To find the equation of the graph that represents a set of all points in the plane equidistant from the point [tex]\( F(0, 5) \)[/tex] and the line [tex]\( y = -5 \)[/tex], you can follow these detailed steps:
1. Identify the focus and directrix:
- The focus of the parabola is [tex]\( F(0, 5) \)[/tex].
- The directrix is the line [tex]\( y = -5 \)[/tex].
2. Determine the vertex:
- The vertex of the parabola is the midpoint between the focus and the directrix.
- The focus is at [tex]\( y = 5 \)[/tex] and the directrix is at [tex]\( y = -5 \)[/tex].
- The midpoint of the y-coordinates of the focus and directrix is [tex]\( \frac{5 + (-5)}{2} = 0 \)[/tex].
- Thus, the vertex is at [tex]\( (0, 0) \)[/tex].
3. Determine the value of [tex]\( p \)[/tex]:
- [tex]\( p \)[/tex] is the distance from the vertex to the focus or the directrix.
- The distance from the vertex at [tex]\( y = 0 \)[/tex] to the focus at [tex]\( y = 5 \)[/tex] is [tex]\( 5 \)[/tex].
- Hence, [tex]\( p = 5 \)[/tex].
4. Write the standard form of the equation for the parabola:
- The standard form of a parabola with the vertex at [tex]\( (h, k) \)[/tex] and opening upwards or downwards is [tex]\( (x - h)^2 = 4p(y - k) \)[/tex].
- For our case, [tex]\( h = 0 \)[/tex], [tex]\( k = 0 \)[/tex], and [tex]\( p = 5 \)[/tex].
5. Substitute in the standard form:
- [tex]\( (x - 0)^2 = 4 \times 5 (y - 0) \)[/tex]
- [tex]\( x^2 = 20y \)[/tex]
6. Rewrite the equation in the form [tex]\( y = kx^2 \)[/tex]:
- [tex]\( y = \frac{1}{20} x^2 \)[/tex]
7. Results:
- Thus, the equation for the graph that is equidistant from the point [tex]\( F(0, 5) \)[/tex] and the line [tex]\( y = -5 \)[/tex] is [tex]\( y = \frac{1}{20} x^2 \)[/tex].
So, the correct selection from the given options is:
[tex]\[ y = \frac{1}{20} x^2 \][/tex]
Thus, the answer is [tex]\( y = \frac{1}{20} x^2 \)[/tex].
