Conic Sections: Challenge #8

The function of line $AB$ can be express by

$\frac{x}{9}+\frac{3y}{4}=1$

Proof will noted at bottom of solution.

Then find the point intersect by line $AB$ and the ellipse,

$A(-\frac{9}{5}, \frac{8}{5})$ , $B(\frac{45}{17},\frac{16}{17})$

So area of the quadrilateral PACB will be

$Area=\frac{1}{2} | \begin{matrix} 0 & -\frac{9}{5} & 1 & \frac{45}{17}\\ 0 & \frac{8}{5} & 3 & \frac{16}{17} \end{matrix} |=7$

Proof:

Let point A,B be $(x_1,y_1)$ and $(x_2,y_2)$ ,

Using tangent line function of ellipse, we know

$l_A : \frac{x_1x}{9}+\frac{y_1y}{4}=1$

$l_B : \frac{x_2x}{9}+\frac{y_2y}{4}=1$

And P be intersect point by this two lines so x=1,y=3.

$l_A : \frac{x_1}{9}+\frac{3y_1}{4}=1$

$l_B : \frac{x_2}{9}+\frac{3y_2}{4}=1$

And so the straight line joins $l_A$ and $l_B$ is

$l_AB : \frac{x}{9}+\frac{3y}{4}=1$