Question

Prove that the curve $y=x^4+3x^2+2x$ does not meet the straight line $y=2x-1$ and find the distance between their nearest points.

Answer 1

Let the point on the curve be $(x,x^4+3x^2+2x)$ and the line be $2x-y-1=0$

The distance D will be $D=|\frac{2x-x^4-3x^2-2x-1}{\sqrt{2^2+1^2}}|=\frac{-x^4-3x^2-1}{\sqrt{5}}$

The x-coordinate of the point from where the distance is minimum can be found by differentiating the D

$\frac{d\left(D\right)}{dx}=\frac{1}{\sqrt{5}}(-4x^3-6x)=0$

$x=0$ is a solution containing real numbers.

$D_{min}=|\frac{-0^4-{3.0}^2-1}{\sqrt{5}}|=\frac{1}{\sqrt{5}}$