The locus of intersections of tangents to a hyperbola with perpendiculars drawn from each focus is a circle with radius equal to the semimajor axis.
So, you can take either $A_1(0,4)$ or $A_2(0,6)$ as the intersection point and compute its distance from the center $C$ to get $a$, then use $f^2=a^2+b^2$ to find $b$.
In the same vein, all points symmetric to a focus with respect to tangents to the hyperbola lie on a circle with radius equal to twice the semimajor axis centered on the other focus:
Therefore, you can compute $2a$ by either taking $B_1(2,4)$ and finding its distance from $(4,6)$ or taking $B_2(-4,6)$ and finding its distance from $(-2,4)$. From there you can again recover $b$ via the equation $f^2=a^2+b^2$.
With the semiaxis lengths, center and transverse axis in hand, producing an equation for the hyperbola should be a simple matter. Note that for this equation, you really need $a^2$ and $b^2$ and not $a$ and $b$ themselves, so you can avoid taking any square roots in the above calculations.
In your comments to the question, you say that you are given the two foci (we’ll call them $F_1$ and $F_2$) and a vertex $V$. There are several ways to proceed with these data.
A direct approach is to use the definition of a hyperbola as the set of points with a constant difference of distances from two fixed points (the foci). Since you have a point on the hyperbola, you know what this difference $d$ must be: it’s the absolute value of $\|V-F_1\|-\|V-F_2\|$. So, an equation of the hyperbola is $(\|P-F_1\|-\|P-F_2\|)^2=(\|V-F_1\|-\|V-F_2\|)^2$. You’ll need to do some work to eliminate the square roots after you substitute coordinates for all of the points.
Another approach is to use the focus-directrix form of the equation, in which the ratio of distances to a focus and a line called the directrix is a constant called the eccentricity. The eccentricity is easily computed from the given data. The center of the hyperbola is the midpoint of the foci, from which we get $$e=\frac c a = {\|F_2-F_1\|/2\over \|V-(F_1+F_2)/2\|}$$ which you can simplify as necessary. With the eccentricity in hand, you can find the directrix and set up the equation of the hyperbola. This is a bit more work than the first approach, but has the advantage that the resulting square roots are easier to eliminate.
For a hyperbola, the eccentricity $e=\sec\theta$, where $\theta$ is the angle that the asymptotes make with the transverse axis. This suggests yet another approach: If the equations of the asymptotes are $ax+by+c=0$ and $Ax+By+C=0$, then an equation of the hyperbola can be written as $(ax+by+c)(Ax+By+C)=d$ for some constant $d$. Given the eccentricity and the foci, you can find the equations of the two asymptotes and adjust $d$ so that the given vertex satisfies the equation.
However, once you’ve gone as far as computing the eccentricity $e$ and the semi-major axis length $a$, the most straightforward way to go is to compute $b$ and write down the equation ${x^2\over a^2}-{y^2\over b^2}=1$ of the hyperbola in standard position and then rotate and translate it. This is quite easy to do in (homogeneous) matrix form. The matrix of the equation in standard position is simply $$M=\pmatrix{\frac1{a^2}&0&0\\0&-\frac1{b^2}&0\\0&0&-1},$$ i.e., the equation is $[x,y,1]M[x,y,1]^T=0$. The required center is just the midpoint of the two foci, so the necessary translation is easy to compute. The rotation matrix is also easy to construct if you remember that the columns of a transformation matrix are the images of the basis vectors, so the first column will be a unit vector in the direction of the hyperbola’s transverse axis, and the second column is just the first rotated by 90°, which can be found by swapping components and negating one of them.
Best Answer
The hyperbola given by the equation $xy=c^2>0$ is your prime example of a rectangular hyperbola. Let's rotate it by some arbitrary angle $\phi$! This rotation maps points $(x,y)$ to new points $(x',y')$, whereby $$(x,y)^\top=\left[\matrix{\cos\phi&-\sin\phi\cr\sin\phi&\cos\phi\cr}\right](x',y')^\top\ .$$ This means that the coordinates of the preimage $(x,y)$ are obtained from the coordinates $(x',y')$ of the image by $$x=\cos\phi\> x'-\sin\phi\> y',\qquad y=\sin\phi\> x'+\cos\phi\> y'\ .$$ If $(x,y)$ is lying on the model hyperbola then $xy=c^2$. This implies that the coordinates of the image point $(x',y')$ have to satisfy $$(\cos\phi\> x'-\sin\phi\> y')(\sin\phi\> x'+\cos\phi\> y')=c^2\ .$$ This can be rewritten as $$\cos\phi\sin\phi(x'^2-y'^2)+(\cos^2\phi-\sin^2\phi)x'y'=c^2\ .$$ Leaving away the primes here gives the result of the question.
$$\Rightarrow (x^2-y^2)\sin\theta+2xy\cos\theta=2c^2$$