Conics!

To find the conics using the general form $ax^2 + bxy + cy^2 + dx + ey = -1$ .

To show the conic going thru the points $(0,\dfrac{4}{5}) ,(1,\dfrac{4}{5}), (\dfrac{\sqrt{5} - 1}{2},0), (0,\dfrac{-4}{5})$ , and $(\dfrac{1 - \sqrt{5}}{2},0)$ 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 $y = f(x)$ :

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

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

To show the conic going thru the points $(1,\dfrac{4}{5}), (\dfrac{\sqrt{5} - 1}{2},0), (0,\dfrac{-4}{5})$ , and $(\dfrac{1 - \sqrt{5}}{2},0)$ and $(-1,\dfrac{4}{5})$ is an hyperbola.

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

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

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

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

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

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

$a = \dfrac{-1}{x_{0}^2}, d = b = 0, c = \dfrac{25}{32} x_{0}, e = \dfrac{5}{8}(\dfrac{1}{x_{0}^2}) \implies 25x_{0}^3y^2 - 32x^2 + 20y = -32x_{0}^2 \implies$ $\boxed{25x_{0}^3(y + \dfrac{2}{5x_{0}^3})^2 - 32x^2 = \dfrac{4 - 32x_{0}^5}{x^3}}$ 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 $y >= \dfrac{-4}{5}$ .

Let $g(x) = (\dfrac{2}{5x_{0}^3})(\sqrt{8x_{0}^3x^2 + 1 - 8x_{0}^5} - 1)$

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

For $A_{1}$ :

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

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

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

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

$A_{1}^{*} = \dfrac{2}{5x_{0}^3}(\dfrac{\sqrt{8x_{0}^3 + 1 - 8x_{0}^5}}{2} + \dfrac{1}{2}x_{0} - 1 + (\dfrac{1 - 8x_{0}^5}{2\sqrt{8x_{0}^3}}) * \ln|\dfrac{\sqrt{8x_{0}^3 + 1 - 8x_{0}^5} + \sqrt{8x_{0}^3}}{1 + \sqrt{8x_{0}^5}}|) \approx \boxed{.150469}$

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

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

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

$\implies A_{1} = A_{1}^{*} - A_{1}^{**} = \boxed{.01473}$

Similarly for $A_{2}$ :

$A_{2}^{*} = \int_{0}^{x_{0}} g(x) dx$ = $-\dfrac{4x_{0}^2}{5\sqrt{4x_{0}^2 - 1}}\arcsin(\dfrac{\sqrt{4x_{0}^2 - 1}}{2x_{0}}) \approx \boxed{-.264261}$

$A_{2}^{**} = \int_{0}^{x_{0}} f(x) dx = \dfrac{2}{5x_{0}^3}(\dfrac{-x_{0}}{2} + (\dfrac{1 - 8x_{0}^5}{2\sqrt{8x_{0}^3}})\ln|\dfrac{1 + \sqrt{8x_{0}^5}}{\sqrt{1 - 8x_{0}^5}}|) \approx \boxed{-.308239}$

$\implies A_{2} = A_{2}^{*} - A_{2}^{**} = \boxed{.043978}$

$\implies$ The desired Area $A = A_{1} + A_{2} = \boxed{0.058708}$