Area between Conic and a Curve

To show the conic is a hyperbola.

Using the point $(1,\dfrac{4}{5})$ we obtain (1): $a + \dfrac{16}{25}b + c + \dfrac{4}{5}d = -1$

Using the point $(0,\dfrac{-4}{5})$ we obtain (2): $\dfrac{16}{25}b - \dfrac{4}{5}d = -1$

Using the point $(\dfrac{\sqrt{5} - 1}{2},0)$ we obtain (3): $(\dfrac{\sqrt{5} -1}{2})^2a - \dfrac{\sqrt{5} - 1}{2}c = -1$

Using the point $(-\dfrac{\sqrt{5} - 1}{2},0)$ we obtain (4): $(\dfrac{\sqrt{5} -1}{2})^2a - \dfrac{\sqrt{5} - 1}{2}c = -1$

Using the point $(-1,\dfrac{4}{5})$ we obtain (5): $a + \dfrac{16}{25}b - c + \dfrac{4}{5}d = -1$

(3) and (4) $\implies a = -(\dfrac{2}{\sqrt{5} - 1})^2$ and $c = 0 \implies$

$\dfrac{16}{25}b + \dfrac{4}{5}d = \dfrac{2}{\sqrt{5} - 1}$

$\dfrac{16}{25}b - \dfrac{4}{5}d = -1$

$\implies b = \dfrac{25}{32}(\dfrac{3 - \sqrt{5}}{\sqrt{5} - 1}) = \dfrac{25}{32}(\dfrac{\sqrt{5} - 1}{2})$ and $d = \dfrac{5}{8}(\dfrac{2}{\sqrt{5} - 1})^2$

Letting $x_{0} = \dfrac{\sqrt{5} - 1}{2} \implies a = \dfrac{-1}{x_{0}^2}, b = \dfrac{25}{32}x_{0}, c = 0$ and $d = \dfrac{5}{8}(\dfrac{1}{x_{0}^2}) \implies$

$25x_{0}^3y^2 - 32x^2 + 20y = -32x_{0}^2 \implies$ $\boxed{25x_{0}^3(y + \dfrac{2}{5x_{0}^3})^2 - 32x^2 = \dfrac{4 - 32x_{0}^5}{x^3}}$ which is a hyperbola.

To find the desired area we need to solve for $y = g(x)$ .

Solving for $y$ we obtain: $y = (\dfrac{2}{5x_{0}^3})(\sqrt{8x_{0}^3x^2 + 1 - 8x_{0}^5} - 1)$ , where $y >= \dfrac{-4}{5}$ .

Using the symmetry about the $y$ axis let $y = g(x)$ and $f(x) = \dfrac{4}{5}(\sqrt{x} - \sqrt{1 - x^2})$ .

Let $A_{1} = \int_{x_{0}}^{1} g(x) - f(x) dx$ and $A_{2} = \int_{0}^{x_{0}} f(x) - g(x) dx$ so that the desired area $A = 2(A_{1} + A_{2})$ .

For $A_{1}$ :

For $A^{*} = \int_{x_{0}}^{1} g(x)$ :

Let $\sqrt{8x_{0}^3} x = \sqrt{1 - 8x_{0}^5}\tan(\theta) \implies dx = \sqrt{\dfrac{1 - 8x_{0}^5}{8x_{0}^3}}$

$\implies A^{*} = \int_{x_{0}}^{1} g(x) dx =$ $\dfrac{2}{5x_{0}^3} * (\dfrac{1 - 8x_{0}^5}{\sqrt{8x_{0}^3}}\int_{\theta_{1}}^{\theta_{2}} \sec^3(\theta) d\theta - \int_{x_{0}}^{1} dx)$

Using integration by parts and evaluating the integral and simplifying we obtain:

$A^{*} = \dfrac{2}{5x_{0}^3}(\dfrac{\sqrt{8x_{0}^3 + 1 - 8x_{0}^5}}{2} + \dfrac{1}{2}x_{0} - 1 + (\dfrac{1 - 8x_{0}^5}{2\sqrt{8x_{0}^3}}) * \ln|\dfrac{\sqrt{8x_{0}^3 + 1 - 8x_{0}^5} + \sqrt{8x_{0}^3}}{1 + \sqrt{8x_{0}^5}}|)$

For $A^{**} = \dfrac{4}{5} \int_{x_{0}}^{1} (\sqrt{1 - x^2} - \sqrt{x}) dx$ Let $x = \sin(\theta) \implies dx = \cos(\theta) \implies$

$A^{**} = \dfrac{4}{5}(\dfrac{\pi}{4} - \dfrac{1}{2}\arcsin(x_{0}) - \dfrac{1}{2}x_{0}\sqrt{1 - x_{0}^2} + \dfrac{2}{3}x_{0}^{\dfrac{3}{2}} - \dfrac{2}{3})$

$A_{1} = A^{*} + A^{**} \approx \boxed{.043741}$

Similarly For $A_{2} = = \int_{0}^{x_{0}} f(x) - g(x) dx$ we obtain:

$A_{2} = \dfrac{4}{5}(\dfrac{2}{3}x_{0}^{\dfrac{3}{2}} - \dfrac{1}{2}\arcsin(x_{0}) - \dfrac{1}{2}x_{0}\sqrt{1 - x^2}) - \dfrac{2}{5x_{0}^3}(\dfrac{-x_{0}}{2} + (\dfrac{1 - 8x_{0}^5}{2\sqrt{8x_{0}^3}})\ln|\dfrac{1 + \sqrt{8x_{0}^5}}{\sqrt{1 - 8x_{0}^5}}|) \approx \boxed{.106526}$

$\therefore$ The desired area is $A = 2(A_{1} + A_{2}) = \boxed{.300534}$ .

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Note: Using the conic $ax^2 + bxy + cy^2 + dx + ey = - 1$ with the points above gives us:

$a + \dfrac{4}{5}b + \dfrac{16}{25}c + d + \dfrac{4}{5}e = -1$

$a - \dfrac{4}{5}b + \dfrac{16}{25}c - d + \dfrac{4}{5}e = -1$

$(\dfrac{\sqrt{5} - 1}{2})^2a + \dfrac{\sqrt{5} - 1}{2}d = -1$

$(\dfrac{\sqrt{5} - 1}{2})^2a - \dfrac{\sqrt{5} - 1}{2}d = -1$

$\dfrac{16}{25}c - \dfrac{4}{5}e = - 1$

Again, letting $x_{0} = \dfrac{\sqrt{5} - 1}{2}$ and solving the system we obtain:

$a = \dfrac{-1}{x_{0}^2}, d = 0, c = \dfrac{25}{32} x_{0}, e = \dfrac{5}{8}(\dfrac{1}{x_{0}^2})$ and $b = 0$ which results in the same hyperbola we had above with a zero coefficient for the $xy$ term.