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Q.
The shortest distance between the parabolas $y^{2}=4 x$ and $y^{2}=2 x-6$ is
Conic Sections
Solution:
Normal to $y^{2}=4 x$ at $\left(m^{2}, 2 m\right)$ is
$y+m x-2 m-m^{3}=0$
Normal to $y^{2}=2(x-3)$ at $\left(1 / 2 t^{2}+3, t\right)$ is
$y+t(x-3)-t-\frac{1}{2} t^{3}=0$
Both are same if $-2 m-m^{3}=-4 m-1 / 2 m^{3}$
$\Rightarrow m=0, \pm 2$
So, points will be $(4,4)$ and $(5,2)$
Hence, shortest distance will be $=\sqrt{1+4}=\sqrt{5}$.