Question

The minimum radius vector of the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$ ls

• A. $a+b$
• B. $a-b$
• C. $\sqrt{a^2+b^2}$
• D. $\sqrt{a+b}$

Answer 1

The correct option is A $a+b$

lf r is the radius vector, $s=r^2=x^2+y^2$ is the square root= the radlus vector. As $\left(x_{,}y\right)$ lles on
$\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$
$s=x^2+\frac{b^2{x}^2}{x^2-a^2}=x^2+b^2+\frac{a^2{b}^2}{x^2-a^2}$
$\therefore$ $\frac{ds}{dx}=2x-\frac{2x{a}^2{b}^2}{(x^2-a^2{)}^2}=2x\{1-\frac{a^2{b}^2}{({\times }^2-a^2{]}^2}\}$
$\therefore$ $\frac{d^{2}s}{d{x}^2}=2\{1-\frac{a^2{b}^2}{(x^2-a^2{)}^2}\}+\frac{a^2{b}^{2}x}{(x^2-a^2{)}^3}$
$\frac{ds}{dx}=0\Rightarrow (x^2-a^2{)}^2=a^2{b}^2\Rightarrow x^2=a^2+ab$
Then $\frac{d^{2}s}{d{x}^2}>0$ $\therefore$ $s$ ls minimum when $x^z=a^2+$ ab
$\therefore$ minimum radius vector $=\sqrt{(a^2+ab)+\frac{b^2(a^2+ab)}{ab}}$
$=\sqrt{(a+b{)}^2}=a+b$