Explanation

(c)
Given curve is
$\ \ y = e^{2x} + x^{2}$
At $\ \ x = 0,\ y = 1$
$\therefore$ Any point on the curve is
$\ \frac{dy}{dx} = 2e^{2x} + 2x$
Slope of normal at (0,1)$= - \frac{1}{2 + 0} = - \frac{1}{2}$
$\therefore Equation\ of\ normal\ is$
$\ \ y - 1 = - \frac{1}{2}(x - 0)$
$\Rightarrow \ 2y - 2 = - x$
$\Rightarrow \ \ x + 2y - 2 = 0$
Required distance=$\left| \frac{0 + 0 - 2}{\sqrt{1 + 4}} \right|\ \ \$
$\ \ = \frac{2}{\sqrt{5}}\$