Finding the equation of a hyperbola, knowing the equations of its asymptotes and a point on it Find the equation of the hyperbola, given that transverse axis parallel to the $x$-axis, equations of asymptotes are $4x + y - 7 = 0$ and $3x - y - 5 = 0$ and the hyperbola passes through point $(4,4)$.
How could I solve this problem?
 A: $(4x+y-7)(3x-y-5)=k,$ and it passes through $(4,4)$: $(4\cdot 4+4-7)(3\cdot 4-4-5)=k$ or $k=39,$ so your equation is $$(4x+y-7)(3x-y-5)=39.$$
Edit Expanding to the form $12x^2-xy-y^2-41x+2y-4=0,$ it is a hyperbola since $B^2-4AC=(-1)^2-4\cdot 12\cdot (-1)=37>0.$
To get the canonical form it is convenient to find the center. In maxima CAS: solve([diff(12*x^2-x*y-y^2-41*x+2*y-4,x),diff(12*x^2-x*y-y^2-41*x+2*y-4,y)],[x,y]);. It is $(\frac{12}{7},\frac17).$ (To check expand $12(x-12/7)^2-(x-12/7)(y-1/7)-(y-1/7)^2-39=0$).
Now rotate by $\theta=\frac{\arctan{\frac1{13}}}{2}$ since $\tan{2\theta}=\frac{B}{C-A}$ with the rotation \begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} eliminates the $xy$ term.
The eigenvalues of the matrix $\begin{pmatrix}A&B/2\\B/2&C\end{pmatrix}=\begin{pmatrix}12&-1/2\\-1/2&-1\end{pmatrix}$ reveal the form $\frac{\sqrt{170}+11}{2} x'^2-\frac{\sqrt{170}-11}{2}y'^2=39.$ And the canonical form is $$(\frac{x'}{\frac{2\cdot 39}{\sqrt{170}+11}})^2-(\frac{y'}{\frac{2\cdot 39}{\sqrt{170}-11}})^2=1.$$
