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A secant OPQ is drawn through the centre $\mathrm{O(0,0)}$ of two concentric circles of radii a and $\mathrm{b(a<b)}$ to meet the circles in P and Q. Lines through P parallel to y-axis and through Q parallel to x-axis meet at the point R. Then the locus of R is an ellipse touching both the circle, if the foci of this ellipse lie on the inner circle then its eccentricity is
Option: 1
$1 / 2$

Option: 2
$1 / \sqrt{3}$

Option: 3
$1 / \sqrt{2}$

Option: 4
$\frac{1}{2 \sqrt{2}}$

Answers (1)

best_answer

$\mathrm{x^2+y^2=a^2}$ inner $....(1)$

$\mathrm{x^2+y^2=b^2}$ outer $....(2)$

$\mathrm{\therefore \quad P\, \, is \, \, (a \cos \theta, a \sin \theta)}$

$\mathrm{Q\, \, is \, \, (b \cos \theta, b \sin \theta)}$

$\therefore \quad$ The point R has x of P and y of Q

$\mathrm{\therefore \quad R \, \, is \, \, x=a \cos \theta, y=b \sin \theta}$

$\therefore$ Locus of R is $\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a<b}$

It is an ellipse whose foci lie on y-axis at $\mathrm{(0, b e)}$ and $\mathrm{(0,-b e)}$

if the foci lie on inner circle $\mathrm{x^2+y^2=a^2}$ then $\mathrm{b^2 e^2=a^2}$ but

$\mathrm{ a^2=b^2(1-e)^2}$

or $\mathrm{ b^2 e^2=b^2\left(1-e^2\right) }$

$\mathrm{\therefore 2 e^2=1 or e=1 / \sqrt{2}}$

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manish painkra

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