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$The ratio of the area enclosed by the locus of the midpoint of P S and area of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 is$
$(P be any point on the ellipse and S be iti focus)$

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detailed solution

Correct option is D

Mid point of PS is (h, k)∴ h=acos⁡θ+ae2⇒cos⁡θ=2h−aea and k=bsin⁡θ2 Eliminating θ, we get locus as (2h−ae)2a2+4k2b2=1⇒ x−ae22a24+y2b24=1which is ellipse having enclosed area ∴ Area =πa2⋅b2=πab4 Hence, required ratio =14

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The ratio of the area enclosed by the locus of the midpoint of PS and area of the ellipse x2a2+y2b2=1 is ( P be any point on the ellipse and S be iti focus)

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$The ratio of the area enclosed by the locus of the midpoint of P S and area of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 is$
$(P be any point on the ellipse and S be iti focus)$