Conic Mania 2 !

For the conic $ax^2 + bxy + cy^2 + dx + ey = -16$ .

For $(0,4):\implies 4c + e = -4$

For $(0,-4):\implies 4c - e = -4$

For $(2,0):\implies 2a + d = -8$

For $(-2,0):\implies 2a - d = -8$

For $(1,4):\implies a + 4b + 16c + d + 4e = -16$

$\implies c = -1, e = 0, a = -4, d = 0$ and $b = 1 \implies -4x^2 + xy - y^2 = -16$ or $4x^2 - xy + y^2 = 16$ which is an ellipse since $b^2 - 4ac > 0$ .

Solving for $y$ we obtain:

$y = \dfrac{1}{2}(\pm\sqrt{64 - 15x^2} + x)$

Using the portion of the ellipse above the line $y = \dfrac{x}{2} \implies \boxed{y_{1}(x) = \dfrac{1}{2}(\sqrt{64 - 15x^2} + x)}$ .

For the circle in (2) $(x - a)^2 + (y - b)^2 = r^2$ .

For (1) $(0,4):\implies a^2 + 16 - 8b + b^2 = r^2$

For (2) $(0,5):\implies a^2 + 25 - 10b + b^2 = r^2$

Subtracting (1) from (2) we obtain: $9 - 2b = 0 \implies b = \dfrac{9}{2}$ .

For (3) $(-1,3) \implies 1 + 2a + a^2 + 9 - 6b + b^2 = r^2$

For (1) $(0,4):\implies a^2 + 16 - 8b + b^2 = r^2$

Subtracting (1) from (2) we obtain: $a + b = 3$ and using $b = \dfrac{9}{2} \implies a = -\dfrac{3}{2}$

$\implies r^2 = \dfrac{5}{2} \implies (x + \dfrac{3}{2})^2 + (y - \dfrac{9}{2})^2 = \dfrac{5}{2}$ .

Solving for $y$ we obtain:

$y = \pm\sqrt{\dfrac{5}{2} - (x + \dfrac{3}{2})^2} + \dfrac{9}{2}$

Using the portion of the circle below the line $y = \dfrac{9}{2} \implies \boxed{y_{2}(x) = -\sqrt{\dfrac{5}{2} - (x + \dfrac{3}{2})^2} + \dfrac{9}{2}}$

You can check that $(-1,3)$ satisfies the ellipse above.

From graph above the ellipse and the circle in (2) intersect at $(-1,3), (0,4)$ .

For $\displaystyle\int_{-1}^{0} y_{1}(x) dx$ :

Let $I_{1} = \displaystyle\dfrac{1}{2}\int_{-1}^{0} \sqrt{64 - 15x^2} dx$ .

Let $\sqrt{15}x = 8\sin(\theta) \implies dx = \dfrac{8}{\sqrt{15}}\cos(\theta) d\theta \implies$

$\dfrac{32}{\sqrt{15}}\displaystyle\int \cos^2(\theta) d\theta = \dfrac{16}{\sqrt{15}}\displaystyle\int (1 + \cos(2\theta)) d\theta = \dfrac{16}{\sqrt{15}}(\theta + \dfrac{1}{2}\sin(2\theta)) =$ $\dfrac{16}{\sqrt{15}}(\arcsin(\dfrac{\sqrt{15}x}{8}) + \dfrac{\sqrt{15}x\sqrt{64 - 15x^2}}{64})$

$\implies I_{1}(x)|_{-1}^{0} = \dfrac{16}{\sqrt{15}}(\arcsin(\dfrac{\sqrt{15}}{8}) + \dfrac{7\sqrt{15}}{64})$

$\implies \displaystyle\int_{-1}^{0} y_{1}(x) dx = \dfrac{16}{\sqrt{15}}\arcsin(\dfrac{\sqrt{15}}{8}) + \dfrac{7}{4} - \dfrac{1}{4} = \dfrac{16}{\sqrt{15}}\arcsin(\dfrac{\sqrt{15}}{8}) + \dfrac{3}{2} \approx = \boxed{3.587736}$

For $\displaystyle\int_{-1}^{0} y_{2}(x) dx$ :

Let $I_{2} = -\displaystyle\int_{-1}^{0} \sqrt{\dfrac{5}{2} - (x + \dfrac{3}{2})^2} dx$

Let $x + \dfrac{3}{2} = \sqrt{\dfrac{5}{2}}\sin(\theta) \implies dx = \sqrt{\dfrac{5}{2}}\cos(\theta) d\theta$

$-\dfrac{5}{2}\displaystyle\int \cos^2(\theta) d\theta = -\dfrac{5}{4}\displaystyle\int (1 + \cos(2\theta)) d\theta = -\dfrac{5}{4}(\theta + \dfrac{1}{2}\sin(2\theta)) = -\dfrac{5}{4}(\arcsin(\dfrac{x + \dfrac{3}{2}} {\sqrt{\dfrac{5}{2}}}) + \dfrac{2}{5}(x + \dfrac{3}{2})(\sqrt{\dfrac{5}{2} - (x + \dfrac{3}{2})^2}))$

$\implies I_{2}(x)|_{-1}^{0} = -\dfrac{5}{4}(\arcsin(\dfrac{3}{\sqrt{10}}) - \arcsin(\dfrac{1}{\sqrt{10}}))$

$\implies \displaystyle\int_{-1}^{0} y_{2}(x) dx = -\dfrac{5}{4}(\arcsin(\dfrac{3}{\sqrt{10}}) - \arcsin(\dfrac{1}{\sqrt{10}})) + \dfrac{9}{2} \approx \boxed{3.340881}$

$\implies \boxed{A_{1} = \displaystyle\int_{-1}^{0} y_{1}(x) - y_{2}(x) dx = 0.246855}$ .

For the circle in (3) $(x - a)^2 + (y - b)^2 = r^2$ .

From the circle in (2) using pts $(0,4)$ and $0,5)$ we had $b = \dfrac{9}{2}$ .

For (4) $(1,4): \implies 1 - 2a + a^2 +16 - 8b + b^2 = r^2$

For (1) $(0,4):\implies a^2 + 16 - 8b + b^2 = r^2$

Subtracting (1) from (4) we obtain $a = \dfrac{1}{2} \implies r^2 = \dfrac{1}{2} \implies$

$(x - \dfrac{1}{2})^2 + (y - \dfrac{9}{2})^2 = \dfrac{1}{2}$ .

Solving for $y$ we obtain:

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

Using the portion of the circle below the line $y = \dfrac{9}{2} \implies \boxed{y_{3}(x) = -\sqrt{\dfrac{1}{2} - (x - \dfrac{1}{2})^2} + \dfrac{9}{2}}$ .

From graph above the ellipse and the circle in (3) intersect at $(0,4), (1,4)$ .

Changing the limits of integration for the ellipse $y_{1}(x)$ above $\implies \displaystyle\int_{0}^{1} y_{1}(x) dx = \dfrac{16}{\sqrt{15}}\arcsin(\dfrac{\sqrt{15}}{8}) + 2 \approx \boxed{4.087736}$ .

For $\displaystyle\int_{0}^{1} y_{3}(x) dx$ :

Let $I_{3} = -\displaystyle\int_{0}^{1} \sqrt{\dfrac{1}{2} - (x + \dfrac{1}{2})^2} dx$

Let $x - \dfrac{1}{2} = \dfrac{1}{\sqrt{2}}\sin(\theta) \implies dx = \dfrac{1}{\sqrt{2}}\cos(\theta) d\theta \implies$

$I_{3} = -\dfrac{1}{2}\displaystyle\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \cos^2(\theta) d\theta = -\dfrac{1}{4}\displaystyle\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} (1 + \cos(2\theta)) d\theta =$ $-\dfrac{1}{4}(\theta + \dfrac{1}{2}\sin(2\theta))|_{\frac{-\pi}{4}}^{\frac{\pi}{4}} = -\dfrac{\pi + 2}{8}$

$\implies \displaystyle\int_{0}^{1} y_{3}(x) dx = -\dfrac{\pi + 2}{8} + \dfrac{9}{2} = \dfrac{34 - \pi}{8} \approx \boxed{3.857301}$

$\implies \boxed{A_{2} = \displaystyle\int_{0}^{1} y_{1}(x) - y_{3}(x) dx = 0.230435}$

Translating the center of the circles $(x + \dfrac{3}{2})^2 + (y - \dfrac{9}{2})^2 = \dfrac{5}{2}$ and $(x - \dfrac{1}{2})^2 + (y - \dfrac{9}{2})^2 = \dfrac{1}{2}$ to $(0,0)$ and $(0,2)$ respectively we obtain:

$x^2 + y^2 = \dfrac{5}{2}$

$x^2 + (y - 2)^2 = \dfrac{1}{2}$ .

The circles above intersect at $(\dfrac{1}{2},\dfrac{3}{2})$ and $(-\dfrac{1}{2},\dfrac{3}{2})$ .

Solving for $y$ in both circles above we obtain:

$y_{4}(x) = \sqrt{\dfrac{5}{2} - x^2}$ above the line $y = 0$

                                   and,

$y_{5}(x) = -\sqrt{\dfrac{1}{2} - x^2} + 2$ below the line $y = 2$ .

$A_{3} = \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} y_{4}(x) - y_{5}(x) dx$ .

For $I_{4} = \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} y_{4}(x) dx$

Let $x = \sqrt{\dfrac{5}{2}}\sin(\theta) \implies dx = \sqrt{\dfrac{5}{2}}\cos(\theta) d\theta \implies$

$\dfrac{5}{2}\displaystyle\int \cos^2(\theta) d\theta = \dfrac{5}{4}\displaystyle\int (1 + \cos(\theta)) d\theta = \dfrac{5}{4}(\theta + \sin(\theta)\cos(\theta)) =$ $\dfrac{5}{4}(\arcsin(\sqrt{\dfrac{2x}{5}}) + \dfrac{1}{5}(\sqrt{2}x\sqrt{5 - 2x^2}))$

$\implies I_{4}(x)|_{-\frac{1}{2}}^{\frac{1}{2}} = \dfrac{5}{2}\arcsin(\dfrac{1}{\sqrt{10}}) + \dfrac{3}{4}$

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

Let $x - \dfrac{1}{2} = \dfrac{1}{\sqrt{2}}\sin(\theta) \implies dx = \dfrac{1}{\sqrt{2}}\cos(\theta) d\theta \implies$

$I_{5} = -\dfrac{1}{2}\displaystyle\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \cos^2(\theta) d\theta -2 = -\dfrac{1}{4}\displaystyle\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} (1 + \cos(2\theta)) d\theta - 2 =$ $-\dfrac{1}{4}(\theta + \dfrac{1}{2}\sin(2\theta))|_{\frac{-\pi}{4}}^{\frac{\pi}{4}} - 2 =$ $-\dfrac{\pi}{8} + \dfrac{7}{4}$

$\implies \boxed{A_{3} = \dfrac{5}{2}\arcsin(\dfrac{1}{\sqrt{10}}) + \dfrac{3}{4} + \dfrac{\pi}{8} - \dfrac{7}{4} = \dfrac{5}{2}\arcsin(\dfrac{1}{\sqrt{10}}) + \dfrac{\pi - 8}{8} \approx 0.197075}$

$$

$\implies A_{1} + A_{2} + A_{3} = \boxed{0.674365}$ .

$$

Note: You could have used the initial circles $(x + \dfrac{3}{2})^2 + (y - \dfrac{9}{2})^2 = \dfrac{5}{2}$ and $(x - \dfrac{1}{2})^2 + (y - \dfrac{9}{2})^2 = \dfrac{1}{2}$ and solved for $x_{1}(y) = \sqrt{\dfrac{5}{2} - (y - \dfrac{9}{2})^2} - \dfrac{3}{2} \text{(right of} \:\ x = -\dfrac{3}{2})$ and $x_{2}(y) = -\sqrt{\dfrac{1}{2} - (y - \dfrac{9}{2})^2} + \dfrac{1}{2}\text{(left of} \:\ x = \dfrac{1}{2})$ , then $A_{3} = \displaystyle\int_{4}^{5} x_{1}(y) - x_{2}(y) \:\ dy$ .

I translated the two circles for the sole reason of not being able to shade in the region using functions of y in Geogebra. By translating the two circles the region $R_{3}$ in the graph above results in the same area as the initial region.

I hope this didn't cause any confusion.