Intersection of a cone and a plane. I need a proof that the intersection of a cone with a plane parallel to the cone's axis is a hyperbola.
 A: Cone equation: $x^2+y^2 = z^2 \hspace{3mm}$    (1) 
Plane equation parallel to the cone's axis: $ax+by = c \hspace{3mm} $  (2)
If you isolate the variable $y$ in equation (2) and substitute in (1), you will find a equation in the form
$$\frac{z^2}{d_z^2} -\frac{(x-x_0)^2}{d_x^2} = 1,$$ where $x_{0}, d_x$ and $ d_y$ are constants.
Remark 1: if $b=0$, isolate $x$ and you have a similar result.
Remark 2: For simplicity, I have considered  the vertex of the cone in $(0,0,0)$. The general case is just a translation/rotation.
A: Define the cone as points $(x,y,z)$ on the surface defined by the equation
$$
z^2=x^2+y^2
$$
and make the cut $y=\sqrt k$. Then
$$
z^2-x^2=k
$$
If the plane is not of the form $y=\sqrt k$ we can always rotate the situation around the cone's axis (using the symmetry of the cone) to obtain such a situation.

Comparing $z^2-x^2=k$ to the $y=\frac{a}{x}$ version of a hyperbola, one can make a change of coordinates:
We can rotate $(x,z)$ in the $xz$-coordinate system by an angle of $45$ degrees clockwise to obtain $(x',z')=\frac{\sqrt 2}{2}(x+z,-x+z)$. This will make $(x',z')$ describe the same curve as $(x,z)$, but with the axis of symmetry rotated from the $z$-axis to the line $x'=z'$:

Considering the inverse rotation we see that $(x,z)=\frac{\sqrt 2}{2}(x'-z',x'+z')$. So with this the equation $z^2-x^2=k$ becomes
$$
\left(\tfrac{\sqrt 2}2(x'+z')\right)^2-\left(\tfrac{\sqrt 2}2(x'-z')\right)^2=k
$$
and since $\left(\frac{\sqrt2}2\right)^2=\frac{1}{2}$ and $(x'+z')^2-(x'-z')^2=4x'z'$ it follows that
$$
2x'z'=k\iff z'=\frac{0.5k}{x'}
$$
which is an equation of the form $y=\frac{a}{x}$. The figure above shows as an example how $z^2-x^2=4$ transforms into $z'=\frac{2}{x'}$ after being rotated 45 degrees.
