Question

The distance of the point on $y=x^4+3x^2+2x$ which is nearest to the line $y=2x-1$ is

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

Answer 1

The correct option is C

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

Explanation for the correct option:

let $\left(x_0,y_0\right)$ be the nearest point to the line $y=2x-1$

The tangent line at this point will be parallel to the line $y=2x-1$

Therefore:

${\overline{)\frac{dy}{dx}}}_{x_0}=$slope of line$y=2x-1$$=2$

$$\begin{gathered} {\overline{)\frac{dy}{dx}}}_{x_0}=4{x_0}^3+6x_0+2 \\ \Rightarrow 4{x_0}^3+6x_0+2=2 \\ \Rightarrow 4{x_0}^3+6x=0 \\ \Rightarrow x_0\left(4{x_0}^2+6\right)=0 \end{gathered}$$

$\Rightarrow x_0=0$ (since $4{x_0}^2+6>0$)

$\Rightarrow y_0=0$

Distance between a point $\left(x_1,y_1\right)$ from the line $ax+by+c=0$ is $\frac{\mathbf{|}\mathbf{a}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}\mathbf{b}{\mathbf{y}}_{\mathbf{1}}\mathbf{+}\mathbf{c}\mathbf{|}}{\sqrt{\left(a^2+b^2\right)}}$

The distance $\left(0,0\right)$ from the line $y=2x-1$ is $\frac{0-0+1}{\sqrt{1+4}}$

$=\frac{1}{\sqrt{5}}$

Hence, option (C) is the correct answer