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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?

Blue
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Anas Yasser
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    Welcome to MSE! Could you please show your efforts? – user0102 Jun 15 '20 at 03:49
  • believe me I tried hard to solve this problem but I failed – Anas Yasser Jun 15 '20 at 04:04
  • @AnasYasser: Even when you don't know how to solve a problem, you can provide useful context. For instance, if this is a textbook exercise, what topic(s) and/or results were covered in the chapter? If this is a contest problem or online challenge, what's the target audience, and what level of sophistication is expected from the solution? ... The more you can say about the problem —showing that you're engaged in finding its solution— the better; simply posting an isolated problem statement tends to give the impression that you're just looking for someone to do your homework for you. – Blue Jun 15 '20 at 06:09
  • ok I got your point of view and I appreciate it – Anas Yasser Jun 15 '20 at 07:10
  • Let me prove my good intentions , I know that the general equation of the hyperbola is x^2/a^2 -y^2/b^2 when the transverse axis is parallel to the x-axis ,and when the center of the hyperbola is the origin,from solving the two equations we can get the center point of the hyperbola ,and from their slope we can get the values of a&b hence the equation is formed , my proplem here is that when I substitute with point given in the question ,it doesn`t achieve the equation, so that I restored to share it on this site – Anas Yasser Jun 15 '20 at 07:22

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$(4x+y-7)(3x-y-5)=k,$ 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 $$(\frac{x'}{\frac{2\cdot 39}{\sqrt{170}+11}})^2-(\frac{y'}{\frac{2\cdot 39}{\sqrt{170}-11}})^2=1.$$

Jan-Magnus Økland
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