Finding triangle of maximal area given two vertices and condition on internal bi-sector 
$A=(2,5), B=(5,11)$ and a point $P$ moves such that internal bi-sector of $\angle APB$ passes through $(4,9).$ The maximum area of $ \triangle APB\;$ is __?

My attempt:
I found that $(4,9)$ lies on the line segment of $AB.$ Then I thought that due to the internal bi-sector condition it must have something to do with ellipse. Because, the two triangles formed by connecting focii of ellipse of perimeter are a congruent. So, if $P$ is a point on perimeter and $Q= (4,9),$ then  $\angle APQ= \angle BPQ.$
I initially thought the optimum would happen when $ \angle ABP = \angle BAP =45^{\circ},$ but seems like that leads to a situation where the constraint is not obeyed.
So, how exactly do I get the optimum area while holding the constraint true?



Since the point $D=(4,9)$ belongs to $AB$,
it is a foot of the bisector of $\angle ACB$,
and a known expression for it is
\begin{align}
D&=\frac{aA+bB}{a+b}
\tag{1}\label{1}
.
\end{align}
With known coordinates we have
\begin{align}
(4,9)&=
\left(\frac{2a+5b}{a+b},\, \frac{5a+11b}{a+b} \right)
\tag{2}\label{2}
\end{align}
or a system
\begin{align} 
\frac{2a+5b}{a+b}&=4
\tag{3}\label{3}
,\\
\frac{5a+11b}{a+b}&=9
\tag{4}\label{4}
\end{align}
with solution $b=2a$.
Expression for the area of $\triangle ABC$ squared is
\begin{align} 
S_2(a,b,c)&=
\tfrac1{16}\,(4a^2b^2-(a^2+b^2-c^2)^2)
\tag{5}\label{5}
,
\end{align}
so for $b=2a$ we must have
\begin{align} 
S_2(a,2a,c)&=
-\tfrac9{16}\,(a^2)^2+\tfrac58\,c^2\,a^2-\tfrac1{16}\,c^4
\tag{6}\label{6}
.
\end{align}
It's easy to find that
\begin{align} 
\max_a S_2(a,2a,c)&=
S_2(a,2a,c)\Big|_{a=5}
=225
,
\end{align}
hence
the maximal area of such triangle is $15$.

