Question

Let $\ell$ be a line which is normal to the curve $y =2 x ^{2}+ x +2$ at a point $P$ on the curve. If the point $Q (6,4)$ lies on the line $\ell$ and $O$ is origin, then the area of the triangle $OPQ$ is equal to_______

Answer 1

$y=2 x^{2}+x+2$

$\frac{ dy }{ dx }=4 x +1$
Let $P$ be $( h , k )$, then normal at $P$ is
$y - k =-\frac{1}{4 h +1}( x - h )$
This passes through $Q (6,4)$
$\therefore 4- k =-\frac{1}{4 h +1}(6- h ) $
$\Rightarrow(4 h +1)(4- k )+6- h =0$
Also $k =2 h ^{2}+ h +2$
$\therefore(4 h +1)\left(4-2 h ^{2}- h -2\right)+6+ h =0 $
$\Rightarrow 4 h ^{3}-3 h ^{2}+3 h -8=0$
$\Rightarrow h =1, k =5$
Now area of $\triangle OPQ$ will be $=\frac{1}{2} \begin{vmatrix}1 & 0 & 0 \\ 1 & 1 & 5 \\ 1 & 6 & 4\end{vmatrix}=13$