Do you need vector calculus or linear algebra to acquire a general conic's parameters? As can be seen from this question, for the general conic section $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, the slope of the principal axis is $$\tan \theta=\frac{B}{A-C+(\mathrm{sgn}\;{B})\sqrt{B^2+(A-C)^2}}$$ and the eccentricity(if it's a hyperbola or an ellipse) is $$\epsilon=\left(\sqrt{\frac12-\frac{(\mathrm{sgn}\;\Delta)(A+C)}{2\sqrt{B^2+(A-C)^2}}}\right)^{-1}$$
I'd like to prove them both myself, but I suspect they both require either vector calculus or linear algebra, or similar topics at that level. I don't have the slightest grasp over either, so it'd be great to know if both can be proved by using simple cartesian geometry.
I'm not asking for the proof; if it requires either of the above, I'm happy to leave it for when I know them, and I can prove it myself otherwise. I just need to know what proving it entails.
 A: This can indeed be done with simple geometry.
(Note: if it's a parabola - which also has no centre - you can factor out the equation into the standard form to find the axis and tangent straightaway. It may not be so easy with central conics.)
First, the centre of the conic is shifted to the origin. This is done by changing $x$ and $y$ to $x-h$ and $y-k$. $(h, k)$ is the centre, and both coordinates are obtained by making it so that there are no linear terms. (This is because the centre is defined as the point which bisects all chords passing through it. If the centre is the origin, and say $(x, y)$ lies on the conic, then $(-x, -y)$ also has to lie on it, since that's the other end of the bisected chord. For that to happen, the conic equation must have no linear terms.) The equation is thus simplified to the form $Ax^2+By^2+Cxy=1$.
As in @robjohn's comment, the conic can be rotated about the origin to settle down with the $x-$ And $y-$ axes as its axes, as in the standard form. This is done as detailed in this answer, by using the matrix $\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta &  \cos\theta
\end{pmatrix}
$ to change the variables.
The equation is then further simplified into this form, without $xy$ terms; $$x^2(A\cos^2 \theta + B\sin^2 \theta +C\cos\theta\sin\theta)+y^2(A\sin^2\theta+B\cos^2 \theta-C\cos\theta\sin\theta)=1$$
(Since the coefficient of $xy$ has to be 0 when the conic is in its standard form
$2\sin\theta\cos\theta(B-A)+C(\cos^2\theta-\sin^2\theta)=0$)
From the last two equations, $\theta$ can be obtained, which gives us the slope of the axis. Tidying up the equation into the standard form will also yield the eccentricity  easily.
Edit: As in the comment below, $\frac{C}{A-B}=\frac{\sin(2\theta)}{\cos(2\theta)}=\tan(2\theta)$. This can be used to figure the slope out pretty quick.
