If $PQ$ subtends right angle at the centre of ellipse then find $\frac{1}{OP^2}+\frac{1}{OQ^2}. $

Question

$PQ$ 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 $\frac{a^2}{\cos^2\alpha}$.

Answer 1

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 $\large\color{red}{\frac{1}{a^2}+\frac{1}{b^2}}$.

Answer 2

Using your initial work, we have:$$\tan(\alpha)\tan(\beta)=-\frac{a^2}{b^2}\tag{1}$$I then did some rearrangement as follows: $$\begin{align} \frac{1}{a^2\cos^2(\alpha)+b^2\sin^2(\alpha)}&=\frac{\frac{1}{\cos^2(\alpha)}}{b^2(\frac{a^2}{b^2}+\tan^2(\alpha))}\\ &=\frac{\sec^2(\alpha)}{b^2(\frac{a^2}{b^2}+\tan^2(\alpha))}\\ &=\frac{1+\tan^2(\alpha)}{b^2(-\tan(\alpha)\tan(\beta)+\tan^2(\alpha))}\text{ (using (1) above)}\\ &=\frac{1+\tan^2(\alpha)}{b^2\tan(\alpha)(\tan(\alpha)-\tan(\beta))}\tag{2} \end{align}$$ Similarly we can show that: $$\begin{align} \frac{1}{a^2\cos^2(\beta)+b^2\sin^2(\beta)}&=\frac{1+\tan^2(\beta)}{b^2\tan(\beta)(\tan(\beta)-\tan(\alpha))}\\ &=-\frac{1+\tan^2(\beta)}{b^2\tan(\beta)(\tan(\alpha)-\tan(\beta))}\tag{3} \end{align}$$ Now we can combine (2) and (3) to get: $$\begin{align}\require{cancel} \frac{1}{a^2\cos^2(\alpha)+b^2\sin^2(\alpha)}+\frac{1}{a^2\cos^2(\beta)+b^2\sin^2(\beta)}&=\frac{1+\tan^2(\alpha)}{b^2\tan(\alpha)(\tan(\alpha)-\tan(\beta))}-\frac{1+\tan^2(\beta)}{b^2\tan(\beta)(\tan(\alpha)-\tan(\beta))}\\ &=\frac{\tan(\beta)(1+\tan^2(\alpha))-\tan(\alpha)(1+\tan^2(\beta))}{b^2\tan(\alpha)\tan(\beta)(\tan(\alpha)-\tan(\beta))}\\ &=\frac{\tan(\beta)+\tan(\beta)\tan^2(\alpha)-\tan(\alpha)-\tan(\alpha)\tan^2(\beta)}{b^2\tan(\alpha)\tan(\beta)(\tan(\alpha)-\tan(\beta))}\\ &=\frac{\tan(\alpha)\tan(\beta)(\tan(\alpha)-\tan(\beta))-(\tan(\alpha)-\tan(\beta))}{b^2\tan(\alpha)\tan(\beta)(\tan(\alpha)-\tan(\beta))}\\ &=\frac{(\tan(\alpha)\tan(\beta)-1)\cancel{(\tan(\alpha)-\tan(\beta))}}{b^2\tan(\alpha)\tan(\beta)\cancel{(\tan(\alpha)-\tan(\beta))}}\\ &=\frac{(\tan(\alpha)\tan(\beta)-1)}{b^2\tan(\alpha)\tan(\beta)}\\ &=\frac{-\frac{a^2}{b^2}-1}{b^2(-\frac{a^2}{b^2})}\\ &=\frac{\frac{-a^2-b^2}{b^2}}{-a^2}\\ &=\frac{a^2+b^2}{a^2b^2}\\ &=\frac{1}{a^2}+\frac{1}{b^2} \end{align}$$