Question

(Algebra 2 Conics):

Answer 1

$\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ 4p(y- k)=(x- h)^2 \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ -\cfrac{1}{4}(y+2)^2=x-7\implies -\cfrac{1}{4}[y-(-2)]^2=x-7 \\[2em] [y-(-2)]^2=-4(x-7)\implies [y-(\stackrel{k}{-2})]^2=4(\stackrel{p}{-1})(x-\stackrel{h}{7})$

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.