Question

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

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

Answer 1

Answer: (D)

$1/\sqrt{5}$

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 this distance is $S=\frac{x^4+3x^2+1}{\sqrt{5}}$

$\Rightarrow \frac{dS}{dx}=\frac{4x^3+6x}{\sqrt{5}}$

For $\frac{dS}{dx}=0\Rightarrow x=0$

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 $x=0,$ and minimum is $\frac{1}{\sqrt{5}}$