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Q.

If $p_{1} a n d p_{2}$ are the length of the perpendiculars from origin on the tangent and normal drawn to the curve $x^{2 / 3} + y^{2 / 3} = 6^{2 / 3}$ , respectively then find the value of $4 p_{1}^{2} + p_{2}^{2} =$

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answer is 36.

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Detailed Solution

$x = 6 co s^{3} θ, y = 6 \sin^{3} θ \frac{d y}{d x} = - \tan θ T : x \sin θ + y \cos θ = 6 \sin θ \cos θ N : x \cos θ - y \sin θ = 6 \cos 2 θ P_{1} = 6 \sin θ \cos θ \Rightarrow 2 P_{1} = 6 \sin 2 θ P_{2} = 6 \cos 2 θ 4 P_{1}^{2} + P_{2}^{2} = 36$

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