Question

If any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ intercepts equal length ${}^{\prime }{l}^{\prime }$ on the axes, then $l$ is equal to

• A. $a^2+b^2$
• B. $\sqrt{a^2+b^2}$
• C. $(a^2+b^2{)}^2$
• D. None of these

Answer 1

Answer: (B)

$\sqrt{a^2+b^2}$

The equation of tangent to the given ellipse at point $P(a\cos \theta ,b\sin \theta )$ is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$
Intercept of line on the axes are $\frac{a}{\cos \theta }$ and $\frac{b}{\sin \theta }$
Given that, $\frac{a}{\cos \theta }=\frac{b}{\sin \theta }=l$
$\Rightarrow \cos \theta =\frac{a}{l}$
and $\sin \theta =\frac{b}{l}$
$\Rightarrow {\cos }^2\theta +{\sin }^2\theta =\frac{a^2}{l^2}+\frac{b^2}{l^2}=1$
$\Rightarrow l^2=a^2+b^2\therefore l=\sqrt{a^2+b^2}$.