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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.

MyWorld
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    All non-degenerate parabolas are affinely equivalent. Pick a convenient one and apply a suitable transformation. Another possible approach: a parabola is tangent to the line at infinity at the intersection with its axis. Also, have you considered that the solution is not unique? You have two points, a tangent and one more constraint on the coefficients from the requirement that the curve be a parabola. That’s not enough information to determine a conic uniquely. – amd Jun 13 '19 at 18:08
  • I found (chosing $b=1$) that a solution is $ax^2+2yz+2b''xy=0$. Does this seem correct? – MyWorld Jun 13 '19 at 18:59
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    No. The presence of the cross term in $xy$ means that the conic is not aligned with the coordinate axes. – amd Jun 13 '19 at 19:05
  • My bad. I took $(0,1,0)$ at infinity. The tangent line in this point is given by $\frac{\partial \phi} {\partial y}(x, y, z) =0$, or $a'y+bz+b''x=0$. This line has to correspond with $z=0$, so $b\ne 0$ and $a' =b''=0$. So a solution is, chosing $b=1$, given by $ax^2+2yz+2b'xz=0$. I hope this is correct. – MyWorld Jun 13 '19 at 19:33
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    Could be. What does this look like with $z=1$? – amd Jun 13 '19 at 19:57
  • Looks like you’ve gotten far enough to write up your own answer to this question. – amd Jun 15 '19 at 01:40

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