Shortest distance must be along common normal



Let a point on curve has x coordinate = h

Then y coordinate :

h² = 2y

$$ \Rightarrow $$ y = $${{{h^2}} \over 2}$$

So point is = (h, $${{{h^2}} \over 2}$$)

m₁ (slope of line x $$-$$ y = 1) = 1

$$ \Rightarrow $$ slope of perpendicular line = $$-$$1

Slope of the perpendicular line on the curve x² = 2y,

$${m_2} = {{2x} \over 2} = x \Rightarrow {m_2} = h $$

$$ \therefore $$ Slope of normal = $$ - {1 \over h}$$

$$ - {1 \over h}$$ = $$-$$1 $$ \Rightarrow $$ h = 1

So point is $$\left( {1,{1 \over 2}} \right)$$

$$D = \left| {{{1 - {1 \over 2} - 1} \over {\sqrt {1 + 1} }}} \right| = {1 \over {2\sqrt 2 }}$$