The distance between the origin and the normal to the curve
$y = e^{2 x}$ to $x^{2}$ at x = 0 is

\frac{2}{\sqrt{3}}

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\frac{2}{\sqrt{5}}

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\frac{2}{\sqrt{7}}

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n o n e o f t h e s e

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Solution

The correct option is B $\frac{2}{\sqrt{5}}$
$\frac{d y}{d x} = 2 e^{2 x} + 2 x \Rightarrow (\frac{d y}{d x})_{x = 0} = 2 e^{\circ} + 0 = 2$
Slope of normal $= - \frac{1}{2}, a t x = 0, y = 1$
Equation of normal, $y - 1 = - \frac{1}{2} (x - 0)$
$\Rightarrow x + 2 y - 2 = 0. I t s d i s t a n c e f r o m (0, 0) = \frac{2}{\sqrt{5}}$

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