Answer:
[tex]\textsf{C.} \quad x=\dfrac{1}{6}(y-15)^2-\dfrac{5}{2}[/tex]
Step-by-step explanation:
As the directrix is vertical, the parabola is sideways.
Conic form of a sideways parabola with a horizontal axis of symmetry:
[tex](y-k)^2=4p(x-h)\quad \textsf{where}\:p\neq 0[/tex]
- Vertex = (h, k)
- Focus = (h + p, k)
- Directrix: x = (h - p)
- Axis of symmetry: y = k
If p > 0, the parabola opens to the right, and if p < 0, the parabola opens to the left.
Given:
- Focus: (-1, 15)
- Directrix: x = -4
Therefore:
- k = 15
- h + p = -1
- h - p = -4
Add h + p = -1 to h - p = -4 to eliminate p:
⇒ 2h = -5
⇒ h = -5/2
⇒ p = 3/2
Substituting the found values into the formula:
[tex]\implies (y-15)^2=4\left(\dfrac{3}{2}\right)\left(x+\dfrac{5}{2}\right)[/tex]
[tex]\implies (y-15)^2=6\left(x+\dfrac{5}{2}\right)[/tex]
[tex]\implies \dfrac{1}{6}(y-15)^2=x+\dfrac{5}{2}[/tex]
[tex]\implies x=\dfrac{1}{6}(y-15)^2-\dfrac{5}{2}[/tex]
