## Task B-4.5.

The ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{16}=1$ and the parabola $y^{2}=2 p x$ intersect at the point $T(4 \sqrt{3}, 2)$. What is the area of the triangle $T F_{E} F_{P}$, if $F_{E}$ is the focus of the ellipse located on the positive part of the $x$-axis, and $F_{P}$ is the focus of the parabola?
## Solution.

First, determine the unknown quantities for the ellipse:

$$
\begin{array}{r}
\frac{(4 \sqrt{3})^{2}}{a^{2}}+\frac{4}{16}=1, \quad a^{2}=64 \\
e=\sqrt{64-16}=4 \sqrt{3}, \quad F_{E}(4 \sqrt{3}, 0)
\end{array}
$$

Then for the parabola:

$$
4=2 p \cdot 4 \sqrt{3}, \quad p=\frac{\sqrt{3}}{6}, \quad F_{P}\left(\frac{\sqrt{3}}{12}, 0\right)
$$

The area of the triangle is

$$
P=\frac{4 \sqrt{3}-\frac{\sqrt{3}}{12}}{2} \cdot 2=\frac{47 \sqrt{3}}{12}
$$