The maximum area of OAPF

$O = (0,0)\;,\quad F = (0,1)\;,\quad A = (3,0)\newline \text{Let the parametric point on the ellipse be } P(3\,cos(\theta), \sqrt{10}\,sin(\theta))\newline \text{Area of }\triangle OAF \text{ is constant and can be calculated as :}$

$\newline \large area(\triangle OAF) = \frac{1}{2}\cdot OA\cdot OF = \frac{1}{2}\cdot3\cdot1 = \frac{3}{2} \hspace{20pt}\dots\;\textbf{Eq. 1}$

$\newline \text{Now the area of quadrilateral }OAPF\text{ will be maximum when the area of }\triangle PAF\text{ will be maximum. Since the length }FA \text{ is constant, we have to}\newline\text{find the point }P\text{ such that }P\text{ is at maximum distance from }FA.\newline \text{Equation of line }FA \text{:}\newline\hspace{100pt} \large\frac{y-1}{x-0} = \frac{0-1}{3-0}$ $\newline\Rightarrow x + 3y - 3 = 0\newline \text{Distance of point }P\text{ from the line }FA\text{ can be calculated as: }$

$PD = \Large\frac{|3cos(\theta) + 3\sqrt{10}sin(\theta) - 3|}{\sqrt{1^2+3^2}} = \large\frac{3}{\sqrt{10}}\cdot|cos(\theta) + \sqrt{10}sin(\theta) - 1|$

$\newline\large\Rightarrow PD = \frac{3}{\sqrt{10}}\cdot\mid\sqrt{11}\Big(\frac{1}{\sqrt{11}}cos(\theta) + \frac{\sqrt{10}}{\sqrt{11}}sin(\theta)\Big) - 1\mid$

$\large\Rightarrow PD = \frac{3}{\sqrt{10}}\cdot\mid\sqrt{11}sin(\theta + \alpha) - 1\mid\quad\quad \text{where }cos(\alpha) = \frac{\sqrt{10}}{\sqrt{11}}\Rightarrow \alpha = 17.55\degree$

$\text{The value }(\theta + \alpha)\text{ can vary from }\alpha \text{ to }90\degree + \alpha\text{. Hence, the value of }sin(\theta + \alpha) \text{ have maximum value at }\theta + \alpha = 90\degree\text{. So, }$

$PD_{max} = \large\frac{3}{\sqrt{10}}\cdot\mid\sqrt{11} - 1\mid = \frac{3}{\sqrt{10}}\cdot(\sqrt{11} - 1)$

$\large area(\triangle PAF) = \frac{1}{2}\cdot AF\cdot PD_{max} = \frac{1}{2}\cdot \sqrt{10}\cdot \frac{3}{\sqrt{10}}\cdot(\sqrt{11} - 1) = \frac{3}{2}(\sqrt{11} - 1)\hspace{20pt}\dots\;\textbf{Eq. 2}$

$\text{Adding }\textbf{Eq. 1}\text{ and }\textbf{Eq. 2} \text{ we get }$

$\large area(OAPF) = A = \frac{3\sqrt{11}}{2}\Rightarrow 10000A = 15000\sqrt{11} = 49749.37 \Rightarrow \lfloor10000A\rfloor = \boxed{49749}$

Let the $x$ -coordinate of $P$ be $p$ . Since $\frac{x^2}{9} + \frac{y^2}{10} = 1$ , the $y$ -coordinate of $P$ in the first quadrant is $\sqrt{10 - \frac{10}{9}p^2}$ , so that P is ( $p$ , $\sqrt{10 - \frac{10}{9}p^2}$ ). Since $\overrightarrow{FA} = (3, -1)$ and $\overrightarrow{FP} = (p, \sqrt{10 - \frac{10}{9}p^2} - 1)$ , $\triangle FAP$ has an area of $\frac{1}{2} \begin{bmatrix} 3 & -1 \\ p & \sqrt{10 - \frac{10}{9}p^2} - 1 \end{bmatrix} = \frac{3}{2}\sqrt{10 - \frac{10}{9}p^2} - \frac{3}{2} + \frac{1}{2}p$ .

Since $\triangle FAO$ has an area of $\frac{1}{2} \cdot 3 \cdot 1 = \frac{3}{2}$ , quadrilateral $OAPF$ has an area of $A = \frac{3}{2}\sqrt{10 - \frac{10}{9}p^2} - \frac{3}{2} + \frac{1}{2}p + \frac{3}{2} = \frac{3}{2}\sqrt{10 - \frac{10}{9}p^2} + \frac{1}{2}p$ .

The derivative $A' = -\frac{5p}{3\sqrt{10 - \frac{10}{9}p^2}} + \frac{1}{2}$ , which is a decreasing function and therefore has a maximum when $A' = 0$ , which is at $p = \frac{3\sqrt{11}}{11}$ , so that $A$ has a maximum value of $A = \frac{3}{2}\sqrt{10 - \frac{10}{9}(\frac{3\sqrt{11}}{11})^2} + \frac{1}{2}(\frac{3\sqrt{11}}{11}) = \frac{3\sqrt{11}}{2}$ . Therefore, $\lfloor 10000A \rfloor = \boxed{49749}$ .