Area of a Region

Using the equations of rotation to rotate the x and y axis we have:

$$

${\bf x = x' cos\theta - y' sin\theta }$

${\bf y = x' sin\theta + y' cos\theta }$

Replacing the equations of rotation in the original equation and simplifying we obtain: $$

${\bf (4 cos^2\theta + 12 cos\theta sin\theta + 9 sin^2\theta)\:x'^2 + (4 sin^2\theta - 12 cos\theta sin\theta + 9 cos^2\theta)\:y'^2 + }$

${\bf (12\sqrt{13} sin\theta + 8\sqrt{13} cos\theta)\:x' + (12\sqrt{13} cos\theta - 8\sqrt{13} sin\theta) \:y' + (5 sin(2\theta) + 12 cos(2\theta))\:x'y' }$ ${\bf - 65 = 0 }$

To eliminate the ${\bf x'y'}$ term let ${\bf 5 sin(2\theta) + 12 cos(2\theta) = 0 \implies }$

${\bf tan(2\theta) = \frac{-12}{5} \implies }$ ${\bf \frac{2 tan\theta}{1 - tan^2\theta} = \frac{-12}{5} \implies }$

${\bf 6 tan^2\theta - 5 tan\theta - 6 = 0 \implies tan\theta = \frac{5 +- 13}{12} }$

Choosing the positive value for ${\bf tan\theta \implies tan\theta = \frac{3}{2} \implies }$ ${\bf cos\theta = \frac{2}{\sqrt{13}}\: and \:sin\theta = \frac{3}{\sqrt{13}} \implies }$

${\bf x = \frac{2}{\sqrt{13}}\: x' - \frac{3}{\sqrt{13}} y' }$ ${\bf \:and \:y = \frac{3}{\sqrt{13}}\:x' + \frac{2}{\sqrt{13}}\:y' }$

$$

Replacing the equations of rotation in the original equation and simplifying we obtain: $$

${\bf 169x'^2 + 676x' - 845 = 0 \implies x'^2 + 4x' - 5 = 0 \implies (x' + 5)(x' - 1) = 0 }$

${\bf \therefore }$ we have two parallel lines ${\bf x' = 1\: and \: x' = -5 }$ in the ${\bf x'y' }$ plane. $$

To find parallel lines in the ${\bf xy }$ plane: $$

For ${\bf x' = 1: }$ $$

${\bf y' = 0 \implies (\frac{2}{\sqrt{13}},\:\frac{3}{\sqrt{13}}) }$ $$

${\bf y' = 1 \implies (-\frac{1}{\sqrt{13}},\:\frac{5}{\sqrt{13}}) \implies \:slope\: m = -\frac{2}{3} \implies }$ $$ ${\bf y - \frac{5}{\sqrt{13}} = -\frac{2}{3}(x + \frac{1}{\sqrt{13}}) \implies }$ $$ ${\bf -2\sqrt{13}\:x - 3\sqrt{13}\:y + 13 = 0 }$ $$

and for ${\bf x' = -5:}$

${\bf y' = 0 \implies (-\frac{10}{\sqrt{13}},\:-\frac{15}{\sqrt{13}}) }$ $$

${\bf y' = 1 \implies (-\frac{13}{\sqrt{13}},\:-\frac{13}{\sqrt{13}}) \implies \:slope\: m = -\frac{2}{3} \implies }$ $$ ${\bf -2x - 3y - 5\sqrt{13} = 0 }$

${\bf \therefore }$ The two parallel lines in the ${\bf xy }$ plane are: $$

${\bf -2\sqrt{13}\:x - 3\sqrt{13}\:y + 13 = 0 }$ or ${\bf y = -\frac{2}{3} x + \frac{\sqrt{13}}{3} }$

${\bf -2x - 3y - 5\sqrt{13} = 0 }$ or ${\bf y = -\frac{2}{3} x - \frac{5\sqrt{13}}{3} }$ $$

The Area A ${\bf = 2 * \sqrt{13} \int_{-a}^{a} dx = }$ ${\bf 2 * \sqrt{13} \:x|_{-a}^{a} }$ = ${\bf 4 * \sqrt{13} * a = }$ ${\bf m * \sqrt{n} * a \implies m + n = 17. }$