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)

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