Question

1. write an equation for a parabola in which the set of all points in the plane are equidistant from the focus and line. f(-2, 0); x = 2

options:
x = \frac{1}{4}y^2
x = -\frac{1}{4}y^2
x = -\frac{1}{8}y^2
x = \frac{1}{8}y^2

Explanation:

Step1: Identify parabola parameters

Focus $F(-2,0)$, directrix $x=2$. For a horizontal parabola, the standard form is $x = \frac{1}{4p}y^2$, where $p$ is the distance from vertex to focus.

Step2: Calculate value of $p$

The vertex is the midpoint between focus and directrix. The x-coordinate of vertex is $\frac{-2+2}{2}=0$, so vertex is $(0,0)$. $p$ is the x-coordinate of focus: $p=-2$.

Step3: Substitute $p$ into standard form

$$x = \frac{1}{4(-2)}y^2$$
Simplify the coefficient: $\frac{1}{4(-2)} = -\frac{1}{8}$

Answer:

$\boldsymbol{x = -\frac{1}{8}y^2}$ (corresponding to the option: $\boldsymbol{x = -\frac{1}{8}y^2}$)