Question

The minimum distance between the curves $x^2+y^2+4x+16y+66=0$ and $y^2=8x$ is

• A. $3\sqrt{2}$ units
• B. $5\sqrt{2}$ units
• C. $4\sqrt{2}-2$ units
• D. $4\sqrt{2}+2$ units

Answer 1

The correct option is A $3\sqrt{2}$ units

Given, equation of parabola is $y^2=8x$
$\Rightarrow 2y{y}^{\prime }=8\Rightarrow y^{\prime }=\frac{4}{y}$
Slope of normal to the parabola at $P(2t^2,4t)$ is
$\frac{-1}{y_P^{\prime }}=\frac{-4t}{4}=-t$
For shortest distance, the normal must pass through centre of the given circle i.e., $C(-2,-8)$
$\Rightarrow -t=\frac{4t+8}{2t^2+2}$
$\Rightarrow 2t^3+6t+8=0\Rightarrow 2(t+1)(t^2-t+4)=0$
$\Rightarrow t=-1$ or $t^2-t+4=0$ $\text{(not possible as}D<0)$
$\Rightarrow t=-1$
$\therefore P\equiv (2,-4)$

Hence, shortest distance
$=CP-r$
$=\sqrt{(2+2{)}^2+(8-4{)}^2}-\sqrt{2}$
$=4\sqrt{2}-\sqrt{2}=3\sqrt{2}$