Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Find the equation of the parabola with its focus at [tex](6, 2)[/tex] and its directrix [tex]y = 0[/tex].

A. [tex]y = \frac{1}{4}(x-6)^2 + 1[/tex]
B. [tex]y = -\frac{1}{4}(x-6)^2 + 1[/tex]
C. [tex]y = 4(x-6)^2 + 1[/tex]
D. [tex]y = \frac{1}{4}(x-1)^2 + 6[/tex]

Sagot :

To find the equation of the parabola with its focus at [tex]\( (6, 2) \)[/tex] and its directrix [tex]\( y = 0 \)[/tex], we need to follow these steps:

1. Identify the vertex:
- The vertex of a parabola is equidistant from its focus and its directrix.
- Since the focus is at [tex]\( (6, 2) \)[/tex] and the directrix is [tex]\( y = 0 \)[/tex], the vertex will lie exactly in the middle of the focus and directrix along the [tex]\( y \)[/tex]-axis.
- The [tex]\( x \)[/tex]-coordinate of the vertex will be the same as the [tex]\( x \)[/tex]-coordinate of the focus, which is [tex]\( 6 \)[/tex].
- The [tex]\( y \)[/tex]-coordinate of the vertex can be found by averaging the [tex]\( y \)[/tex]-coordinates of the focus and the directrix. So,
[tex]\[ y_{\text{vertex}} = \frac{2 + 0}{2} = 1 \][/tex]
- Therefore, the vertex is at [tex]\( (6, 1) \)[/tex].

2. Determine the value of [tex]\( p \)[/tex]:
- [tex]\( p \)[/tex] is the distance between the vertex and the focus or the directrix.
- Since the directrix is [tex]\( y = 0 \)[/tex] and the vertex is at [tex]\( y = 1 \)[/tex], the distance [tex]\( p \)[/tex] is:
[tex]\[ p = 1 \][/tex]

3. Formulate the equation in the standard form:
- The standard form for the equation of a parabola opening upwards or downwards is:
[tex]\[ (x - h)^2 = 4p(y - k) \][/tex]
where [tex]\( (h, k) \)[/tex] is the vertex. Since [tex]\( p \)[/tex] is positive, the parabola opens upwards.
- In our case, [tex]\( h = 6 \)[/tex], [tex]\( k = 1 \)[/tex], and [tex]\( p = 1 \)[/tex]. So the equation becomes:
[tex]\[ (x - 6)^2 = 4 \cdot 1 \cdot (y - 1) = 4(y - 1) \][/tex]
Simplifying for [tex]\( y \)[/tex]:
[tex]\[ (x - 6)^2 = 4(y - 1) \][/tex]
[tex]\[ y = \frac{(x - 6)^2}{4} + 1 \][/tex]
Hence, the equation is:
[tex]\[ y = \frac{1}{4}(x - 6)^2 + 1 \][/tex]

4. Compare with the given options:

Option A: [tex]\( y = \frac{1}{4}(x - 6)^2 + 1 \)[/tex]

Option B: [tex]\( y = -\frac{1}{4}(x - 6)^2 + 1 \)[/tex]

Option C: [tex]\( y = 4(x - 6)^2 + 1 \)[/tex]

Option D: [tex]\( y = \frac{1}{4}(x - 1)^2 + 6 \)[/tex]

From the above, we see that Option A matches our equation.

Therefore, the correct answer is:

A) [tex]\( y = \frac{1}{4}(x - 6)^2 + 1 \)[/tex]