Question

Find the standard form of the equation of the parabola with a focus at (-2, 0) and a directrix at x = 2.

Answer 1

Answer:

$y^2=\frac{1}{8}x$

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

$(x-h)=4p(y-k)^2$

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:

$(x-0)=4(2)(y-0)^2$

Simplify that a bit:

$x=8y^2$

Solving for y^2:

$y^2=\frac{1}{8}x$