Write an equation for a parabola in which the set of all points in the plane are equidistant from the focus and line. F(0, –5); y = 5

well, "the set of all points in the plane equidistant from the focus and a line", is referring to the focus point of the parabola and the directrix line.

bearing in mind that, both fellows are at a distance "p" from the vertex, that puts the vertex right in the middle of them.

notice, the focus point is below the directrix, meaning, is a vertical parabola, and is also opening downwards, like in the picture below.

from -5 to 5 over the y-axis, there are 10 units, so the "p" distance is 5, so that puts the vertex right at the origin.

since the parabola is opening downwards, "p" is negative, thus -5.

[tex]\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{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}\\\\ -------------------------------\\\\ \begin{cases} h=0\\ k=0\\ p=-5 \end{cases}\implies 4(-5)(y-0)=(x-0)^2 \\\\\\ -20y=x^2\implies y=-\cfrac{1}{20}x^2[/tex]