Answer:
[tex]y^2=\frac{1}{8}x[/tex]
Step-by-step explanation:
The focus lies on the x axis and the directrix is a vertical line through x = 2. The parabola, by nature, wraps around the focus, or "forms" its shape about the focus. That means that this is a "sideways" parabola, a "y^2" type instead of an "x^2" type. The standard form for this type is
[tex](x-h)=4p(y-k)^2[/tex]
where h and k are the coordinates of the vertex and p is the distance from the vertex to either the focus or the directrix (that distance is the same; we only need to find one). That means that the vertex has to be equidistant from the focus and the directrix. If the focus is at x = -2 y = 0 and the directrix is at x = 2, midway between them is the origin (0, 0). So h = 0 and k = 0. p is the number of units from the vertex to the focus (or directrix). That means that p=2. We fill in our equation now with the info we have:
[tex](x-0)=4(2)(y-0)^2[/tex]
Simplify that a bit:
[tex]x=8y^2[/tex]
Solving for y^2:
[tex]y^2=\frac{1}{8}x[/tex]
Answer: x = -1/8y^2
Step-by-step explanation
Focus: (-2,0)
Directrix: x=2
It meets the criteria.
