Proving equations for conic sections? How can we prove that the equations for conic sections are, indeed, sections of a cone? My guess is that it involves some sort of equality with the quadric surface equation for a cone, but I can't quite figure it out.
 A: Set the origin at the vertex of the cone, and the $z$ axis along the axis of the cone. Then the cone has equation 
$$z^2=a(x^2+y^2)$$
Now, if you intersect the cone to a plane, the plane will have equation $ax+by+cz=d$. If you solve this for $x,y$ or $z$, depending on their coefficients not being zero, you will get an equation in two letters of degree at most $2$. That equation is either degenerated, or if quadratic, it represents one of the 4 conics.
Now, to prove that any second degree equation in $x,y$ produces a conic:
Consider the equation 
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
where at least one of $A,B,C$ is non-zero.
Consider the quadratic form 
$$Ax^2+Bxy+Cy^2$$
by orthogonally diagonalizing the corresponding matrix, you get a change of variable $(u,v)^T=P(x,y)^T$ such that the equation becomes
$$\lambda_1 u^2+ \lambda_2 v^2+ Gu+Hv+I=0$$
Note that the change of variable is given by an orthonormal matrix, therefore it corresponds to a rotation and/or reflection. If we can show that this corresponds to a conic section we are done. 
By completing the square, we get a new change of variable $u'=u- \alpha, v'=v=\beta$, which corresponds to a translation, such that we get
$$\lambda_1 u'^2+\lambda_2 v'^2 =J$$
Now, all you have to do is prove that this represents the intersection between a cone $$z^2=a(x^2+y^2)$$ and a plane of the form $z= bx$ or $z=cy$ or $z=d$, for the right values of $a,b,c$.
