Check the picture below.
now, let's keep in mind that, the vertex is half-way between the focus point and the directrix, it's a "p" distance from each other.
since this horizontal parabola is opening to the left-hand-side, "p" is negative, notice in the picture, "p" is 2 units, and since it's negative, p = -2.
its vertex is half-way between those two guys, so that puts the vertex at (-5, 3)
[tex]\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ 4p(y- k)=(x- h)^2 \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=7\\ p=-2 \end{cases}\implies 4(-2)[x-(-5)]=[y-7]^2 \\\\\\ -8(x+5)=(y-7)^2\implies x+5=\cfrac{(y-7)^2}{-8}\implies \boxed{x=-\cfrac{1}{8}(y-7)^2-5}[/tex]
