How do you write an equation of an ellipse in standard form given minor axis has length 12 and its focal points are at (1, 2) and (4,-2)?

Oct 4, 2016

$592 \left(x - 5\right) x + 24 x y - 60 y + 585 {y}^{2} - 82844 = 0$

Explanation:

How do you write an equation of an ellipse in standard form given minor axis has length $12$ and its focal points are at ${f}_{1} = \left(1 , 2\right)$ and ${f}_{2} = \left(4 , - 2\right)$?

The ellipse is slanted by a rotation of

$\theta = \arctan \left(2 + 2 , 1 - 4\right)$

it's center is at ${p}_{0} = \left\{{x}_{0} , {y}_{0}\right\} = \frac{{f}_{1} + {f}_{2}}{2}$ so deffining a rotation matrix

$R = \left(\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right)$ so

$R = \left(\begin{matrix}\frac{4}{5} & \frac{3}{5} \\ - \frac{3}{5} & \frac{4}{5}\end{matrix}\right)$

$M = \left(\begin{matrix}\frac{1}{a} ^ 2 & 0 \\ 0 & \frac{1}{b} ^ 2\end{matrix}\right)$

and calling $p = \left(x , y\right)$ we have

$\left(p - {p}_{0}\right) {R}^{T} \cdot M \cdot R \left(p - {p}_{0}\right) = 1$

as the generic ellipse equation, with certer in ${p}_{0}$ and main axes $a , b$.

$c = \frac{\left\lVert {f}_{1} - {f}_{2} \right\rVert}{2}$

${a}^{2} = {b}^{2} + {c}^{2}$

Here $b = 12$ and $c = \frac{5}{2}$ so $a = \sqrt{{12}^{2} + {\left(\frac{5}{2}\right)}^{2}} = \frac{\sqrt{601}}{2}$

The final equation is

$592 \left(x - 5\right) x + 24 x y - 60 y + 585 {y}^{2} - 82844 = 0$