Matrix powers and hyperbola

(We're in $$\mathbb{R}^2$$) How to find hyperbola equation, that has symmetry axis crossing (0,0) point and for $$n=1,2,\ldots$$ points $${\begin{pmatrix} 4 & 3 \\ 1 & 1 \end{pmatrix}}^n \begin{pmatrix} -3 \\ 4 \end{pmatrix}$$ lie on that hyperbola, also how to show that all those points lie on it?

By those points I mean $$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ for $$n=1$$, $$\begin{pmatrix} 3 \\ 1 \end{pmatrix}$$ for $$n=2$$, etc.

So far I've tried diagonalization of $$\begin{pmatrix} 4 & 3 \\ 1 & 1 \end{pmatrix}$$, but it's very messy and I don't know where it could get me to be honest. I guess that hyperbola can be rotated.

Edit: So knowing that in $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ we got $$E=F=0$$ and choosing $$F=-1$$ we have

$$Ax^2 + Bxy + Cy^2 = 1$$ and substituting $$x=0$$, $$y=1$$ we have $$C=1$$, again substituting $$x=3$$, $$y=1$$ we have $$B = -3A$$ and again $$x=15$$, $$y=4$$ we finally obtain $$15^2A -15\cdot 4 \cdot 3A = -15$$, so $$15A-12A=-1$$, thus $$A=-\frac{1}{3}$$ and $$B=1$$,

so our curve is $$-\frac{1}{3}x^2 + xy + y^2 - 1 = 0$$

Now we can easly check what curve is it using quadratic forms. Let $$Q(X) = -\frac{1}{3}x^2 + xy + y^2$$, then our curve is $$Q(X) - 1 = 0$$

Now $$\mathrm{m}(Q) = \begin{pmatrix} -\frac{1}{3} & 1 \\ 1 & 1 \end{pmatrix}$$, so

$$\chi_{\mathrm{m}(Q)}(x) = x^2 + \frac{2}{3}x - \frac{4}{3}$$,

$$\lambda = \frac{-1-\sqrt{13}}{3}$$, $$\mu= \frac{-1+\sqrt{13}}{3}$$,

thus $$Q(X) = PDP^{-1} = P \begin{pmatrix} \frac{-1-\sqrt{13}}{3} & 0 \\ 0 & \frac{-1+\sqrt{13}}{3} \end{pmatrix} P^{-1}$$, so in new base we have

$$\frac{-1-\sqrt{13}}{3} (x')^2 + \frac{-1+\sqrt{13}}{3}(y')^2 - 1 = 0$$, so

$$\frac{(x')^2}{\left ( \sqrt{\frac{3}{\sqrt{13} + 1}} \right )^2} - \frac{(y')^2}{\left ( \sqrt{\frac{3}{\sqrt{13} - 1}} \right )^2} = -1$$ and our curve is hyperbola.

• Well, there’s a brute-force solution: generate the first five points and use the fact that five points in general position define a conic. – amd Feb 8 at 1:38
• @amd Sure, but how to do that and why five points and not four for example? This task shouldn't be that hard I guess, I mean it was on my exam on very first university term. – chandx Feb 9 at 3:24
• It’s five points because the general conic equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ has five degrees of freedom. Plug the points that you’ve generated into this equation and solve the resulting linear system for the unknown coefficients. Doesn’t sound very hard to me. – amd Feb 9 at 3:53
• Come to think of it, you’re given that the hyperbola is centered at the origin, so $D=E=0$ and you can choose $F=-1$, so three points will suffice. – amd Feb 9 at 4:28
• @amd Okay, I updated my question, is it okay now? – chandx Feb 11 at 16:23