What's the Conics? What's the Area ?

Let $ax^2 + bxy + cy^2 + dx + y = 0$

Using the points $(0,1), (\dfrac{1}{2},\dfrac{1}{2}), (\dfrac{-1}{2},\dfrac{1}{2}), (\dfrac{1}{4},0)$ and $(0,0)$ we obtain the system:

$a + b + c + 2d + 2 = 0$

$a - b + c - 2d + 2 = 0$

$c + 1 = 0$

$a + 4d = 0$

$\implies a + c + 2 = 0, c = -1, a = -4d \implies d = \dfrac{1}{4} \implies a = -1$ and $b = \dfrac{-1}{2}$ $\implies 4x^2 + 2xy + 4y^2 - x - 4y = 0$ , since $b^2 - 4ac < 0 \implies$ the conic is an ellipse..

To find the area we need to solve for $y$ obtaining:

$y = \dfrac{1}{4}(\pm\sqrt{4 - 15x^2} - (x - 2))$

We only need $g(x) = \dfrac{1}{4}(-\sqrt{4 - 15x^2} - (x - 2))$ , which is the portion of the ellipse below the line $y = \dfrac{2 - x}{4}$ .

For the parabola $y = ax^2 + bx + c$ that goes thru the points $(\dfrac{2}{\sqrt{15}}, \dfrac{\sqrt{15} - 1}{2\sqrt{15}}), (\dfrac{-2}{\sqrt{15}}, \dfrac{\sqrt{15} + 1}{2\sqrt{15}})$ and $(0,0)$ we obtain:

$c = 0$

$\dfrac{4}{15}a + \dfrac{2}{\sqrt{15}}b =\dfrac{\sqrt{15} -1}{2\sqrt{15}}$

$\dfrac{4}{15}a - \dfrac{2}{\sqrt{15}}b =\dfrac{\sqrt{15} +1}{2\sqrt{15}}$

$\implies a = \dfrac{15}{8}$ and $b = \dfrac{-1}{4} \implies y = \dfrac{15x^2 - 2x}{8}$ .

To find the points of intersection for the two curves let $f(x) = g(x) \implies 2x - 15x^2 = 2(\sqrt{4 - 15x^2} + x - 2) \implies (4 - 15x^2)^2 = 4(4 - 15x^2) \implies$

$15x^2(15x^2 - 4) = 0 \implies x = 0, x = \pm\dfrac{2}{\sqrt{15}}$ .

The area $A = \int_{\dfrac{-2}{\sqrt{15}}}^{0} f(x) - g(x) dx + \int_{0}^{\dfrac{-2}{\sqrt{15}}} f(x) - g(x) dx ) = \dfrac{1}{8}\int_{\dfrac{-2}{\sqrt{15}}}^{\dfrac{2}{\sqrt{15}}} 15x^2 - 2x + 2(\sqrt{4 - 15x^2} + x - 2) dx$

Letting $\sqrt{15}x = 2\sin(\theta) \implies dx = \dfrac{2}{\sqrt{15}}\cos(\theta) \implies$

$A = \dfrac{1}{8}(\dfrac{4\pi}{\sqrt{15}} - \dfrac{16}{\sqrt{15}} + \dfrac{16}{3\sqrt{15}}) = \dfrac{3\pi - 8}{6\sqrt{15}} =$ $\dfrac{3\pi - 2^3}{2 * 3\sqrt{3 * 5}} = \dfrac{b\pi - a^b}{ab\sqrt{bc}}$ $\implies a + b + c = \boxed{10}$ .