Defining the foci of "slanted" ellipse equation

Question

How to define the foci ($F_1,F_2$) coordinates of the slanted ellipse $x^2+4xy+9y^2=9$?

Answer 1

Your ellipse is centered at the origin, hence in order to find its foci it is enough to find its vertices, or the direction of its axis. To do that, we may compute the stationary points of the quadratic form $q(x,y)=x^2+4xy+9y^2$ under the constraint $x^2+y^2=1$ through the method of Lagrange's multipliers. $\nabla q = 2\lambda(x,y) $ leads to the system $$\left\{\begin{eqnarray*}x+2y &=& \lambda x \\ 2x+9y&=&\lambda y \end{eqnarray*} \right.\tag{1}$$ hence the values of $\lambda$ associated with the stationary points are the eigenvalues of the matrix $$\begin{pmatrix}1 & 2 \\ 2 & 9 \end{pmatrix} \tag{2}$$ i.e. $5\pm 2\sqrt{5}$, and $(1)$ gives that the directions of the axis are $(1,2+\sqrt{5})$ for the major axis and $(1,2-\sqrt{5})$ for the minor axis. It follows that a vertex on the minor axis is given by $$ V_m = \left(3\cdot\frac{\sqrt{5}-2}{\sqrt{10}},\frac{3}{\sqrt{10}}\right) \tag{3}$$ and a vertex on the major axis is given by: $$ V_M = \left(3\cdot\frac{\sqrt{5}+2}{\sqrt{10}},-\frac{3}{\sqrt{10}}\right) \tag{4}$$ so: $$ a^2=\|V_M\|^2 = 3+\frac{6}{\sqrt{5}},\qquad b^2 = \|V_m\|^2 = 3-\frac{6}{\sqrt{5}} \tag{5} $$ and $c^2 = a^2-b^2 = \frac{12}{\sqrt{5}}$. At last, $$ F_1, F_2 = \pm \frac{c}{a}V_M\tag{6} $$ hence: $$\boxed{ F_1, F_2 = \color{red}{\left(\pm 3\sqrt{\frac{2}{5}(\sqrt{5}+2)},\mp 3\sqrt{\frac{2}{5}(\sqrt{5}-2)}\right)}} \tag{7} $$

Answer 2

$x^2 + 4 x y + 9 y^2 = 9$ or $\begin{pmatrix}x &y\end{pmatrix}\begin{pmatrix}1&2\\2&9\end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix}=9$

Let's rotate by standard theory. The eigenvalues are $\lambda=5\pm 2\sqrt{5}$ and $P=\begin{pmatrix}\frac1{\sqrt{10+4\sqrt{5}}}&-\frac1{\sqrt{10-4\sqrt{5}}}\\\frac{2+\sqrt{5}}{\sqrt{10+4\sqrt{5}}}&\frac{-2+\sqrt{5}}{\sqrt{10-4\sqrt{5}}}\end{pmatrix}$ so that ${\bf x'}^tP^tAP{\bf x'}=\begin{pmatrix}x' &y'\end{pmatrix}\begin{pmatrix}5-2\sqrt{5}&0\\0&5+2\sqrt{5}\end{pmatrix}\begin{pmatrix}x' \\y'\end{pmatrix}=9$, that is $\frac{x'^2}{\frac{9}{5-2\sqrt{5}}}+\frac{y'^2}{\frac{9}{5+2\sqrt{5}}}=1$ which makes the major axis $a=\frac{3}{\sqrt{5-2\sqrt{5}}}$ and the minor axis $b=\frac{3}{\sqrt{5+2\sqrt{5}}}$. The distance from the center $(0,0)$ to the foci is $c=\sqrt{a^2-b^2} = \sqrt{ \frac{9}{5-2\sqrt{5}}-\frac{9}{5+2\sqrt{5}}}=\frac{6}{\sqrt[4]{5}}$. Now we just transform back $(0,\pm \frac{6}{\sqrt[4]{5}})$: $\begin{pmatrix}\frac1{\sqrt{10+4\sqrt{5}}}&-\frac1{\sqrt{10-4\sqrt{5}}}\\\frac{2+\sqrt{5}}{\sqrt{10+4\sqrt{5}}}&\frac{-2+\sqrt{5}}{\sqrt{10-4\sqrt{5}}}\end{pmatrix}\begin{pmatrix}0\\\pm\frac{6}{\sqrt[4]{5}}\end{pmatrix}=(\mp\frac{6}{\sqrt[4]{5}\sqrt{10-4\sqrt{5}}},\pm\frac{6(-2+\sqrt{5})}{\sqrt[4]{5}\sqrt{10-4\sqrt{5}}})\approx(\mp 3.905,\pm 0.922)$, which are the same as Jack's.