JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Application of Derivatives

Differentiating $y=2x^2+x+2$ gives slope $4x+1$, so the normal at a point $P(h,2h^2+h+2)$ has slope $-\frac{1}{4h+1}$. Requiring this normal to pass through $Q(6,4)$ and substituting the coordinates of $P$ leads to $4h^3+3h^2-3h-4=0$, whose only real root is $h=1$; the quadratic factor $4h^2+7h+4$ has negative discriminant. Hence $P=(1,5)$. With $O$ at the origin, the area of triangle $OPQ$ is $\frac{1}{2}|x_P{y}_Q-x_Q{y}_P|=\frac{1}{2}|1\cdot 4-6\cdot 5|=\frac{1}{2}\cdot 26=13$.