Question

The minimum distance of origin from the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$ is $(a>0,b>0)$

• A. a-b
• B. a+b
• C. 2a+2b
• D. 2(a-b)

Answer 1

The correct option is B a+b

Let $(asec\theta ,bcosec\theta )$ be a point on the curve, then its diastance from the origin

$=\sqrt{a^{2}se{c}^2\theta +b^{2}cose{c}^2\theta }$

$\therefore f\left(\theta \right)=a^{2}se{c}^2\theta +b^{2}cose{c}^2\theta$

$=a^2+a^{2}ta{n}^2\theta +b^2+b^{2}co{t}^2\theta$

$=a^2+b^2+a^{2}ta{n}^2\theta +b^{2}co{t}^2\theta$

$\ge a^2+b^2+2\sqrt{a^2{b}^2}=(a+b{)}^2$

$\therefore$ minimum value of $f\left(\theta \right)=(a+b{)}^2$

$\therefore$ minimum value of $\sqrt{a^{2}se{c}^2\theta +b^{2}cose{c}^2\theta }$ is $a+b$