The shortest distance between the line x - y = 1 and the curve x ^...

Shortest distance between curves is always along common normal.

Let $P\left({x}_{0}, {y}_{0}\right)$ be any point on the parabola ${x}^{2} = 2y$.

Clearly, slope of line $x - y = 1$ is $1$.

Hence, slope of perpendicular line is $- 1$.

$\dfrac{dy}{dx}P =$Slope of tangent at $P$

Therefore, $\dfrac{- 1}{\dfrac{dy}{dx}P} =$Slope of normal$= - 1$

$⇒ \dfrac{dy}{dx}P = 1$

$⇒ {x}_{0} = 1 ∴ {y}_{0} = \dfrac{1}{2}$

$⇒ P\left(1, \dfrac{1}{2}\right)$

$∴$ Shortest distance $= \left|\dfrac{1 - \dfrac{1}{2} - 1}{\sqrt{{1}^{2} + {1}^{2}}}\right| = \dfrac{1}{2\sqrt{2}}$

option \left(3\right)