<s>[INST] Finding the foci and vertices of an ellipse. How would you find the foci and vertices of the following ellipse:
$$\frac{2x^2}{15} +\frac{8y^2}{45} -\frac{2\sqrt3}{45}xy=1?$$
 [/INST] Consider a rotation:
$$
x=X\cos\alpha-Y\sin\alpha,
\qquad
y=X\sin\alpha+Y\cos\alpha
$$
Then your ellipse becomes
$$
6(X\cos\alpha-Y\sin\alpha)^2+8(X\sin\alpha+Y\cos\alpha)^2
-2\sqrt{3}(X\cos\alpha-Y\sin\alpha)(X\sin\alpha+Y\cos\alpha)=45
$$
The term in $XY$ has coefficient
$$
4\cos\alpha\sin\alpha
-2\sqrt{3}\cos^2\alpha+2\sqrt{3}\sin^2\alpha
$$
which you want to be vanishing; dividing by $\cos^2\alpha$ we get
$$
2\sqrt{3}\tan^2\alpha+4\tan\alpha-2\sqrt{3}=0
$$
so
$$
\tan\alpha=\frac{-2+4}{2\sqrt{3}}=\frac{1}{\sqrt{3}}
\quad\text{or}\quad
\tan\alpha=\frac{-2-4}{2\sqrt{3}}=-\sqrt{3}
$$
So we can take $\alpha=\pi/6$ and the equation of the ellipse becomes
$$
X^2(6\cos^2\alpha+8\sin^2\alpha-2\sqrt{3}\cos\alpha\sin\alpha)
+Y^2(6\sin^2\alpha+8\cos^2\alpha+2\sqrt{3}\cos\alpha\sin\alpha)=45
$$
that is
$$
5X^2+9Y^2=45
$$
or
$$
\frac{X^2}{9}+\frac{Y^2}{5}=1
$$
Find foci and vertices, then use the inverse rotation.
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