Finding the foci and vertices of an ellipse. How would you find the foci and vertices of the following ellipse:
$$\frac{2x^2}{15} +\frac{8y^2}{45} -\frac{2\sqrt3}{45}xy=1?$$
 A: Notice first of all that your ellipse is centered at $O=(0,0)$ (because if $(x,y)$ belongs to the ellipse, then also $(-x,-y)$ belongs to it). 
To find the axes of the ellipse, notice that if $P=(x,y)$ is a vertex, then the tangent at $P$ is perpendicular to $PO$, that is $y'(y/x)=-1$. You can compute $y'$ by differentiating the equation of the ellipse:
$$
{4\over15}x+{16\over45}yy'-{2\sqrt3\over45}y-{2\sqrt3\over45}xy'=0,
$$
whence:
$$
y'={2\sqrt3y-12x\over16y-2\sqrt3x}.
$$
The above condition $y'(y/x)=-1$ implies then that the coordinates of a vertex are related by:
$$
{y\over x}={-1\pm2\over\sqrt3}.
$$
Plugging that into the ellipse equation you can get the coordinates of the vertices, and then of course those of the foci.
A: 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.
