Question bank/JEE Main/Mathematics

Find the shortest distance between line y=x-2 & the parabola
y=x^2+3x+2.

A$\frac{5}{\sqrt{2}}$B$\frac{3}{\sqrt{2}}$C$\frac{4}{\sqrt{2}}$D$\frac{6}{\sqrt{2}}$

Show answer and solution▾

$\left( \frac{dy}{dx} \right)_{f} = 1\det{P\left( x_{1},y_{1} \right)}$
$P\left( X_{1},y_{1} \right)\$ij $\left( \frac{dy}{dx} \right)_{f} = 1$
$x_{1} = - 1$
$y_{1} = 0$
$P( - 1,0)$
Shortest distance = distance of (-1,0) from the line y=x - 2
U;wure nwjh (-1,0)

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx