Explanation

(b)
Let radius vector is $r$
$\therefore\ \ r^{2} = x^{2} + y^{2}$
$\Rightarrow \ \ r^{2} = \frac{a^{2}y^{2}}{y^{2} - b^{2}} + y^{2}\ \ \ \ \left( \because\ \frac{a^{2}}{x^{2}} + \frac{b^{2}}{y^{2}} = 1 \right)$
For minimum value of $r$
$\frac{d\left( r^{2} \right)}{dy} = 0\ \ \Rightarrow \ \frac{- 2yb^{2}a^{2}}{\left( y^{2} - b^{2} \right)^{2}} + 2y = 0$
$\Rightarrow \ \ \ \ y^{2} = b(a + b)$
$\therefore\ \ \ x^{2} = a(a + b)$
$\Rightarrow \ \ \ r^{2} = (a + b)^{2}$
$\Rightarrow \ \ r = a + b$