Question

The distance between the point $(1,1)$ and the tangent to the curve $y = e^{2x} + x^2$ drawn at the point $x = 0$ is

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

Answer 1

Answer: (C)

Putting $x = 0$ in $y = e^{2x} + x^2$ we get $y = 1$
$\therefore $ The given point is $P\left(0, 1\right)$
$y = e^{2x} + x^{2} \quad\cdots\left(1\right)$
$\Rightarrow \frac{dy}{dx} = 2e^{2x} + 2x$
$\Rightarrow \left[\frac{dy}{dx}\right]_{P} = 2$
$\therefore $ Equation of tangent at $P$ to $\left(1\right)$ is
$y - 1 = 2\left(x -0 \right)$
$\Rightarrow 2x - y + 1 =0 \quad \cdots \left(2\right)$
$\therefore $ Required distance = Length of $\bot$ from $\left(1,1\right)$ to $\left(2\right)$
$= \frac{2-1+1}{\sqrt{4+1}} = \frac{2}{\sqrt{5}}$.