If the normal to the curve x^{2/3} + y^{2/3} | vedclass

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(B) The equation of the curve is $x^{2/3} + y^{2/3} = a^{2/3}$.Parametric equations are $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$.The derivative

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$y \cos \phi - x \sin \phi = a \cos 2 \phi$

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(B) The equation of the curve is $x^{2/3} + y^{2/3} = a^{2/3}$.Parametric equations are $x = a \cos^3 	heta$ and $y = a \sin^3 	heta$.The derivative $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{3a \sin^2 	heta \cos 	heta}{-3a \cos^2 	heta \sin 	heta} = -	an 	heta$.The slope of the tangent is $m_T = -	an 	heta$.The slope of the normal is $m_N = \frac{-1}{m_T} = \cot 	heta$.Given that the normal makes an angle $\phi$ with the $X$-axis,its slope is $	an \phi$.Thus,$	an \phi = \cot 	heta = 	an(\frac{\pi}{2} - 	heta)$,which implies $\phi = \frac{\pi}{2} - 	heta$,or $	heta = \frac{\pi}{2} - \phi$.Substituting $	heta$ into the parametric coordinates: $x = a \cos^3(\frac{\pi}{2} - \phi) = a \sin^3 \phi$ and $y = a \sin^3(\frac{\pi}{2} - \phi) = a \cos^3 \phi$.The equation of the normal at $(a \sin^3 \phi, a \cos^3 \phi)$ with slope $	an \phi$ is:$y - a \cos^3 \phi = 	an \phi (x - a \sin^3 \phi)$$y \cos \phi - a \cos^4 \phi = x \sin \phi - a \sin^4 \phi$$y \cos \phi - x \sin \phi = a (\cos^4 \phi - \sin^4 \phi)$$y \cos \phi - x \sin \phi = a (\cos^2 \phi - \sin^2 \phi)(\cos^2 \phi + \sin^2 \phi)$$y \cos \phi - x \sin \phi = a \cos 2 \phi$.

If the normal to the curve $x^{2/3} + y^{2/3} = a^{2/3}$ makes an angle $\phi$ with the $X$-axis,then the equation of that normal is

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