Question

If tan $\theta_1 \tan \theta_2 =-\frac{a^2}{b^2}$ then the chord joining two points $\theta_1$ and $ \theta_2$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ will subtend a right angle at

• A. focus
• B. centre
• C. end of the major axis
• D. end of the minor axis

Answer 1

Answer: (B)

Given $tan \,\theta_{1} tan\,\theta_{2} = -\frac{a^{2}}{b^{2}} ,$
Let $P\left(\theta_{1}\right), Q\left(\theta_{2}\right)$ be two points on the ellipse.
Slope of $CP, CQ$ are $\frac{b\, sin\, \theta_{1}}{a \,cos\, \theta_{1}} ;\frac{ b \,sin \,\theta_{2}}{a \,cos\, \theta_{2}} $
i.e., $\frac{b}{a} tan\, \theta_{1}, \frac{b}{a} \,tan\,\theta_{2} $
Product of the slopes $= \frac{b^{2}}{a^{2}} tan\, \theta_{1} tan\,\theta_{2} $
$ = \frac{b^{2}}{a^{2}}\left(-\frac{a^{2}}{b^{2}}\right) = -1 $
$ \therefore $ chord will subtend a right angle at the centre.