Answer:
-36
Step-by-step explanation:
![\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}\\\\ -\cfrac{1}{4}(y+2)^2=x-7\implies -\cfrac{1}{4}[y-(-2)]^2=x-7 \\[2em] [y-(-2)]^2=-4(x-7)\implies [y-(\stackrel{k}{-2})]^2=4(\stackrel{p}{-1})(x-\stackrel{h}{7})](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%204p%28x-%20h%29%3D%28y-%20k%29%5E2%20%5C%5C%5C%5C%204p%28y-%20k%29%3D%28x-%20h%29%5E2%20%5Cend%7Barray%7D%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20vertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20-%5Ccfrac%7B1%7D%7B4%7D%28y%2B2%29%5E2%3Dx-7%5Cimplies%20-%5Ccfrac%7B1%7D%7B4%7D%5By-%28-2%29%5D%5E2%3Dx-7%20%5C%5C%5B2em%5D%20%5By-%28-2%29%5D%5E2%3D-4%28x-7%29%5Cimplies%20%5By-%28%5Cstackrel%7Bk%7D%7B-2%7D%29%5D%5E2%3D4%28%5Cstackrel%7Bp%7D%7B-1%7D%29%28x-%5Cstackrel%7Bh%7D%7B7%7D%29)
so h = 7, k = -2, meaning the vertex is at (7, -2).
the squared variable is the "y", meaning is a horizontal parabola.
the "p" distance is negative, for a horizontal parabola that means, it's opening towards the left-hand-side.
we know the focus and directrix are "p" units away from the vertex, and we know the parabola is opening horizontally towards the left-hand-side.
the focus is towards it opens 1 unit away, at (6, -2).
the directrix is on the opposite direction, 1 unit away, at (8, -2), namely x = 8.
The LCM of 34 using prime factorisation method is 2 and 17