I have the hyperbola $$x^2 + \frac{2}{\sqrt{3}} x y - y^2 = 1$$ and I want to find the foci, but the only resources I can find that talk about finding the foci require the formula to be in standard form, $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
I don't think I can convert my hyperbola to standard form because it isn't purely horizontal or vertical. How do I go about finding the foci in this case?
Notice first of all that your hyperbola is centred at $O=(0,0)$. If $P=(x,y)$ is a vertex of the hyperbola, then the tangent at $P$ is perpendicular to $PO$, that is $y'=-x/y$. You can compute $y'$ by differentiating the equation of the ellipse: $$ y'={x+y/\sqrt3\over y-x/\sqrt3}. $$ Equating that to $-x/y$ gives: $$ x^2-y^2-2\sqrt3 xy=0, $$ which can be combined with the equation of the hyperbola to obtain $$ P=\left(\pm\sqrt{2\sqrt3+3\over8}, \pm\sqrt{2\sqrt3-3\over8}\right). $$
But yours is a right hyperbola (coefficients of $x^2$ and $y^2$ are opposite), hence we immediately obtain the foci as: $$ F=\sqrt2 P=\left(\pm\sqrt{2\sqrt3+3\over4}, \pm\sqrt{2\sqrt3-3\over4}\right). $$