Question

If the distance of the point on $y=x^4+3x^2+2x$ which is nearest to the line $y=2x-1$ is p, Find $5p^2$.

Answer 1

$y=x^4+3x^2+2x$ and $y=2x-1$

The distance between the points lying on two curves which is shortest.

$D=\frac{x^4+3x^2+2x-2x+1}{\sqrt{{(-2)}^2+{\left(1\right)}^2}}$ ........ (Substitute $y=x^4+3x^2+2xiny=2x-1)$

$D=\frac{x^4+3x^2+1}{\sqrt{5}}$

Since $x^4$ & $x^2$ both are squared terms, $x^4\ge 0$ & $3x^2\ge 0$

$\therefore D_{min}=\frac{0+0+1}{\sqrt{5}}=\frac{1}{\sqrt{5}}So,p=\frac{1}{\sqrt{5}}\therefore {5p}^2=5{\left(\frac{1}{\sqrt{5}}\right)}^2=5\left(\frac{1}{5}\right)=1$