Question

What is the focus of the parabola given by the equation (y − 3)^2 = 8(x − 5)? A. (-7, 3) B. (-5, 3) C. (5, 3) D. (7, 3) E. (3, -7)

Answer 1

$\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} \boxed{4p(x- h)=(y- k)^2} \\\\ 4p(y- k)=(x- h)^2 \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ (y-3)^2=8(x-5)\implies (y-\stackrel{k}{3})^2=4(\stackrel{p}{2})(x-\stackrel{h}{5})\qquad vertex~(5,3)$

let's notice, the squared variable is the "y", and therefore is a horizontal parabola, with a vertex as you see at 5,3.

the coefficient of the y² is positive, meaning it opens to right-hand-side.

the "p" distance is 2 units, since it's opening to the right, is positive 2.

so from the vertex at 5,3 we move to the right 2 units, to land at (7, 3).