Question

The minimum distance from the point $(4,2)$ to the parabola $y^2=8x$ is

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $2$
• D. $3\sqrt{2}$

Answer 1

Answer: (B)

Let $d=$ distance of a point $(x,y)$, of the parabola from the point $(4,2)$
$\therefore D=d^2={\left(x-4\right)}^2+{\left(y-2\right)}^2$
$=x^2+16-8x+y^2-4y+4$
$\left[\because y^2=8x\right]$
$=\frac{y^4}{64}+16-y^2-4y+4$
$=\frac{y^4}{64}-4y+20$
$\therefore \frac{d}{dy}\left(D\right)=\frac{4y^3}{64}-4$ and $\frac{d^2}{d{y}^2}\left(D\right)=\frac{12y^2}{64}>0$
$\therefore D$ is Min when $\frac{d\left(D\right)}{dy}$
$\Rightarrow \frac{4y^3}{64}-4=0$
$\Rightarrow y^3=64$
$\Rightarrow y=4$.
But $y^2=8x$
$\therefore 16=8x$
$\Rightarrow x=2$.
$\therefore$ reqd. min. distance
$=\sqrt{{\left(2-4\right)}^2+{\left(4-2\right)}^2}$
$=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$