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

\frac{1}{\sqrt{5}}

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

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\frac{- 1}{\sqrt{5}}

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

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Solution

The correct option is A $\frac{1}{\sqrt{5}}$
$y = e^{2 x} + x^{2}$
At $x = 0 \Rightarrow y = 1$
$\frac{d y}{d x} = 2 e^{2 x} + 2 x$
Slope of tangent at (0,1) $= 2$
Equation of tangent through (0,1) is
$y - 1 = 2 (x - 0)$
$\Rightarrow y - 2 x - 1 = 0$
Distance of tangent from origin $= | \frac{- 1}{\sqrt{1 + 4}} | = \frac{1}{\sqrt{5}}$

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Similar questions

y = e^{2 x} + x^{2}

x = 0

= e^{2 x} + x^{2}

=

y = e^{2 x} + x^{2}

x = 0

y = x^{2} - x + 1

x_{1} = 0, x_{2} = - 1

x_{3} = \frac{5}{2}

(1, 1)

y = e^{2 x} + x^{2}

y

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