QJEEM21Ma10d0: JEE Main 2021 Mathematics - Class 11 JEEM21Ma10d0 Solution: The shortest distance between the line x y = 1 and the curve x…

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

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