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Select the correct answer.

Given the focus [tex]$(-1,15)$[/tex] and the directrix [tex]$x=-4$[/tex], what is the equation of the parabola?

A. [tex]$x=-\frac{1}{6}(y-15)^2-\frac{5}{2}$[/tex]
B. [tex]$x=\frac{1}{6}(y+15)^2+\frac{5}{2}$[/tex]
C. [tex]$x=\frac{1}{6}(y-15)^2-\frac{5}{2}$[/tex]
D. [tex]$y=\frac{1}{6}(x-15)^2-\frac{5}{2}$[/tex]


Sagot :

To determine the equation of a parabola given the focus [tex]\((-1, 15)\)[/tex] and the directrix [tex]\(x = -4\)[/tex], we need to use the properties of parabolas. Here's the step-by-step solution:

1. Identify the Vertex:
- The vertex lies equidistant from the focus and the directrix. The x-coordinate of the vertex is halfway between [tex]\(-1\)[/tex] (focus) and [tex]\(-4\)[/tex] (directrix).
- Midpoint formula for x-coordinate: [tex]\(x_{vertex} = \frac{-1 + (-4)}{2} = \frac{-5}{2} = -2.5\)[/tex].

The y-coordinate of the vertex remains the same as the y-coordinate of the focus, which is 15.
- Therefore, the vertex is at [tex]\(\left(-2.5, 15\right)\)[/tex].

2. Determine the Equation Form:
- Since the directrix is vertical ([tex]\(x = -4\)[/tex]), the parabola opens horizontally.
- The equation of a parabola that opens horizontally is in the form [tex]\((y - k)^2 = 4p(x - h)\)[/tex], but for our convenience (and the form of the given options), we'll use the form [tex]\(x = a(y - k)^2 + h\)[/tex], where [tex]\((h, k)\)[/tex] is the vertex.

3. Calculate Parameters [tex]\(a\)[/tex] and [tex]\(h\)[/tex]:
- From the vertex [tex]\(\left(-2.5, 15\right)\)[/tex], we have [tex]\(h = -2.5\)[/tex] and [tex]\(k = 15\)[/tex].
- The distance [tex]\(p\)[/tex] is the distance from the vertex to the focus, which is the same as the distance from the vertex to the directrix.
- Distance [tex]\(p = |-1 - (-4)| / 2 = 1.5\)[/tex].

4. Find the Value of [tex]\(a\)[/tex]:
- [tex]\(a = \frac{1}{4p} = \frac{1}{6}\)[/tex].

5. Form the Equation:
- Substituting [tex]\(a = \frac{1}{6}\)[/tex], [tex]\(h = -2.5\)[/tex], and [tex]\(k = 15\)[/tex] into the horizontal parabola formula:
[tex]\[ x = \frac{1}{6}(y - 15)^2 - 2.5. \][/tex]

6. Choose the Correct Option:
- The equation matches option C precisely.

Therefore, the correct choice is:

[tex]\[ \boxed{C. \ x = \frac{1}{6}(y - 15)^2 - \frac{5}{2}}. \][/tex]