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: $$

${\bf (9 cos^2\theta - 6 cos\theta sin\theta + sin^2\theta)x'^2 + -2(4 sin(2\theta) + 3 cos(2\theta))x'y' +}$

${\bf (9 sin^2\theta + 6 cos\theta sin\theta + cos^2\theta)y'^2 - 12\sqrt{10}(cos\theta + 3 sin\theta)x' + 12\sqrt{10}(sin\theta - 3 cos\theta)y' = 0. }$

To eliminate the ${\bf x'y' }$ term set ${\bf 4 sin(2\theta) + 3 cos(2\theta) = 0 \implies tan(2\theta) = -\frac{3}{4} }$

${\bf \implies tan2\theta = \frac{2 tan\theta}{1 - tan^2\theta} = -\frac{3}{4} \implies }$

${\bf 3 tan^2\theta - 8 tan\theta - 3 = 0 \implies tan\theta = \frac{8 \:\pm \:10}{6} }$

$$

Choosing the positive value of ${\bf tan\theta \implies tan\theta = 3 \implies cos\theta = \frac{1}{\sqrt{10}} }$ and ${\bf sin\theta = \frac{3}{\sqrt{10}} \implies }$

${\bf x = \frac{1}{\sqrt{10}} x' - \frac{3}{\sqrt{10}} y'}$

$$

${\bf y = \frac{3}{\sqrt{10}} x' + \frac{1}{\sqrt{10}} y' }$

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

${\bf \frac{9}{10}(x'^2 - 6 x'y' + 9 y'^2) - \frac{6}{10}(3 x'2 - 8 x'y' - 3 y'2) + \frac{1}{10}(9 y'^2 + 6 x'y' + y'^2) }$

${\bf - 12(x' - 3 y') - 36(3 x' + y') = 0 }$

${\bf \implies y'2 = 12 x' }$ $$

So we have a parabola in the x'y' system with focus ${\bf (3,0) }$ and directrix ${\bf x' = -3. }$

${\bf -\sqrt{10}\:x - 3\sqrt{10}\:y + 30 = 0 \implies y = -\frac{1}{\sqrt{3}}\:x + \sqrt{10} }$ $$

Two point on this line are ${\bf (0,\sqrt{10}) }$ and ${\bf (3\sqrt{10},0) }$ $$

Rotating the point ${\bf (0,\sqrt{10}) }$ we have: $$

${\bf 0 = \frac{1}{\sqrt{10}} x' - \frac{3}{\sqrt{10}} y' }$ $$

${\bf \sqrt{10} = \frac{3}{\sqrt{10}} x' + \frac{1}{\sqrt{10}} y' \implies (x' = 3,y' = 1) }$

Similarily, rotating the point ${\bf (3\sqrt{10},0) \implies (x' = 3,y' = -9) }$

$$

${\bf \therefore }$ In the ${\bf x'y'}$ system we have the line ${\bf x' = 3, \:}$ and the curve ${\bf y'^2 = 12 x' }$ and the line ${\bf x' = 3 }$ intersect at ${\bf (3,6) }$ and ${\bf (3,-6) \implies }$ $$

The area A ${\bf = \int_{-6}^{6} 3 - \frac{y'^2}{12} dy' = 3y' - \frac{y'^3}{36}|_{-6}^{6} = 24.}$