The locus of the point which divides the double ordinates of the ellipse $$x2a2+y2b2=1$$ in the ratio $$1$$:$$2$$ internally is

Correct option is 1x2a2+9y2b2=1

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$The locus of the point which divides the double ordinates of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 in the ratio 1 : 2 internally is$

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x2a2+9y2b2=1

x2a2+9y2b2=19

9x2a2+9y2b2=1

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detailed solution

Correct option is A

Let P≡(acos⁡θ,bsin⁡θ),Q≡(acos⁡θ,−bsin⁡θ) . PR:RQ=1:2∴ h=acos⁡θ or cos⁡θ=ha----(1) and k=b3sin⁡θ or sin⁡θ=3kb----(2)On squaring and adding (i) and (ii), we getx2a2+9y2b2=1

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The locus of the point which divides the double ordinates of the ellipse x2a2+y2b2=1 in the ratio 1:2 internally is

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$The locus of the point which divides the double ordinates of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 in the ratio 1 : 2 internally is$