac{\pi}{4}\right)$ on the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2^{\frac{2}{3}}$ is

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