My attempt:
General equation $$ ax^2+a'y^2+a''z^2+2byz+2b'xz+2b''xy=0.$$ (conic section is the intersection of this conic with the plane $z=1$).
Expressing that $(0,0,1$) lies on this conic: $a''=0$. So we have $\phi(x,y,z)=ax^2+a'y^2+2byz+2b'xz+2b''xy=0$.
The axis is parallel to $x=0$. A parabola has a unique axis that corresponds to the eigenvector of $\lambda = a+a'$. (eigenvector of the matrix $\begin{pmatrix} a&b'' \\ b''&a'\end{pmatrix}$, eigenvalue $0$ corresponds to the line at infinity, which is not an affine line and therefore not an axis).
Directions $(x_1,y_1,0), (-y_1,x_1,0)$ are perpendicular to each other so that they satisfy $ b''x_1^2+(a'-a)x_1y_1-b''y_1^2=0.$
Now I'm stuck and I honestly feel like this is a lot of extra unnecessary work. Can someone either help me with my approach, or is there an easier way to find the required equation?
Thanks.