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

$$\begin{gathered} \frac{{\mathrm{a}}^2}{{\mathrm{x}}^2}+\frac{{\mathrm{b}}^2}{{\mathrm{y}}^2}=1 \\ \mathrm{x}=\mathrm{a} \sec \mathrm{\theta } \\ \mathrm{y}=\mathrm{ba} \sec \mathrm{\theta } \\ \mathrm{satisfy} \mathrm{given} \mathrm{eqns}. \mathrm{radius} \mathrm{vector} \mathrm{is} \end{gathered}$$
$\begin{array}{rcl}\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}& =& \sqrt{{\mathrm{a}}^2{\sec }^2\mathrm{\theta }+{\mathrm{b}}^2 {\mathrm{cosec}}^2\mathrm{\theta }}\\ & =& \sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2{\tan }^2\mathrm{\theta }+{\mathrm{b}}^2+{\mathrm{b}}^2{\cos }^2\mathrm{\theta }}\end{array}$

$$\begin{gathered} \mathrm{A}.\mathrm{M}\ge \mathrm{G}.\mathrm{M}. \\ \frac{{\mathrm{a}}^2{\tan }^2\mathrm{\theta }+{\mathrm{b}}^2{\cos }^2\mathrm{\theta }}{2}\ge \sqrt{{\mathrm{a}}^2{\tan }^2\mathrm{\theta }.{\mathrm{b}}^2{\cos }^2\mathrm{\theta }} \\ {\mathrm{a}}^2{\tan }^2\mathrm{\theta }+{\mathrm{b}}^2{\cot }^2\mathrm{\theta }\ge 2\mathrm{ab} \\ \min . \mathrm{radius} \mathrm{vector} \\ =\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}} \\ =\mathrm{a}+\mathrm{b} \end{gathered}$$