<s>[INST] Equation of one branch of a hyperbola in general position Given a generic expression of a conic:
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0$$
is there a way to write an expression for one of the branches as a function of the coefficients? I tried using the quadratic formula to get an expression for $y$:
$$y=\frac{-(Bx+E)\pm \sqrt{(Bx+E)^2 - 4(C)(Ax^2 + Dx + F)}}{2C}$$
but this doesn't always work. Consider:
$$xy=1$$
Here, $A=0, B=1, C=0, D=0, E=0, F=0$, so $y=\frac{\cdot}{0}$, which isn't particularly helpful. In other cases it is not as bad, but still not what I'm looking for. E.g.
$$x^2 - y^2 - 1=0$$
Using the formula above, we get
$$y=\pm \sqrt{x^2-1}$$
which seems nice, but $y=\sqrt{x^2-1}$ it is actually one half of each branch rather than one entire branch, as can be seen here.
http://www.wolframalpha.com/input/?i=plot%28x^2+-+y^2+-+1%3D0%29
http://www.wolframalpha.com/input/?i=plot%28y%3Dsqrt%28x^2-1%29%29
I am trying to draw one of these branches, so I need an ordered set of points along a predefined "grid" of either of the variables. Is it possible to do this?
 [/INST] Following Will Jagy's suggestion, here are some examples:
Example #1
Consider 
$$x^2 - y^2 -1 = 0$$

($A=1$, $B=0$, $C=-1$, $D=0$, $E=0$, $F=-1$). From 
$$p_c = \begin{pmatrix}x_c\\y_c\end{pmatrix} = \begin{pmatrix}\frac{BE-2CD}{4AC-B^2}\\ \frac{DB-2AE}{4AC-B^2}\end{pmatrix}$$
we have
$$p_c = \begin{pmatrix}0\\0\end{pmatrix}$$
Now, the eigenvalues/vectors of the matrix $$\begin{pmatrix}A & \frac{B}{2}\\\frac{B}{2} & C\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$$ are $\lambda_1 = -1$, $\lambda_2 = 1$, $v_1 = \begin{pmatrix}0\\1\end{pmatrix}$, $v_2 = \begin{pmatrix}1\\0\end{pmatrix}$
From the parametric form
$$g(t)=p_c + v_1 cosh(t) + v_2 sinh(t)$$
we have 
$$g(t) = \begin{pmatrix}0\\1\end{pmatrix} cosh(t) + \begin{pmatrix}1\\0\end{pmatrix} sinh(t)$$
which looks like

Now how would we draw the other branch? That is, how do you determine the values of the parameter that constitute each branch?
Example #2
Consider 
$$7x^2 - 3y^2 - 25 = 0$$
($A=7$, $B=0$, $C=-3$, $D=0$, $E=0$, $F=-25$). From 
$$p_c = \begin{pmatrix}x_c\\y_c\end{pmatrix} = \begin{pmatrix}\frac{BE-2CD}{4AC-B^2}\\ \frac{DB-2AE}{4AC-B^2}\end{pmatrix}$$
we have
$$p_c = \begin{pmatrix}0\\\frac{-25}{6}\end{pmatrix}$$
Now, the eigenvalues/vectors of the matrix $$\begin{pmatrix}A & \frac{B}{2}\\\frac{B}{2} & C\end{pmatrix} = \begin{pmatrix}7 & 0 \\ 0 & -3 \end{pmatrix}$$ are $\lambda_1 = -3$, $\lambda_2 = 7$, $v_1 = \begin{pmatrix}0\\1\end{pmatrix}$, $v_2 = \begin{pmatrix}1\\0\end{pmatrix}$
From the parametric form
$$g(t)=p_c + v_1 cosh(t) + v_2 sinh(t)$$
we have 
$$g(t) = \begin{pmatrix}0\\ \frac{-25}{6}\end{pmatrix} + \begin{pmatrix}0\\1\end{pmatrix} cosh(t) + \begin{pmatrix}1\\0\end{pmatrix} sinh(t)$$
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