Question

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

• A. $\frac{4}{\sqrt{5}}$
• B. $\frac{3}{\sqrt{5}}$
• C. $\frac{2}{\sqrt{5}}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{5}}$

Distance of any point $(x,y)$ from $y=2x-1$ is: $\left|\frac{y-2x+1}{\sqrt{5}}\right|.$
If $(x,y)$ is on $y=x^4+3x^2+2x$ then the distance $S=\frac{x^4+3x^2+1}{\sqrt{5}}$
$\frac{dS}{dx}=\frac{4x^3+6x}{\sqrt{5}}$
$\Rightarrow \frac{dS}{dx}=0$
$\Rightarrow x=0$
Also, $S^{\prime }\left(x\right)<0$ for $x<0$ and $S^{\prime }\left(x\right)>0$ for $x>0$
Thus $S$ is minimum when $x=0$
$\Rightarrow S_{min}=\frac{1}{\sqrt{5}}$