Hello Miguel,
One of the wonderful properties of parabolic curves is that for any point on the parabola, this point will be equidistant from the focus (a point) and the directrix (a line perpendicular to the axis of symmetry).
Let's let the parabola's axis of symmetry be the $y$-axis and its vertex be at the origin (don't worry, we can use a vertical translation to put the focus at the origin after we are done). So, if the focal point is $(0,F)$, then the directrix must be the line $y=-F$.
Thus, for some point $(x,y)$ on the parabola, we must have the square of the distance from this point to the focus being equal to the square of the distance from this point to the directrix. So, we may state:
$$(x-0)^2+(y-F)^2=(x-x)^2+(y+F)^2$$
Simplify:
$$x^2+(y-F)^2=(y+F)^2$$
Expand:
$$x^2+y^2-2Fy+F^2=y^2+2Fy+F^2$$
Collect like terms:
$$x^2=4Fy$$
Solve for $y$:
$$y=\frac{1}{4F}x^2$$
Now, if we wish to shift this curve vertically, so that the focus is at the origin, we may write:
$$y=\frac{1}{4F}x^2-F$$
Now, in the given problem, we are told the feedhorn (the focal point) is nine inches from the vertex, and so our cross-section becomes:
$$y=\frac{1}{4\cdot9}x^2-9=\frac{1}{36}x^2-9$$
