What's the Conic Two?

To show the conic is an ellipse.

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

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

Using the point $(0,\dfrac{-4}{5})$ we obtain (3): $\dfrac{16}{25}c - \dfrac{4}{5}e = -1$

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

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

Using (4) and (5) we obtain: $a = -(\dfrac{2}{\sqrt{5} - 1})^2$ and $d = 0$ .

Using (2) and (3) we obtain: $c = \dfrac{-25}{16}$ and $e = 0$ .

Using (1) $\implies b = \dfrac{5}{4} (\dfrac{2}{\sqrt{5} - 1})^2$ .

Let $x_{0} = \dfrac{\sqrt{5} - 1}{2} \implies a = \dfrac{-1}{x_{0}^2}, b = \dfrac{5}{4}(\dfrac{1}{x_{0}^2}), c = \dfrac{-25}{16}$ , and $d = e = 0 \implies \boxed{16x^2 - 20xy + 25x_{0}^2y^2 = 16x_{0}^2}$ and $b^2 - 4ac < 0 \implies$ we have an ellipse.

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

Solving for $y$ we obtain $y = \dfrac{2}{5x_{0}^2}(\pm\sqrt{4x_{0}^4 - (4x_{0}^2 - 1)x^2} + x)$ .

We want the ellipse below the line $y = \dfrac{2}{5x_{0}^2}x$ which is $g(x) = \dfrac{2}{5x_{0}^2}(-\sqrt{4x_{0}^4 - (4x_{0}^2 - 1)x^2} + x)$

Using the given $f(x) = \dfrac{4}{5}(\sqrt{x} - \sqrt{1 - x^2}) \implies$

$A_{1} = \int_{x_{0}}^{1} g(x) - f(x) dx$ and $A_{2} = \int_{0}^{x_{0}} f(x) - g(x) dx$ .

For $A_{1}$ :

Let $A^{*} = \int_{x_{0}}^{1} \dfrac{2}{5x_{0}^2}(-\sqrt{4x_{0}^4 - (4x_{0}^2 - 1)x^2} + x) dx$

Let $\sqrt{4x_{0}^2 - 1} x = 2x_{0}^2 \sin(\theta) \implies dx = \dfrac{2x_{0}^2}{\sqrt{4x_{0}^2 - 1}} d\theta$

$\implies A^{*} = \dfrac{2}{5x_{0}^2}(-\dfrac{2x_{0}^4}{\sqrt{4x_{0}^2 - 1}} * (\arcsin(\dfrac{\sqrt{4x_{0}^2 - 1}x}{2x_{0}^2}) - \dfrac{x}{2}\sqrt{4x_{0}^4 - (4x_{0}^2 - 1)x^2}) + \dfrac{x^2}{2})|_{x_{0}}^{1} \approx \boxed{.135739}$ .

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

Let $x = \sin(\theta) \implies dx = \cos(\theta) d\theta \implies$

$A^{**} = \dfrac{4}{5}(\dfrac{\pi}{4} - \dfrac{1}{2}\arcsin(x_{0}) - \dfrac{1}{2}x_{0}\sqrt{1 - x_{0}^2} + \dfrac{2}{3} x_{0}^{\dfrac{3}{2}} - \dfrac{2}{3}) \approx \boxed{-.106728}$ .

$A_{1} = A^{*} + A^{**} \approx \boxed{.029011}$

Similarly, for $A_{2}$ we obtain:

$A_{2} = \dfrac{4}{5}(\dfrac{2}{3}x_{0}^{\dfrac{3}{2}} - \dfrac{1}{2}\arcsin(x_{0}) - x_{0}\sqrt{1 - x_{0}^2})) + \dfrac{4x_{0}^2}{5\sqrt{4x_{0}^2 - 1}}\arcsin(\dfrac{\sqrt{4x_{0}^2 - 1}}{2x_{0}}) \approx \boxed{.062548}$

$\implies$ The desired area is $A = A_{1} + A_{2} = \boxed{0.091559}$ .