Question

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

• A. $2$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{2}{\sqrt{5}}$
• D. $\frac{1}{2}$
• E. $\frac{1}{\sqrt{5}}$

Answer 1

The correct option is C

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

Explanation for the correct option:

Finding the distance from the origin and the normal to the curve:

Step 1: Finding slope of the curve at a given point:

Given $y=e^{2x}+x^2$ at $x=0$

Substitute $x=0$ in $y=e^{2x}+x^2$ to obtain y-coordinate.

$$\begin{gathered} \Rightarrow y=e^{2\left(0\right)}+{\left(0\right)}^2 \\ =1 \end{gathered}$$

So, point $(0,1)$ lie on the curve.

Now, Differentiate with $y$ respect to '$x$' to obtain slope.

$$\begin{gathered} \frac{dy}{dx}=2e^{2x}+2x \\ {\left(\frac{dy}{dx}\right)}_{x=0}=2 \end{gathered}$$

Thus, the slope of the curve at $x=0$ is $2$.

Step 2: Finding the equation of the normal using slope:

The normal to the curve is perpendicular to the curve at point $x=0$

Therefore, Slope of the normal is given by

$$\begin{gathered} m=\frac{-1}{slopeofcurveatx=0} \\ m=\frac{-1}{2} \end{gathered}$$

Equation of normal at $(0,1)$ is

$$\begin{gathered} y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-1=\frac{-1}{2}(x-0) \\ \Rightarrow 2y+x=2 \end{gathered}$$

Step 3: Finding distance of line from origin:

The required distance from origin,

$$\begin{gathered} \begin{array}{rcl}& =& \left|\frac{2y+x-2}{\sqrt{a^2+b^2}}\right|\text{at}(0,0)\end{array} \\ =\left|\frac{2\left(0\right)+0-2}{\sqrt{2^2+1^2}}\right| \\ =\left|\frac{-2}{\sqrt{5}}\right| \\ =\frac{2}{\sqrt{5}} \end{gathered}$$

Hence, option (C) is the correct answer.