The general conic section is given by $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0.$$
In the case where $B\neq0$, $C=0$, we have $$Ax^2+Bxy+Dx+Ey+F\overset{1}=0,$$ with $B^2-4AC=B^2>0$, so that this is a hyperbola.
How do we rewrite $\overset{1}=$ into the standard (or canonical) form?
(Ideally, please just show the algebra and do not make use of other results from conic sections without also proving them.)
Basically, the idea is to rotate your axes so that the given curve is a hyperbola in the usual sense. In order to do this, we let
$$x=X\cos \theta+Y\sin\theta \; \; \; \; \; \; \; \; \; \; \; \; \; y=-X\sin\theta+Y\cos\theta$$
where $(X,Y)$ are the coordinates in the rotated axes.
For ease of notation, I shall define
$$c=\cos\theta \; \; \; \; \; \; \; \; \; \; \; \; \; s = \sin\theta$$
Plugging this in, we get
$$(Ac^2-Bsc)X^2+(2Asc+Bc^2-Bs^2)XY+(As^2+Bsc)Y^2+(Dc-Es)X+(Ds+Ec)Y=-F$$
With a hyperbola, we want to make sure that the $XY$ term disappears. i.e., we want to choose $\theta$ such that the coefficient of $XY$, which is $2Asc+Bc^2-Bs^2$, is $0$.
\begin{align} \ & 2Asc+Bc^2-Bs^2=0 \\ \ \implies & A\sin(2\theta)+B\cos(2\theta)=0 \\ \ \implies & \theta=-\frac 12 \tan^{-1} \frac BA \end{align}
With this particular value of $\theta$, you work out what $\sin \theta$ and $\cos \theta$ are, and plug them back into the equation above to get something in the from of $$PX^2+QY^2+RX+SY=-F$$
which implies that
$$P \biggl(X+\frac{R}{2P} \biggr)^2-Q \biggl(Y+\frac{S}{2Q} \biggr)^2=\frac{R^2}{4P}+\frac{S^2}{4Q}-F$$
where $P,Q$ should be positive.
Then, via a shift in the axes and some rearrangements, we get the required hyperbola.
Probably easier to understand with an example.
Say we have the equation
$$x^2-\sqrt 3xy+\sqrt 3x+y-\frac 73=0$$
Then as before, we let
$$x=X\cos \theta+Y\sin\theta \; \; \; \; \; \; \; \; \; \; \; \; \; y=-X\sin\theta+Y\cos\theta$$
$$c=\cos\theta \; \; \; \; \; \; \; \; \; \; \; \; \; s = \sin\theta$$
Plug this in to get
$$(c^2+\sqrt 3sc)X^2+(2sc-\sqrt 3c^2+\sqrt 3s^2)XY+(s^2 - \sqrt 3sc)Y^2+(\sqrt 3c-s)X+(\sqrt 3s+c)Y=\frac 73$$
We want to chose $\theta$ such that $2sc-\sqrt 3c^2+\sqrt 3s^2=0$, so
\begin{align} \ & 2sc-\sqrt 3c^2+\sqrt 3s^2=0 \\ \ \implies & \sin (2\theta)- \sqrt 3 \cos (2\theta)=0 \\ \ \implies & 2\theta = \frac \pi 3\\ \ \implies & \theta = \frac \pi 6 \end{align}
If $\theta = \frac \pi 6$ then $\cos \theta = \frac{\sqrt 3}{2}$ and $\sin\theta = \frac 12$, so
$$\biggl(\frac 34+\frac 34 \biggr)X^2+\biggl(\frac 14 - \frac 34 \biggr)Y^2+\biggl(\frac 32 -\frac 12 \biggr)X+ \biggl(\frac{\sqrt 3}{2}+\frac{\sqrt 3}{2} \biggr)Y=\frac 73$$ $$\frac 32 X^2- \frac 12 Y^2+X+\sqrt 3 Y=\frac 73$$ $$\frac 32 \biggl(X+\frac 13 \biggr)^2- \frac 12 \bigl(Y-\sqrt 3 \bigr)^2=1$$
And hence $$\frac{\hat X ^2}{(\sqrt {2/3})^2}- \frac{\hat Y ^2}{(\sqrt 2)^2}=1$$
where $\hat X = X+\frac 13$ and $\hat Y = Y-\sqrt 3$
