More Conics!

For the conic:

$(5,4): 25a + 100b + 16c + 5d + 4e = -288$

$(-5,4): 25a - 100b + 16c - 5d + 4e = -288$

$(3,0): 9a + 3d = -288$

$(-3,0): 9a - 3d = -288$

$(0,-4): 4c - e = -72$

$\implies d = b = 0, a = -32, c = 7$ and $e = 100$

$\implies -32x^2 + 7y^2 + 100y = -288$ which is a hyperbola since $b^{2} - 4ac < 0$ .

Solving for $y$ we obtain:

$y = \dfrac{2}{7}(\pm\sqrt{121 + 56x^2} - 25)$ .

For region $R_{1}:$ $y_{1}(x) = \dfrac{2}{7}(\sqrt{121 + 56x^2} - 25)$ , where $y \geq -4$ .

For region $R_{2}:$ $y^{*}_{1}(x) = \dfrac{2}{7}(-\sqrt{121 + 56x^2} - 25)$ , where $y \leq -\dfrac{12}{7}$ .

For circle:

(1) $(5,4): 25 - 10a + a^2 + 16 - 8b + b^2 = r^2$

(2) $(5,4): 25 + 10a + a^2 + 16 + 8b + b^2 = r^2$

Subtracting (2) from (1) $\implies a = 0$ .

(3) $(-10,-5): 125 + 10b + b^2 = r^2$

(1) $41 - 8b + b^2 = r^2$

Subtracting (1) from (3) $\implies b = -\dfrac{14}{3}$

$\implies r^2 = \dfrac{901}{9} \implies x^2 + (y + \dfrac{14}{3})^2 = \dfrac{901}{9}$

Solving for $y$ we obtain:

$y = \dfrac{1}{3}(\pm\sqrt{901 - 9x^2} - 14)$

For region $R_{1}:$ $y_{2}(x) = \dfrac{1}{3}(\sqrt{901 - 9x^2} - 14)$ above the line $y = -\dfrac{14}{3}$ .

For region $R_{2}:$ $y^{*}_{2}(x) = \dfrac{1}{3}(-\sqrt{901 - 9x^2} - 14)$ below the line $y = -\dfrac{14}{3}$ .

For region $R_{1}:$ $A_{1} = \displaystyle\int_{-5}^{5} y_{2}(x) - y_{1}(x) dx$ .

Let $I_{1} = \displaystyle\int_{-5}^{5} y_{2}(x)$

For $\dfrac{1}{3}\displaystyle\int_{-5}^{5} \sqrt{901 - 9x^2} dx$ :

Let $3x = \sqrt{901}\sin(\theta) \implies dx = \dfrac{\sqrt{901}}{3}\cos(\theta) d\theta \implies$

$\displaystyle\int_{-5}^{5} \sqrt{901 - 9x^2} dx = \dfrac{901}{9}(\arcsin(\dfrac{3x}{\sqrt{901}}) + \dfrac{3x\sqrt{901 - 9x^2}}{901})|_{-5}^{5} = \dfrac{901}{9}\arcsin(\dfrac{15}{\sqrt{901}}) + \dfrac{130}{3}$

$\implies I_{1} = \displaystyle\int_{-5}^{5} y^{2}(x) dx = \dfrac{901}{9}\arcsin(\dfrac{15}{\sqrt{901}}) - \dfrac{10}{3} \approx \boxed{49.0526409}$ .

Let $I_{2} = \displaystyle\int_{-5}^{5} y_{1}(x) dx$ .

For $\dfrac{2}{7}\displaystyle\int_{-5}^{5} \sqrt{121 + 56x^2} dx$

Let $\sqrt{56}x = 11\tan(\theta) \implies dx = \dfrac{11}{\sqrt{56}}\sec^2(\theta) \implies$

$\dfrac{2}{7}\displaystyle\int \sqrt{121 + 56x^2} dx = \dfrac{242}{7\sqrt{56}}\displaystyle\int \sec^3(\theta) d\theta$ .

Let $u = \sec(\theta), du = \sec(\theta)\tan(\theta) d\theta, dv = \sec^2(\theta) d\theta, v = \tan(\theta) \implies$

$\displaystyle\int \sec^2(\theta) d\theta = \dfrac{1}{2}(\sec(\theta)\tan(\theta) + \ln|\sec(\theta) + \tan(\theta))|)$

$\implies \dfrac{2}{7}\displaystyle\int_{-5}^{5} \sqrt{121 + 56x^2} dx = \dfrac{121}{7\sqrt{56}}(\dfrac{56x\sqrt{121 + 56x^2}}{121} + \ln(\dfrac{\sqrt{121 + 56x^2} + \sqrt{56}x}{11}))|_{-5}^{5} =$

$\dfrac{390}{7} + \dfrac{121}{7\sqrt{56}}\ln(\dfrac{39 + 5\sqrt{56}}{39 - 5\sqrt{56}})$

$\implies I_{2} = \displaystyle\int_{-5}^{5} y_{1}(x) dx = \dfrac{121}{7\sqrt{56}}\ln(\dfrac{39 + 5\sqrt{56}}{39 - 5\sqrt{56}}) - \dfrac{110}{7} \approx \boxed{-6.7597045}$

$\implies A_{1} = \boxed{55.8123454}$

For region $R_{2}:$

To find the $x$ points of intersection solve:

$\dfrac{1}{3}(\sqrt{901 - 9x^2} + 14) = \dfrac{2}{7}(\sqrt{121 + 56x^2} + 25) \implies$

$7\sqrt{901 - 9x^2} = 6\sqrt{121 + 56x^2} + 52 \implies 44149 - 441x^2 = 2016x^2 + 624\sqrt{121 + 56x^2}$

$\implies 37089 - 2457x^2 = 624\sqrt{121 + 56x^2} \implies 6036849x^4 - 182255346x^2 + 37089^2 = 21805056x^2 + 47114496 \implies$

$6036849x^4 - 20406040x^2 + 1328479425 = 0$

Let $w = x^2 \implies w^2 = x^4 \implies 6036849w^2 - 20406040w + 1328479425 = 0 \implies$ $w^2 - \dfrac{2738}{81}w + \dfrac{17825}{81} = 0 \implies u = 25, u = \dfrac{713}{81} \implies$ $x = \pm 5, x = \pm \dfrac{\sqrt{713}}{9}$

$x = \pm 5$ does not satisfy the above equation $\implies x = \pm \dfrac{\sqrt{713}}{9}$ .

$A_{2} = \displaystyle\int_{-\frac{\sqrt{713}}{9}}^{\frac{\sqrt{713}}{9}} y^{*}_{1}(x) - y^{*}_{2}(x) \:\ dx$ .

Let $I^{*}_{1} = \displaystyle\int_{-\frac{\sqrt{713}}{9}}^{\frac{\sqrt{713}}{9}} y^{*}_{1}(x) dx$

Using the antiderivative above for $y_{1}(x) \implies$

$I^{*}_{1} = -\dfrac{121}{7\sqrt{56}}(\dfrac{56x\sqrt{121 + 56x^2}}{121} + \ln(\dfrac{\sqrt{121 + 56x^2} + \sqrt{56}x}{11})) - \dfrac{50}{7}|_{-\frac{\sqrt{713}}{9}}^{\frac{\sqrt{713}}{9}} =$ $-\dfrac{1346\sqrt{713}}{567} - \dfrac{121}{7\sqrt{56}}\ln(\dfrac{223 + 2\sqrt{9982}}{223 - 2\sqrt{9982}}) \approx \boxed{-70.0951083}$

Let $I^{*}_{2} = \displaystyle\int_{-\frac{\sqrt{713}}{9}}^{\frac{\sqrt{713}}{9}} y^{*}_{2}(x) dx$

Using the antiderivative above for $y_{2}(x) \implies$

$I^{*}_{2} = -(\dfrac{901}{18}\arcsin(\dfrac{3x}{\sqrt{901}}) + \dfrac{x}{6}\sqrt{901 - 9x^2}) - \dfrac{14}{3}x|_{-\frac{\sqrt{713}}{9}}^{\frac{\sqrt{713}}{9}}$

$= -\dfrac{170}{81}\sqrt{713} - \dfrac{901}{9}\arcsin(\dfrac{\sqrt{713}}{3\sqrt{901}}) \approx$ $\boxed{-86.179990}$

$\implies A_{2} = \boxed{16.0848818}$

$\implies A_{1} + A_{2} = \boxed{71.8972272}$