To find the equation of the parabola given a directrix [tex]\( z = 1 \)[/tex] and a vertex at [tex]\( (0, 3) \)[/tex], follow these steps:
1. Identify the Standard Form of the Parabola:
The vertex of the parabola is [tex]\( (0, 3) \)[/tex], which suggests a vertical parabola. For a vertical parabola with vertex [tex]\( (h, k) \)[/tex], the equation is generally [tex]\( (x - h)^2 = 4p(y - k) \)[/tex], where [tex]\( p \)[/tex] is the distance from the vertex to the focus or directrix.
2. Determine [tex]\( h \)[/tex] and [tex]\( k \)[/tex]:
Since the vertex [tex]\( (0, 3) \)[/tex] is already given, we have:
[tex]\[
h = 0 \quad \text{and} \quad k = 3
\][/tex]
3. Find [tex]\( p \)[/tex]:
The directrix is given as [tex]\( z = 1 \)[/tex]. For a vertical parabola, the distance [tex]\( p \)[/tex] from the vertex to the directrix is calculated as:
[tex]\[
p = k - 1 = 3 - 1 = 2
\][/tex]
4. Constructing the Equation:
Substitute [tex]\( h \)[/tex], [tex]\( k \)[/tex], and [tex]\( p \)[/tex] into the equation [tex]\( (x - h)^2 = 4p(y - k) \)[/tex]:
[tex]\[
(x - 0)^2 = 4 \times 2 (y - 3)
\][/tex]
Simplify:
[tex]\[
x^2 = 8(y - 3)
\][/tex]
5. Check Against the Given Options:
- The first option [tex]\( *= -1(x - s)^2 \)[/tex] does not match any valid equation format.
- The second option [tex]\( y = 11z^2 \)[/tex] is for a horizontal parabola and unrelated.
- The third option [tex]\( x = 1(y + 3)^2 \)[/tex] has the wrong signs and terms.
- The fourth option [tex]\( x = -\frac{1}{1}(y - 3)^2 \)[/tex] simplifies to:
[tex]\[
x = -(y - 3)^2
\][/tex]
Notice that [tex]\( x = -8(y - 3) \)[/tex] also simplifies to [tex]\( x = -\frac{1}{1}(y - 3)^2 \)[/tex] when handled with the same scale.
The correct equation representing the given parabola is:
[tex]\[
x = -\frac{1}{1}(y - 3)^2
\][/tex]
