Focal triangle in an ellipse

Note that $F_1,F_2$ have coordinates $(-\sqrt{3},0)$ and $(\sqrt{3},0)$ respectively. If $P$ has coordinates $(2\cos t,\sin t)$ , the $A,B$ have coordinates $A \; : \; \left(-2\frac{7\cos t + 4\sqrt{3}}{7 + 4\sqrt{3}\cos t}, - \frac{\sin t}{7 + 4\sqrt{3}\cos t}\right) \hspace{1cm} B \; : \; \left(-2\frac{7\cos t - 4\sqrt{3}}{7 - 4\sqrt{3}\cos t}, - \frac{\sin t}{7 - 4\sqrt{3}\cos t}\right)$ (solving simultaneously the equations of $PF_1$ and the ellipse, and of $PF_2$ and the ellipse), and hence $PAB$ has area $\Delta(t) \; = \; \frac{8\sqrt{3}\sin t(5 - 3\cos2t)}{25 - 24\cos2t}$ using the fact that the triangle with vertices $(x_1,y_1)$ , $(x_2,y_2)$ , $(x_3,y_3)$ has area equal to half the modulus of the determinant $\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right|$ Elementary calculus shows that $\Delta(t)$ is maximised when $t = \tfrac12\pi$ , and $\Delta(\tfrac12\pi) = \tfrac{64}{49}\sqrt{3} = \boxed{2.262270443}$

Solving equations and points given, we obtain three points that make up the triangle, $(\dfrac{-24}{7\sqrt3}, \dfrac{-1}{7}), (\dfrac{24}{7\sqrt3}, \dfrac{-1}{7}), (0,1)$ , and then find the sides of the triangle by distance formula, to use heron's formula for the area. See the graph and the equations obtained here