This exercise can be understood as an application of a general result about perimeter bisectors of triangles.
Proposition. Given $\triangle ABC$ with incircle $\bigcirc I$ meeting the edges at $D$, $E$, $F$ as shown. If $F^\prime$ is the point opposite $F$ in $\bigcirc I$, and if $F^{\prime\prime}$ is the point where $\overleftrightarrow{CF^\prime}$ meets $\overline{AB}$, then
$$|\overline{CA}|+|\overline{AF^{\prime\prime}}| = |\overline{CB}|+|\overline{BF^{\prime\prime}}| \tag{$\star$}$$
so that $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$.
Proof of Proposition. Let the perpendicular to $\overline{FF^\prime}$ at $F^\prime$ meet the edges of the triangle at $A^\prime$ and $B^\prime$. By tangent properties of circles, we have
$$\overline{CE}\cong\overline{CD} \qquad \overline{A^\prime E}\cong\overline{A^\prime F^\prime} \qquad \overline{B^\prime D}\cong\overline{B^\prime F^\prime}$$
Consequently, $|\overline{CA^\prime}| + |\overline{A^\prime F^\prime}| = |\overline{CB^\prime}| + |\overline{B^\prime F^\prime}|$, so that $\overline{CF}$ is a perimeter bisector of $\triangle A^\prime B^\prime C$. The Proposition holds by the similarity of $\triangle ABC$ and $\triangle A^\prime B^\prime C$. $\square$
The Proposition has a helpful corollary.
Corollary. Given $\triangle ABC$ with incenter $I$ and perimeter bisector $\overline{CF^{\prime\prime}}$, if $M$ is on $\overline{AB}$ such that $\overline{IM} \parallel \overline{CF^{\prime\prime}}$, then $M$ is the midpoint of $\overline{AB}$.
Proof of Corollary. The points of tangency of the triangle with its incircle separate the perimeter into three pairs of congruent segments, marked $a$, $b$, $c$. Thus, the semi-perimeter of $\triangle ABC$ is $a+b+c$, and since $|\overline{BC}| = b+c$, it follows that $|\overline{BF^{\prime\prime}}| = a = |\overline{AF}|$. Thus, $\overline{FF^{\prime\prime}}$ lies between congruent segments. In $\triangle FF^\prime F^{\prime\prime}$, segment $\overline{IM}$ passes through the midpoint of one side ($\overline{FF^\prime}$) and is parallel to another ($\overline{F^\prime F^{\prime\prime}}$); it necessarily meets the third side ($\overline{FF^{\prime\prime}}$) at its midpoint, which must also be the midpoint of $\overline{AB}$. $\square$
To solve the original problem, it basically suffices to embed the above triangle into an ellipse:
In the above, the ellipse's foci are $C$ and $F^{\prime\prime}$, and $\overline{AB}$ is a chord through the latter. The fundamental nature of ellipses implies that $(\star)$ holds; therefore, $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$. Moreover, the reflection property of ellipses implies that normals at $A$ and $B$ bisect angles $\angle CAF^{\prime\prime}$ and $\angle CBF^{\prime\prime}$; therefore, the intersection of these normals is the incenter of $\triangle ABC$. The result follows by the Corollary. $\square$
Continuing with the same method adopted earlier, we multiply equations $(1)$ and $(2)$ with $\sin(\alpha+\beta)$ and $\sin(\alpha-\beta)$ respectively. Then we subtract the equations, rearrange and simplify to get
$$A-e=\frac{\cos{\alpha}}{\cos{\beta}}\tag{3}$$
We then use this value to obtain the general equation of the chord as
$$\begin{align}\frac{l}{r} &= \left(e+\frac{\cos{\alpha}}{\cos{\beta}}\right)\cdot\cos{\theta}+\frac{\sin{\alpha}}{\cos{\beta}}\cdot\sin{\theta}
\\ &= e\cos{\theta}+\frac{1}{\cos{\beta}}\cdot\left(cos{\alpha}\cdot\cos{\theta}+sin{\alpha}\cdot\sin{\theta
}\right)
\\ &= e\cos{\theta}+\frac{\cos{(\alpha-\theta)}}{\cos{\beta}}
\\ &= e\cos{\theta}+\frac{\cos{(\theta-\alpha)}}{\cos{\beta}} \tag{4}
\end{align}$$
We can go one step further and obtain the equation of a tangent to the conic. For this, we need to set $\beta=0$ which would imply $\cos{\beta}=1$. Thus, the equation of the tangent would be:
$$\frac{l}{r}=e\cos{\theta}+\cos{(\theta-\alpha)}$$
Best Answer
Let $OA$ be a semi-diameter of the ellipse and the tangent at $B$ be parallel to it. It is evident (see figure below) that the area of any other triangle with the same base $OA$ and its third vertex on the ellipse cannot be greater tan the area of triangle $OAB$.
Two semi-diameters like $OA$ and $OB$ are called conjugate semi-diameters and by the second theorem of Apollonius the area of a triangle formed by any two conjugate semi-diameters is always ${1\over2}ab$, where $a$ and $b$ are the semi-axes of the ellipse.
Hence no calculus is needed to answer the given question: maximum area is ${1\over2}ab=4$ and the chord must be drawn so that points $A$ and $B$ are the endpoints of two conjugate semi-diameters. To this end it is useful to know that if $A=(4\cos\alpha,2\sin\alpha)$ then $OB$ is conjugate to $OA$ if $$B=\big(4\cos(\pi/2+\alpha),2\sin(\pi/2+\alpha)\big)=(-4\sin\alpha,2\cos\alpha).$$ The value of $\alpha$ can then be found by imposing line $AB$ to pass through point $P$.