<s>[INST] How to determine the intersection of a plane and a cone? I want to know what is the intersection of the plane $2x+y+z-2=0$ and $x^2+y^2=z^2$. My guess is that it is an ellipse or an hyperbola, but I can't understand how to determine which one.
Subsituting $z=2-2x-y$ in the cone equation, we obtain
$$
5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0
$$
and I can't undertsand whether it is an hyperbola or an ellipse.
Will you please help me figure it out? 
Thanks
 [/INST] Making the change of variables ( a rigid rotation does not alter the form)
$$
x = X\cos\theta-Y\sin\theta\\
y = X\sin\theta+Y\cos\theta
$$
in
$$
f(x,y) = 5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0
$$
we get at
$$
f(x,y)\equiv F(X,Y) = \frac{1}{2} (4 \sin (2 \theta)+3 \cos (2 \theta)+7)X^2+(4 \cos (2 \theta)-3 \sin (2 \theta))X Y+\frac{1}{2} (7-4 \sin (2 \theta)-3 \cos (2 \theta))Y^2-12\cos\theta X+12\sin\theta Y+4=0
$$
and now choosing $\theta$ such that $4\cos(2\theta)-3\sin(2\theta) = 0\to \theta = \frac 12 \arctan(\frac 43)$ we obtain finally
$$
F(X,Y) =6 X^2-\frac{24 X}{\sqrt{5}}+Y^2+\frac{12 Y}{\sqrt{5}}+4 = 0
$$
which is clearly an ellipse. 
NOTE
As can be verified the intersection of the cone $x^2+y^2=z^2$ and the plane $2x+y+z-2=0$ gives
$$
f(x,y) = 4 - 8 x + 3 x^2 - 4 y + 4 x y = 0
$$
and not 
$$
f(x,y) = 5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0
$$
as assumed. In this case with the same rotation the true intersection curve should be
$$
F(X,Y) = 4 X^2-4 \sqrt{5} X-Y^2+4
$$
which is a hyperbola.
</s>