Hyperbola and Ellipse.

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:

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

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

Using $x^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ and the points I chose above to generate the hyperbola.

(1) $(0,-4): 16C - 4E + F = 0 \implies \boxed{F = 4E - 16C}$

(2) $(8,-4): 4B - D = 8 \implies \boxed{D = 4B - 8}$ .

Using (1) and (2) above $\implies$

(3): $(-4,-8): 4B + 12C - E = -12$

(4) $(12,-8): -12B + 12C - E = -12$

(5) $(4,-2): 4B - 6C + E = 8$

(3) + (5) $\implies 4B + 3C =-2$

(3) - (4) $\implies B = 0 \implies C = -\dfrac{2}{3} \implies D = -8 \implies E = 4 \implies F = \dfrac{80}{3}$

$\implies 3x^2 - 2y^2 - 24x + 12y = -80 \implies -3x^2 + 2y^2 + 24x - 12y = 80 \implies$

$2(y - 3)^2 - 3(x - 4)^2 = 50$

Using the given portion of hyperbola below the line $\dfrac{x - 8}{4} \implies$ $y_{h}(x) = 3 - \sqrt{\dfrac{1}{2}(50 + 3(x - 4)^3} \implies A_{h} = \displaystyle\int_{0}^{8} y_{h}(x) dx$ .

For $I(x) = \dfrac{1}{\sqrt{2}}\displaystyle\int \sqrt{50 + 3(x - 4)^2}$

Let $\sqrt{3}(x - 4) = \sqrt{50}\tan(\theta) \implies dx = \sqrt{\dfrac{50}{3}}\sec^2(\theta) d\theta \implies$ $I(\theta) = \dfrac{50}{\sqrt{6}}\displaystyle\int \sec^3(\theta) d\theta$

For $\displaystyle\int \sec^3(\theta) d\theta$

Let $u = \sec(\theta) \implies du = \sec(\theta)\tan(\theta)$ and $dv = \sec^2(\theta) \implies v = \tan(\theta) \implies$ $\displaystyle\int \sec^3(\theta) d\theta = \sec(\theta)\tan(\theta) - \displaystyle\int \sec(\theta) + \displaystyle\int \sec(\theta) d\theta \implies (\displaystyle\int \sec^3(\theta) d\theta = \dfrac{1}{2}(\sec(\theta)\tan(\theta) + \ln|\sec(\theta) + \tan(\theta)|) \implies$

$I(x)|_{0}^{8} = \dfrac{25}{\sqrt{6}}(\dfrac{\sqrt{50 + 3(x - 4)^2}\sqrt{3}(x - 4)}{50} + \ln|\dfrac{\sqrt{50 + 3(x - 4)^2} + \sqrt{3}(x - 4)}{\sqrt{50}}|)|_{0}^{8} = 28 + \dfrac{25}{\sqrt{6}}\ln(\dfrac{7\sqrt{2} + 4\sqrt{3}}{7\sqrt{2} - 4\sqrt{3}})$

$\implies A_{h} = -4 - \dfrac{25}{\sqrt{6}}\ln(\dfrac{7\sqrt{2} + 4\sqrt{3}}{7\sqrt{2} - 4\sqrt{3}}) \approx \boxed{-21.69786}$

The area of the ellipse is $A_{e} = \displaystyle\int_{0}^{8} y_{e}(x) dx$ .

$A_{e} = \dfrac{1}{20}\sqrt{\dfrac{5}{347}}\displaystyle\int_{0}^{8} -\sqrt{2207744 - (347x - 1448)^{2}} + \dfrac{x - 8}{4} dx =$ $\dfrac{1}{4\sqrt{5}\sqrt{347}}\displaystyle\int_{0}^{8} -\sqrt{2207744 - (347x - 1448)^2} + \dfrac{x - 8}{4} dx$ .

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

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

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

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

$A_{e} = -\dfrac{275968}{(347)^{\frac{3}{2}}\sqrt{5}}(\arcsin(\dfrac{347x - 1448}{\sqrt{2207744}}) + \dfrac{347x - 1448}{2207744}\sqrt{2207744 - (347x - 1448)^2})|_{0}^{8} =$

$-\dfrac{275968}{(347)^{\frac{3}{2}}\sqrt{5}}(\arcsin(\dfrac{1328}{2207744}) + \arcsin(\dfrac{1448}{2207744}) + \dfrac{1328\sqrt{444160}}{2207744}$ + $\dfrac{1448\sqrt{111040}}{2207744}) - 8 =$

$-\dfrac{275968}{(347)^{\frac{3}{2}}\sqrt{5}}(\arcsin(\dfrac{83}{28\sqrt{11}}) + \arcsin(\dfrac{181}{56\sqrt{11}}) + \dfrac{513}{34496}\sqrt{1735}) - 8 \approx \boxed{-66.91295}$

$\implies A = \displaystyle\int_{0}^{8} y_{h}(x) - y_{e}(x) dx = A_{h} - A_{e} = \boxed{44.91295}$ .