Question

Let l be a line which is normal to the curve $y = 2x^2 + x \+ 2$ 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 OPQ is equal to _________ .

Answer 1

$\frac {y_1 - 4}{ x_1 - 6} = - \frac {1}{4x_1+1}$

$⇒\frac { 2x^2_1 + x_1 - 2}{x_1 - 6} = - \frac {1}{4x_1+1}$
$= 6 - x_1 = 8x_1^3 + 6x_1^2 - 7x_1 - 2$
$⇒ 8x_1^3 \+ 6x_1^2 – 6x_1 – 8 = 0$
So, $x_1 = 1 ⇒y_1 = 5$
Area = $\frac 12 \begin{vmatrix} 0 & 0 & 1 \\\[0.3em] 6 & 4 & 1 \\\[0.3em] 1 & 5 & 1 \end{vmatrix}$
$= 13$

Hence, the answer is $13$.