Ellipse Mania!

First we set a coordinate system as depicted in the figure: As we can see here the equation of the conic containing five points ${{\left( {{x}_{i}},{{y}_{i}} \right)}_{i=1,2,3,4,5}}$ , is

$\det \left( \begin{matrix} {{x}^{2}} & xy & {{y}^{2}} & x & y & 1 \\ {{x}_{1}}^{2} & {{x}_{1}}{{y}_{1}} & {{y}_{1}}^{2} & {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}}^{2} & {{x}_{2}}{{y}_{2}} & {{y}_{2}}^{2} & {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}}^{2} & {{x}_{3}}{{y}_{3}} & {{y}_{3}}^{2} & {{x}_{3}} & {{y}_{3}} & 1 \\ {{x}_{4}}^{2} & {{x}_{4}}{{y}_{4}} & {{y}_{4}}^{2} & {{x}_{4}} & {{y}_{4}} & 1 \\ {{x}_{5}}^{2} & {{x}_{5}}{{y}_{5}} & {{y}_{5}}^{2} & {{x}_{5}} & {{y}_{5}} & 1 \\ \end{matrix} \right)=0$ In our case, using the points $A$ , $B$ , $C$ , $D$ and $F$ , the equation of the ellipse is

$\det \left( \begin{matrix} {{x}^{2}} & xy & {{y}^{2}} & x & y & 1 \\ 25 & 0 & 0 & -5 & 0 & 1 \\ 9 & 0 & 0 & 3 & 0 & 1 \\ 16 & 0 & 0 & -4 & 0 & 1 \\ 0 & 0 & 144 & 0 & 12 & 1 \\ 9 & 24 & 64 & 3 & 8 & 1 \\ \end{matrix} \right)=0\Leftrightarrow 44{{x}^{2}}-5xy+10{{y}^{2}}+88x-65y-660=0$ The later is of the form
$a{{x}^{2}}+2bxy+c{{y}^{2}}+2dx+2fy+g=0$ with $a=44 \ \ \ b=\dfrac{-5}{2} \ \ \ c=10 \ \ \ d=44 \ \ \ f=\dfrac{-65}{2} \ \ \ g=-660$

The corresponding formulae for the semi-axes ${a}',{b}'$ of the ellipse are ${a}',{b}'=\sqrt{\dfrac{2\left( a{{f}^{2}}+c{{d}^{2}}+g{{b}^{2}}-2bdf-acg \right)}{\pm \left( {{b}^{2}}-ac \right)\left[ \sqrt{{{\left( a-c \right)}^{2}}+4{{b}^{2}}}-\left( a+c \right) \right]}}$ Taking their product and applying the numerical values we have $\begin{aligned} {a}'{b}' & =\dfrac{a{{f}^{2}}+c{{d}^{2}}+g{{b}^{2}}-2bdf-acg}{{{\left( ac-{{b}^{2}} \right)}^{\frac{3}{2}}}} \\ & =\dfrac{44\times {{\left( -\dfrac{65}{2} \right)}^{2}}+10\times {{44}^{2}}+\left( -660 \right){{\left( -\dfrac{5}{2} \right)}^{2}}-2\times \left( -\dfrac{5}{2} \right)\times 44\times \left( -\dfrac{65}{2} \right)-44\times 10\times \left( -660 \right)}{{{\left( 44\times 10-{{\left( -\dfrac{5}{2} \right)}^{2}} \right)}^{\frac{3}{2}}}} \\ & =\dfrac{551936}{347\sqrt{1735}} \\ & =\dfrac{551936}{347\sqrt{347\times 5}} \\ & =\dfrac{551936}{{{347}^{\frac{3}{2}}}\sqrt{5}} \\ \end{aligned}$ Consequently, the area of the ellipse is

$A={a}'{b}'\pi =\dfrac{551936\pi }{{{347}^{\frac{3}{2}}}\sqrt{5}}$ For the answer, $\alpha =551936$ , $\beta =347$ , $\lambda=5$ , thus, $\alpha+\beta+\lambda=\boxed{552288}$ .

Using $ax^2 + bxy + cy^2 + dx + ey = 0$ and the points I chose above to generate the ellipse.

(1) $(0,-4): 4c - e = 0 \implies \boxed{c = \dfrac{e}{4}}$

(2) $(8,4): 16a + 8b + 4c + 2d + e = 0$

(3) $(8,-4): 16a - 8b + 4c + 2d - e = 0$

Subtracting (3) from (2) we obtain: $8b + e = 0 \implies \boxed{b = -\dfrac{e}{8}}$

(4) $(5,8): 25a + 40b + 64c + 5d + 8c = 0$

Replacing $c = \dfrac{e}{4}$ and $b = -\dfrac{e}{8}$ into (2) and (4) $\implies$

$16a + 2d = -e$

$25a + 5d = -19e$

$\implies \boxed{a = \dfrac{11e}{10}}$ and $\boxed{d = -\dfrac{93e}{10}}$

$\implies \dfrac{11}{10}x^2 - \dfrac{1}{8}xy + \dfrac{1}{4}y^2 - \dfrac{93}{10}x + y = 0 \implies$ $44x^2 - 5xy + 10y^2 - 372x + 40y = 0 \implies 10y^2 + 5(x - 8)y + 44x^2 - 372x = 0$

Solving for $y$ we obtain:

$y = \dfrac{x - 8}{4} \pm \dfrac{1}{20}\sqrt{\dfrac{5}{347}}\sqrt{2207744 - (347x - 1448)^{2}}$ .

$y(x) = \dfrac{x - 8}{4} +\dfrac{1}{20}\sqrt{\dfrac{5}{347}}\sqrt{2207744 - (347x - 1448)^2}$ for the portion of the ellipse above the line $y = \dfrac{x - 8}{4}$ .

Setting $2207744 - (347x - 1448)^2 = 0 \implies x = \dfrac{1448 \pm 448\sqrt{11}}{347}$ are the points of intersection of the ellipse and the line $\dfrac{x - 8}{4}$ .

Letting $a = \dfrac{1448 - 448\sqrt{11}}{347}$ and $b =\dfrac{1448 + 448\sqrt{11}}{347}$ the area of the ellipse is $A = 2\displaystyle\int_{a}^{b} y(x) - \dfrac{x - 8}{4} \:\ dx =$
$\dfrac{1}{10}\sqrt{\dfrac{5}{347}}\displaystyle\int_{a}^{b} \sqrt{2207744 - (347x - 1448)^{2}} dx =$ $\dfrac{1}{2\sqrt{5}\sqrt{347}}\displaystyle\int_{a}^{b} \sqrt{2207744 - (347x - 1448)^2} dx$ .

For $I(x) = \displaystyle\int \sqrt{2207744 - (347x - 1448)^2} dx$

Let $347x - 1448 = \sqrt{2207744}\sin(\theta) \implies dx = \dfrac{\sqrt{2207744}}{347}\cos(\theta)$

$\implies I(\theta) = \dfrac{2207744}{347}\displaystyle\int \cos^2(\theta) d\theta = \dfrac{2207744}{694}\displaystyle\int (1 + \cos(2\theta)) d\theta = \dfrac{2207744}{694}(\theta +\sin(\theta)\cos(\theta))$

$\implies I(x) = \dfrac{2207744}{694}(\arcsin(\dfrac{347x - 1448}{\sqrt{2207744}}) + \dfrac{347x - 1448}{2207744}\sqrt{2207744 - (347x - 1448)^2}) \implies$

$A = \dfrac{551936}{(347)^{\frac{3}{2}}\sqrt{5}}(\arcsin(\dfrac{347x - 1448}{\sqrt{2207744}}) + \dfrac{347x - 1448}{2207744}\sqrt{2207744 - (347x - 1448)^2})|_{a}^{b} = \dfrac{1103872}{(347)^{\frac{3}{2}}\sqrt{5}}(\dfrac{\pi}{2}) =$

$\dfrac{551936\pi}{(347)^{\frac{3}{2}}\sqrt{5}} = \dfrac{\alpha\pi}{\beta^{\frac{3}{2}}\sqrt{\lambda}}$ $\implies \alpha + \beta + \lambda = \boxed{552288}$ .