Question

Let $PQ$ be a chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which subtends right angle at the centre $(0,0).$ Then its distance from the centre is equal to

• A. $\frac{ab}{\sqrt{a^2+b^2}}$
• B. $\frac{a}{\sqrt{a^2+b^2}}$
• C. $\frac{b}{\sqrt{a^2+b^2}}$
• D. $\frac{2ab}{\sqrt{a^2+b^2}}$

Answer 1

The correct option is A $\frac{ab}{\sqrt{a^2+b^2}}$

Let $x\cos \alpha +y\sin \alpha =p$ be the chord $PQ.$
Then $p$ is the desired distance.
Homogenizing the equation of the ellipse with the help of this equation, we get the combined equation of $OP$ and $OQ.$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}={\left(\frac{x\cos \alpha +y\sin \alpha }{p}\right)}^2$
$\Rightarrow (\frac{1}{a^2}-\frac{{\cos }^2\alpha }{p^2})x^2+(\frac{1}{b^2}-\frac{{\sin }^2\alpha }{p^2})y^2-\frac{2xy\sin \alpha \cos \alpha }{p^2}=0$
As $OP\perp OQ,$
coefficient of $x^2$ $+$ coefficient of $y^2=0$ (condition of perpendicular lines)
$\frac{1}{a^2}-\frac{{\cos }^2\alpha }{p^2}+\frac{1}{b^2}-\frac{{\sin }^2\alpha }{p^2}=0$
$\Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$
$\Rightarrow p=\frac{ab}{\sqrt{a^2+b^2}}$