The shortest distance between the parabolas y^ 2 = 4x and y^ 2 = 2x...

Shortest distance between two curves occured along the common normal.

$∴$ Normal to $y^{2} = 4x$ at \left(m^{2}, 2m\right)$is$

$y + mx - 2m - m^{3} = 0$

Normal to $y^{2} = 2 \left(x - 3\right)$ at \left(\dfrac{m2}{2} +3, m\right)$is$

$y + m \left(x - 3\right)$ - m - $\dfrac{m3}{2} =$0$</p><p>Both normals are same, if$

$- 2m - m^{3} = - 4m - \dfrac{1}{2} m$^{3} ⇒ m = 0, 2$</p><p>So, points will be$(4,4) and (5,2) or ( 4, - 4 ) and ( 5, - 2 )$.</p><p>Hence, shortest distance will be$

\sqrt{( 1 + 4 )} = \sqrt{5}& unit.</p>