Question

Identify the vertex focus and directrix. X=1/8y^2

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}\\\\ -------------------------------\\\\ x=\cfrac{1}{8}y^2\implies 8x=y^2\implies \stackrel{4p}{8}(x-\stackrel{h}{0})=(y-\stackrel{k}{0})^2 \\\\\\ 4p=8\implies p=2`

first off, notice that the equation has a squared variable of "y", meaning the parabola is a horizontal one, the "p" distance is 2, so "p" is positive, and the vertex is at 0,0, namely the origin.

if we move from the origin 2 units to the right, we'll land on the focus point at (2,0), and if we move 2 units in the opposite direction, we'll land on the directrix, check the picture below.