Question

A tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ cuts the axes in $M$ and $N$. Then the least length of $MN$ is:

• A. $a+b$
• B. $a-b$
• C. $a^2+b^2$
• D. $a^2-b^2$

Answer 1

The correct option is A $a+b$

Given eqn of ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Then eqn of tangent at $a\cos \theta ,b\sin \theta$ is
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$
Now, let the tangent intersects x-axis at M.
So, the coordinates of M is $(a\sec \theta ,0)$
Let the tangent intersects y-axis at N.
So, the coordinates of N is $(0,bcosec\theta )$
Length $MN$ be $d=\sqrt{a^2{\sec }^2\theta +b^{2}cose{c}^2\theta }$
$\Rightarrow d=\sqrt{a^2+a^2{\tan }^2\theta +b^2+b^2{\cot }^2\theta }$
For maximum or minimum,
$d^{\prime }\left(\theta \right)=0$
$\Rightarrow \frac{2a^2\tan \theta {\sec }^2\theta -2b^2\cot \theta cose{c}^2\theta }{2\sqrt{a^2+a^2{\tan }^2\theta +b^2+b^2{\cot }^2\theta }}=0$
$\Rightarrow {\tan }^4\theta =\frac{b^2}{a^2}$
$\Rightarrow {\tan }^2\theta =\frac{b}{a}$
So, at this value of $\theta$, MN is least.
$d=\sqrt{a^2+b^2+ab+ab}=\sqrt{(a+b{)}^2}$
Least length of MN $=a+b$