Let $l$ 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 $l$ and $O$ is origin, then the area of the triangle $O P Q$ is equal to

13.00

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Solution

$\frac{y_{1} - 4}{x_{1} - 6} = - \frac{1}{4 x_{1} + 1}$
$\Rightarrow \frac{2 x_{1}^{2} + x_{1} - 2}{x_{1} - 6} = - \frac{1}{4 x_{1} + 1}$
$\Rightarrow 6 - x_{1} = 8 x_{1}^{3} + 6 x_{1}^{2} - 7 x_{1} - 2$
$\Rightarrow 8 x_{1}^{3} + 6 x_{1}^{2} - 6 x_{1} - 8 = 0$
So, $x_{1} = 1 \Rightarrow y_{1} = 5$

Area $= ∣ ∣ ∣ ∣ \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 6 & 4 & 1 1 & 5 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 13$

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