For this problem I cannot figure out how to find the directix, defined to be perpendicular to the axis of symmetry.
The general point $(x,y)$ on parabola needs to be far from $(-1,0)$ as it is from the line $x=7$. I am confused on how they got $7$ for the directix.
From my understanding, since the focus and vertex lie on the x-axis, the directix must be equidistant between the focus and the directix. So the directix is a line such that the vertex is four units form the focus and four units from the directix. At first glance my thought is that the answer should be $8$ such that the vertex is equidistant midpoint for the focus and the directix.
I tried to conceptualize it, since the directix is perpendicular maybe set up the distance formula for the focus to the general point equal to distance from general point to on the parabola to see why the directix is $7$:
$\sqrt{(x-1)^2 + (y-0)^2} = $
This parabola has a horizontal axis, so we examine the standard form for such a parabola: $(y - k)^{2} = 4p(x - h)^{2}.$ Since the vertex is at $(3, 0),$ we now have $y^{2} = 4p(x - 3).$ To find the value of $p,$ we notice that the focal length is $(-1) - 3 = -4.$ This is the value of $p.$ Our final equation is $\boxed{y^{2} = -16(x - 3)^{2}}.$
If you are also looking for the equation of the directrix, note that the vertex is the midpoint of the segment between the focus and the corresponding point on the directrix. The directrix thus has equation $\boxed{x = 7}.$