Question

The locus of the point of intersection of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the points whose eccentric angles differ by $\frac{\pi }{2}$ is:

• A. $x^2+y^2=a^2$
• B. $x^2+y^2=b^2$
• C. $x^2+y^2=a^2+b^2$
• D. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$

Answer 1

The correct option is D $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$

Let $P(a\cos \theta ,b\sin \theta )$ and $Q\left(a\cos \right(\frac{\pi }{2}+\theta ),b\sin (\frac{\pi }{2}+\theta \left)\right)$ be two points on the ellipse.

The equations of tangents to ellipse at points P and Q are
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

and, $-\frac{x}{a}\sin \theta +\frac{y}{b}\cos \theta =1$

Let $P(h,k)$ be the point of intersection of (i) and (ii). Then,
$\frac{h}{a}\cos \theta +\frac{k}{b}\sin \theta =1$ and $-\frac{h}{a}\sin \theta +\frac{k}{b}\cos \theta =1$

$\Rightarrow {(\frac{h}{a}\cos \theta +\frac{k}{b}\sin \theta )}^2+{(-\frac{h}{a}\sin \theta +\frac{k}{b}\cos \theta )}^2=2$

$\Rightarrow \frac{h^2}{a^2}+\frac{k^2}{b^2}=2$

Hence, the locus of (h, k) is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$.