Question

The minimum distance from the origin to a point on the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$ is

• A. $2ab$
• B. $2(a+b)$
• C. $2a+4b$
• D. $a+b$

Answer 1

The correct option is D $a+b$

Let point on curve be $P(acosec\theta ,bsec\theta )$
distance from origin $OP=\sqrt{a^{2}cose{c}^2\theta +b^{2}se{c}^2\theta }$
$=\sqrt{a^2+a^2{\cot }^2\theta +b^2+b^2{\tan }^2\theta }$
$=\sqrt{a^2+b^2+(a\cot \theta -b\tan \theta {)}^2+2ab}$
$=\sqrt{(a+b{)}^2+(a\cot \theta -b\tan \theta {)}^2}$
So, under the square root we have sum of two square terms, the first of which is a constant and second is a function of $\theta$. The minimum value of second term is $0$.

So, $O{P}_{min.}=a+b$