From equation to area - Part 2

An ellipse in the form of $ax^2+2bxy+cy^2+2dx+2fy+g=0$ has semi-axis lengths of

$a' = \sqrt{\cfrac{2(af^2+cd^2+gb^2-2bdf-acg)}{(b^2-ac)(\sqrt{(a-c)^2+4b^2}-(a+c))}}$

$b' = \sqrt{\cfrac{2(af^2+cd^2+gb^2-2bdf-acg)}{(b^2-ac)(-\sqrt{(a-c)^2+4b^2}-(a+c))}}$

In this question, $a = 1$ , $b = 1$ , $c = 2$ , $d = 3$ , $f = 2$ , and $g = -120$ , which means:

$a' = \sqrt{\cfrac{2(1 \cdot 2^2+2 \cdot 3^2 - 120 \cdot 1^2-2 \cdot 1 \cdot 3 \cdot 2 + 1 \cdot 2 \cdot 120)}{(1^2-1 \cdot 2)(\sqrt{(1-2)^2+4 \cdot 1^2}-(1+2))}} = \sqrt{\cfrac{260}{3 - \sqrt{5}}}$

$b' = \sqrt{\cfrac{2(1 \cdot 2^2+2 \cdot 3^2 - 120 \cdot 1^2-2 \cdot 1 \cdot 3 \cdot 2 + 1 \cdot 2 \cdot 120)}{(1^2-1 \cdot 2)(-\sqrt{(1-2)^2+4 \cdot 1^2}-(1+2))}} = \sqrt{\cfrac{260}{3 + \sqrt{5}}}$

The area of the ellipse is then $A = \pi a'b' = \pi \cdot \sqrt{\cfrac{260}{3 - \sqrt{5}}} \cdot \sqrt{\cfrac{260}{3 + \sqrt{5}}} = 130\pi$ , so $n = \boxed{130}$ .