$PQ$ is a variable chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ . If $PQ$ subtends right angle at the centre of ellipse then find $\frac{1}{OP^2}+\frac{1}{OQ^2}. $
Two points can be taken $(a\cos\alpha,b\sin\alpha),(a\cos\beta,b\sin\beta)$.
As these subtend right angle at origin $$\frac{b^2\tan\alpha .\tan\beta}{a^2}=-1 $$
And we want $$\frac{1}{b^2\sin^2\alpha+a^2\cos^2\alpha}+\frac{1}{b^2\sin^2\beta+a^2\cos^2\beta} $$
I substituted $\tan\alpha$ but it just gave me $\frac{a^2}{\cos^2\alpha}$.
Best Answer
You are on the right track. You know that $\tan(\alpha)\tan(\beta)=-\frac{a^2}{b^2}$, and:
$$\begin{eqnarray*} \frac{1}{OP^2}+\frac{1}{OQ^2} &=& \frac{\frac{1}{\cos^2\alpha}}{b^2\tan^2\alpha+a^2}+\frac{\frac{1}{\cos^2 \beta}}{b^2\tan^2\beta+a^2}\\[0.2cm]&=&\frac{1+\tan^2\alpha}{b^2\tan^2\alpha+a^2}+\frac{1+\tan^2\beta}{b^2\tan^2\beta+a^2}.\end{eqnarray*}$$ Now, by replacing $\tan^2\beta$ with $\frac{a^4}{b^4 \tan^2\alpha}$, the last expression simplifies to $\large\color{red}{\frac{1}{a^2}+\frac{1}{b^2}}$.