Question

If the shortest distance(in units) between the parabolas $y^2=4x$ and $y^2=2x-6$ is $d$ units, then the value of $d^2$ is

Answer 1

Shortest distance occurs along common normal
Equation of normal at $y^2=4x$ having slope $m$ is
$y=mx-2m-m^3$ and point of contact is $(m^2,-2m)$
For $y=mx-2m-m^3=m(x-3)+m-m^3$ to be notrmal to $y^2=2(x-3)$,
$m-m^3=-2\left(\frac{1}{2}\right)m-\frac{1}{2}{m}^3$
$\Rightarrow 2m-\frac{m^3}{2}=0$
$\Rightarrow m=0,\pm 2$
Point of contact for two parabolas are $(m^2,-2m)$ and $(3+\frac{1}{2}{m}^2,-m)$
For $m=0$
Point of contact are $(0,0)$ and $(3,0)$, and shortest distance will be $3$
For $m=2$
Point of contact are $(4,-4)$ and $(5,-2)$, and shortest distance will be $\sqrt{5}$
For $m=-2$
Point of contact are $(4,4)$ and $(5,2)$, and shortest distance will be $\sqrt{5}$
$\therefore$ shortest distance $\left(d\right)=\sqrt{5}$units
$\Rightarrow d^2=5$