Question

Prove that the minimum length of the intercept made by the axes on the tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is equal to $a+b$.

Answer 1

The minimum length of a line segment tangent to the ellipse on the two coordinates oxes is $a+b$

For the minimization of

$f\left(\theta \right)=\frac{a^2}{a^2\theta }+\frac{b^2}{{\sin }^2\theta }$

$\Rightarrow f\left(\theta \right)=a^2.{\sin }^2\theta +b^2.{\csc }^2\theta$

So, $f\left(\theta \right)=a^2(1+{\tan }^2\theta )+b^2(1+{\cot }^2\theta )={\omega }^2+b^2+[a^2{\tan }^2\theta +b^2.{\cot }^2\theta ]\ge a^2+b^2+2ab$

Using $A.M\le G.M$

$\Rightarrow \frac{a^2{\tan }^2\theta +b^2.{\cot }^2\theta }{2}\le \sqrt{a^2.{\tan }^2\theta b^2.{\cot }^2\theta }=ab$

So, $a^2{\tan }^2\theta +b^2.{\cot }^2\theta \ge 2ab$