p_1 and p_2 are the perpendicular distances f | vedclass

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(C) The equation of the curve is $x^{2/3} + y^{2/3} = a^{2/3}$ $\dots(i)$.Any point on the curve can be represented as $(a \cos^3 	heta, a \sin^3 	h

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$5$

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(C) The equation of the curve is $x^{2/3} + y^{2/3} = a^{2/3}$ $\dots(i)$.Any point on the curve can be represented as $(a \cos^3 	heta, a \sin^3 	heta)$.The slope of the tangent is $\frac{dy}{dx} = -\frac{\sin 	heta}{\cos 	heta} = -	an 	heta$.The equation of the tangent is $y - a \sin^3 	heta = -	an 	heta (x - a \cos^3 	heta)$,which simplifies to $x \sin 	heta + y \cos 	heta = a \sin 	heta \cos 	heta = \frac{a}{2} \sin 2	heta$.The perpendicular distance $p_1$ from the origin $(0,0)$ to the tangent is $p_1 = \left| \frac{-\frac{a}{2} \sin 2	heta}{\sqrt{\sin^2 	heta + \cos^2 	heta}} ight| = \frac{a}{2} \sin 2	heta$,so $2p_1 = a \sin 2	heta$ $\dots(ii)$.The slope of the normal is $\cot 	heta$. The equation of the normal is $y - a \sin^3 	heta = \cot 	heta (x - a \cos^3 	heta)$,which simplifies to $x \cos 	heta - y \sin 	heta = a \cos 2	heta$.The perpendicular distance $p_2$ from the origin to the normal is $p_2 = \left| \frac{-a \cos 2	heta}{\sqrt{\cos^2 	heta + \sin^2 	heta}} ight| = a \cos 2	heta$ $\dots(iii)$.Squaring and adding $(ii)$ and $(iii)$ gives $(2p_1)^2 + p_2^2 = a^2 \sin^2 2	heta + a^2 \cos^2 2	heta = a^2$.Thus,$4p_1^2 + p_2^2 = a^2$. Comparing this with $k_1 p_1^2 + k_2 p_2^2 = a^2$,we get $k_1 = 4$ and $k_2 = 1$.Therefore,$k_1 + k_2 = 4 + 1 = 5$.

$p_1$ and $p_2$ are the perpendicular distances from the origin to the tangent and normal drawn at any point on the curve $x^{2/3} + y^{2/3} = a^{2/3}$ respectively. If $k_1 p_1^2 + k_2 p_2^2 = a^2$,then $k_1 + k_2 =$

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