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(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.

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  • $\begingroup$ 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. $\endgroup$ – amd Feb 8 at 1:38
  • $\begingroup$ @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. $\endgroup$ – chandx Feb 9 at 3:24
  • $\begingroup$ 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. $\endgroup$ – amd Feb 9 at 3:53
  • $\begingroup$ 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. $\endgroup$ – amd Feb 9 at 4:28
  • $\begingroup$ @amd Okay, I updated my question, is it okay now? $\endgroup$ – chandx Feb 11 at 16:23

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