Question

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

• A. | a - b |
• B. | a + b |
• C. ab
• D. None of these

Answer 1

The correct option is B | a + b |

Given:

$\text{eqn of}curve:\frac{a^2}{x^2}+\frac{b^2}{y^2}=1(a,b>0)$

let $(\mathrm{asec}\theta ,b\mathrm{cosec}\theta )$ be a point on the curve

then is distance from origin

$=\sqrt{a^2{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta }$

$\therefore f\left(\theta \right)=a^2{\sec }^2\theta +b^2{\mathrm{cosec}}^2\theta$

$=a^2+a^2{\tan }^2\theta +b^2+b^2{\cot }^2\theta$

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

for minimum condition

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

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

$\therefore$ minimum value of (i)

$\sqrt{f\left(0\right)}$

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

$=\mid a+b)$

Ans.- min value $=|a+b|=$ min distance