Conics and lines

Deriving the conic $ax^2 + bxy + cy^2 = 42$ using the points $(\sqrt{3},2\sqrt{3}), (-\sqrt{3},-2\sqrt{3}), (\sqrt{21},0),(-\sqrt{21},0), (\sqrt{\dfrac{27}{7}},2(\sqrt{\dfrac{27}{7}} - 1))$ :

Using the above points we obtain the system:

$21a = 42 \implies a = 2 \implies$ .

$b + 2c = 6$

$(\dfrac{27}{7} - \sqrt{\dfrac{27}{7}})b + (68 - 4\sqrt{\dfrac{27}{7}})c = 120$

$\implies 2(1 - \sqrt{\dfrac{27}{7}})c = 6(\sqrt{\dfrac{27}{7}} - 1) \implies c = -3 \implies b = 12 \implies$

$2x^2 + 12xy - 3y^2 = 42$ which is a hyperbola since $b^2 - 4ac > 0$ .

Solving for $y$ we obtain: $y = \pm\sqrt{\dfrac{14x^2 - 42}{3}} + 2x$ .

Using the symmetry about the line $y = \dfrac{-3}{2}x$ we use $g(x) = -\sqrt{\dfrac{14x^2 - 42}{3}} + 2x$ which is below the $y = 2\sqrt{3}$ where $x \geq \sqrt{3}$ .

Let $f(x) = \dfrac{-3}{2}x + \dfrac{3\sqrt{13}}{2}$ .

$f(x) = g(x) \implies 2\sqrt{\dfrac{14x^2 - 42}{3}} = 7x - 3\sqrt{13} \implies$ $56x^2 - 168 = 147x^2 - 126\sqrt{13}x + 35 \implies 91x^2 - 126\sqrt{13}x + 519 = 0 \implies$ $x = \dfrac{63 \pm 4\sqrt{21}}{7\sqrt{13}}$ .

Let $x_{1} = \dfrac{63 - 4\sqrt{21}}{7\sqrt{13}}$ and $x_{2} = \dfrac{63 + 4\sqrt{21}}{7\sqrt{13}}$

The area $A = 2\int_{x_{1}}^{x_{2}} (f(x) - g(x)) dx = 2\int_{x_{1}}^{x_{2}} (\sqrt{\dfrac{14x^2 - 42}{3}} - \dfrac{7}{2}x + \dfrac{3\sqrt{3}}{2}) dx$

Let $A_{1}(x) = \int \sqrt{\dfrac{14x^2 - 42}{3}} = \sqrt{\dfrac{14}{3}}\int \sqrt{x^2 - 3} dx$

Let $x = \sqrt{3}\sec(\theta) \implies dx = \sqrt{3}\sec(\theta)\tan(\theta) \implies$ $A_{1}(\theta) = \sqrt{42}\int sec^3(\theta) - sec(\theta) d\theta$

For $\int \sec^3(\theta)$ Let $u = \sec(\theta) \implies du = \sec(\theta)\tan(\theta)d\theta$ $dv = sec^2(\theta)d\theta \implies v = tan(\theta) \implies$ $2\int \sec^3(\theta) d\theta = \sec(\theta)\tan(\theta) + \ln|\sec(\theta) + \tan(\theta)| \implies$ $\int \sec^3(\theta) d\theta = \dfrac{1}{2}(\sec(\theta)\tan(\theta) + \ln|\sec(\theta) + \tan(\theta)|)$ $\implies A_{1}(\theta) = \dfrac{\sqrt{42}}{2}(\sec(\theta)\tan(\theta) - \ln|\sec(\theta) + \tan(\theta)|) \implies$
$A_{1} = \dfrac{\sqrt{42}}{2}(\dfrac{x\sqrt{x^2 - 3}}{3} - \ln|\dfrac{x + \sqrt{x^2 - 3}}{\sqrt{3}})|_{x_{1}}^{x_{2}}$ = $\dfrac{\sqrt{42}}{2}$ $(\dfrac{x_{2}\sqrt{x_{2}^2 - 3} - x_{1}\sqrt{x_{1}^2 - 3}}{3})$ + $\ln(\dfrac{x_{1} + \sqrt{x_{1}^2 - 3}}{x_{2} + \sqrt{x_{2}^2 - 3}}) \approx 5.44496 \implies A = 2(5.44496 - 4.83436) = 2(.6106) = \boxed{1.2212}$ .