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)

Centre

Let $P(a\cos {\theta }_1,b\sin {\theta }_2);Q(a\cos {\theta }_1,b\sin {\theta }_2)$ be two points on the ellipse. Then,
$m_1=$ slope of $PO=\frac{b}{a}\tan {\theta }_1$
and $m_2=$ slope of $OQ=\frac{b}{a}\tan {\theta }_2$
Since $m_1{m}_2=\frac{b^2}{a^2}\tan {\theta }_1\tan {\theta }_2=\frac{b^2}{a^2}\times \frac{{(-a)}^2}{b^2}=-1$
Thus $\angle POQ=\frac{\pi }{2}$
Hence, $PQ$ makes a right angle at the centre of the ellipse