How do you prove that a plane intersecting a cone gives an ellipse? I know why it can also give a parabola when the plane's slant is equal to the cone's slant or an hyperbola when the plane's slant is parallel to the cone's axis, but I don't mathematically understand how it can give an ellipse. What I've tried so far is substituting the equation of a plane $z=my-c \ $ into the equation of a cone, resulting in 
\begin{array}{rcl}
\dfrac{(h-my+c)^2}{h^2} &=& \dfrac{x^2+y^2}{r^2} \\[0,4cm]
\iff \dfrac{(h+c)^2-2(h+c)my+m^2y^2}{h^2}  &=& \dfrac{x^2+y^2}{r^2} \\[0,4cm]
\iff \dfrac{h^2+2hc+c^2-2hmy-2cmy+m^2y^2}{h^2} &=& \dfrac{x^2+y^2}{r^2} \\[0,4cm]
\iff r^2\left(1+\dfrac{2c}{h}+\dfrac{c^2}{h^2}-\dfrac{2my}{h}-\dfrac{2cmy}{h^2} + \dfrac{m^2y^2}{h^2}\right) &=& x^2+y^2 \\[0,4cm]
\iff r^2+\dfrac{2cr^2}{h}+\dfrac{c^2r^2}{h^2} - \dfrac{2r^2my}{h} - \dfrac{2r^2cmy}{h^2} + \dfrac{r^2m^2y^2}{h^2} &=& x^2+y^2 \\[0,4cm]
\iff A - By - Cy + Dy^2 &=& x^2+y^2 \\[0,4cm]
\iff \alpha y^2+\beta y + \gamma &=& x^2+y^2 \\[0,4cm]
\end{array}
but I don't see how this last equation can be transformed into an equation of an ellipse. 
Thanks in advance.

The equation transforms into
$$x^2 + (1-\alpha) y^2 - \beta y + \gamma = 0
$$
If $\alpha < 1$ then, by completing the square in the $y$-terms, you will get an equation having the format
$$x^2 + \underbrace{A^2}_{1-\alpha}(y-b)^2 = (\text{constant})
$$
If the constant on the right hand side is positive, you're done because that's an equation of an ellipse.
There's still work left to do, namely verifying that $\alpha < 1$ and that the constant is positive, but that should give you the idea.
