A variable chord $PQ$ of an ellipse centered in $0$ is such that $\angle POQ=\pi/2$. Show that $\dfrac1{OP^2}+\dfrac1{OQ^2}$ is a constant.

Asked Dec 06 '22 at 12:22

Active Dec 06 '22 at 15:09

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$PQ$ is variable chord of the ellipse $x^2 + 4y^2 = 1$.
If $PQ$ subtends a right angle at the center of the ellipse, then $$\dfrac1{OP^2}+\dfrac1{OQ^2}$$ is equal to ? (‘$O$’ being the origin).

edited Dec 06 '22 at 15:09

Jean Marie

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asked Dec 06 '22 at 12:22

user1127900

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Hint: take a parametric representation of the ellipse : $x=\cos(t), y=\frac12 \sin(t)$. Then the orthogonality condition becomes (if $P=M_1$ and $Q=M_2$ : $\cos t_1 \cos t_2+\frac14 \sin t_1 \sin t_2=0$. Up to you... – Jean Marie Dec 06 '22 at 13:43
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Another Hint: you must find $5$. – Jean Marie Dec 06 '22 at 14:12

A variable chord $PQ$ of an ellipse centered in $0$ is such that $\angle POQ=\pi/2$. Show that $\dfrac1{OP^2}+\dfrac1{OQ^2}$ is a constant.

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