I've been trying to prove the canonical form of the hyperbola by myself.
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $
I started from the statement that
A hyperbola may be defined as the curve of intersection between a right circular conical surface and a plane that cuts through both halves of the cone.(Wikipedia)
The conical surface is constructed around the y-vertical axis, with its apex at (0,0,0). I assumed, without loss of generality, that the plane is parallel to the $x$ axis, so, for any point on that plane, we have
$z = a*y+b$, a, b not null
Also, any point lying on the conical surface satisfies
$y^2 = c^2*(x^2+z^2)$, c not null
Combining the 2 equations, we get
$y^2 = c^2x^2+c^2a^2y^2+2c^2aby+c^2b^2$
We can now rewrite it
$y^2(c^2a^2-1)+x^2c^2+2c^2aby=-c^2b^2$
$\frac{y^2(c^2a^2-1)}{-c^2b^2} + \frac{x^2c^2}{-c^2b^2} + \frac{2c^2aby}{-c^2b^2} = 1$
So now we got the sum of $x^2$, $y^2$, and $y$ multiplied by some coefficients equals 1. This does not resemble the canonical form of the hyperbola. The presence of the $y$ is the most notable difference. I know $x$ and $y$ in my equations are in the 3D space, and the canonical form takes them in the 2D plane space, but the 2D equations should only be a scaled version of the 3D ones, because the plane is parallel to the x-axis, isn't it?
$x^2*p_1 + y^2*p_2 + y*p_3 = 1$
I am sure I had overlooked something, but can't figure out what. I appreciate any assistance. Thank you.
If $a \neq 0,$ the cutting plane is parallel to the $x$ axis but not parallel to the $y$ axis. Starting at the point $(x,y,z) = (0,0,b)$ on this plane, and traveling along the line of intersection of the cutting plane and the $y,z$ plane (that is, the line that simultaneously satisfies $z = ay+b$ and $x = 0$), we can move toward the point where this line intersects the $y$ axis or we can move in the opposite direction.
Assuming that you actually have a hyperbola, the intersection with the cone is closer in one direction than the other, that is, one intersection is at $(0,y_1,ay_1+b)$ and the other is at $(0,y_2,ay_2+b)$ with $y_1 \neq -y_2.$ But these two points are the two vertices of the hyperbola, and the center of the hyperbola is halfway between them, at $y_0 = \frac12(y_1 + y_2) \neq 0.$
So the equation that you end up with will be equivalent to something in the form $$ \frac{(y - y_0)^2}{B^2} - \frac{x^2}{A^2} = 1,$$ which is what you get when you translate a hyperbola with the equation $\frac{y^2}{B^2} - \frac{x^2}{A^2} = 1$ so that its center moves from $(0,0)$ to $(0,y_0).$
The extra term in $y$ (without the square) comes from the expansion of $(y - y_0)^2.$
The other thing that may be confusing is that you have set up your cone and plane in such a way as to produce a hyperbola whose axis is in the $y,z$ plane. If you eliminate the $z$ coordinate, the axis of the hyperbola becomes the $y$ axis. But the equation $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ is the equation of a hyperbola whose axis is the $x$ axis.