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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.

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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.

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If you bring everything to the same side you've ended up with something on the form $$ x^2 + Ay^2 + By + C = 0 $$ Hopefully $A$ is positive; otherwise you can't get an ellipse. I haven't checked your computations up until now, but if everything works out you should be able to prove that this is the case exactly when the slant of your plane is flatter than the slant of the cone.

Now complete the square! We have $$ A\bigr(y-\tfrac {-B}{2A}\bigl)^2 = Ay^2 + By + \tfrac{B^2}{4A^2}$$ so your equation above is the same as $$ x^2 + A\bigl (y-\tfrac{-B}{2A}\bigr )^2 = -C - \tfrac{B^2}{4A^2} $$ which should be recognizable as an axis-parallel ellipse centered on $(0,\tfrac{-B}{2A})$.

Since the intersection is intuitively non-empty, $-C-\tfrac{B^2}{4A^2}$ is necessarily non-negative.

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The sides of the cone have slope $\frac hr$ relative to the $x,y$ plane. The plane $z = my - c$ has slope $m.$ If $\lvert m\rvert \geq \left\lvert\frac hr\right\rvert$ then in fact you will not get an ellipse, but rather a parabola or hyperbola. So let's suppose you have set up your cone and plane so that $\lvert m\rvert < \left\lvert\frac hr\right\rvert.$ Let's also suppose that $h$ and $r$ are both non-zero, since otherwise you have a degenerate cone, which does not produce an ellipse.

You found this equation is satisfied along the intersection of the cone and plane: $$r^2+\frac{2cr^2}{h}+\frac{c^2r^2}{h^2} - \frac{2r^2my}{h} - \frac{2r^2cmy}{h^2} + \frac{r^2m^2y^2}{h^2} = x^2+y^2.$$

This is equivalent to $$ 0 = x^2 + \left(1 - \frac{r^2m^2}{h^2}\right)y^2 + \left(\frac{2r^2m}{h} + \frac{2r^2cm}{h^2}\right)y - \left(r^2+\frac{2cr^2}{h}+\frac{c^2r^2}{h^2}\right). \tag 1$$

But since $\lvert m\rvert < \left\lvert\frac hr\right\rvert,$ it follows that $m^2 < \frac{h^2}{r^2},$ and since $h^2$ and $r^2$ are both positive, $\frac{r^2 m^2}{h^2} < 1.$ Therefore $$1 - \frac{r^2m^2}{h^2} > 0,$$ which is the condition you need in order for Equation $(1)$ to be the equation of an ellipse.

A word of caution: all of this proves only that the orthogonal projection of the intersection of the cone and plane onto the $x,y$ plane is an ellipse. It does not immediately prove that the intersection itself is an ellipse. Technically, to complete the proof you have to show that under orthogonal projection onto the $x,y$ plane, the preimage of an ellipse is also an ellipse (or possibly a circle, although you should also be able to show that the preimage is a circle only if $m = 0$).

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