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To determine the equation of the parabola with a focus at [tex]\((-3,0)\)[/tex] and a directrix at [tex]\(x=3\)[/tex], let's follow these steps:
1. Identify the distance between the focus and the directrix:
- The focus is at [tex]\((-3,0)\)[/tex].
- The directrix is the line [tex]\(x = 3\)[/tex].
2. Calculate the midpoint (vertex) between the focus and the directrix:
- The vertex of a parabola is always equidistant from the focus and the directrix.
- The distance between [tex]\(-3\)[/tex] and [tex]\(3\)[/tex] is [tex]\(3 - (-3) = 6\)[/tex].
- The vertex will be at the midpoint of this segment, which is [tex]\(\frac{3 + (-3)}{2} = 0\)[/tex] on the x-axis and the same y-coordinate as the focus and directrix (since they're symmetrical about the y-axis), which is [tex]\(0\)[/tex].
- Therefore, the vertex is at the point [tex]\((0, 0)\)[/tex].
3. Determine the value of [tex]\(p\)[/tex], the distance from the vertex to the focus (or directrix):
- Since the distance between the focus [tex]\((-3,0)\)[/tex] and the vertex [tex]\((0,0)\)[/tex] is [tex]\(3\)[/tex], [tex]\(p = 3\)[/tex].
4. Formulate the standard equation of the parabola:
- For a parabola with a horizontal axis and vertex at [tex]\((h,k) = (0,0)\)[/tex], the equation is [tex]\((y - k)^2 = 4p(x - h)\)[/tex].
- Substituting [tex]\(k = 0\)[/tex], [tex]\(h = 0\)[/tex], and [tex]\(p = 3\)[/tex], we get:
[tex]\[ y^2 = 4p(x) = 4(3)x = 12x. \][/tex]
- Since the parabola opens to the left (as the focus is on the left side of the vertex), [tex]\(p\)[/tex] should be negative. Thus, [tex]\(4p = 4(-3) = -12\)[/tex].
5. Write the final equation:
- Substituting in the negative value for [tex]\(p\)[/tex], the equation becomes:
[tex]\[ y^2 = -12x. \][/tex]
Hence, the correct equation for the parabola is:
[tex]\[ \boxed{y^2 = -12x} \][/tex]
1. Identify the distance between the focus and the directrix:
- The focus is at [tex]\((-3,0)\)[/tex].
- The directrix is the line [tex]\(x = 3\)[/tex].
2. Calculate the midpoint (vertex) between the focus and the directrix:
- The vertex of a parabola is always equidistant from the focus and the directrix.
- The distance between [tex]\(-3\)[/tex] and [tex]\(3\)[/tex] is [tex]\(3 - (-3) = 6\)[/tex].
- The vertex will be at the midpoint of this segment, which is [tex]\(\frac{3 + (-3)}{2} = 0\)[/tex] on the x-axis and the same y-coordinate as the focus and directrix (since they're symmetrical about the y-axis), which is [tex]\(0\)[/tex].
- Therefore, the vertex is at the point [tex]\((0, 0)\)[/tex].
3. Determine the value of [tex]\(p\)[/tex], the distance from the vertex to the focus (or directrix):
- Since the distance between the focus [tex]\((-3,0)\)[/tex] and the vertex [tex]\((0,0)\)[/tex] is [tex]\(3\)[/tex], [tex]\(p = 3\)[/tex].
4. Formulate the standard equation of the parabola:
- For a parabola with a horizontal axis and vertex at [tex]\((h,k) = (0,0)\)[/tex], the equation is [tex]\((y - k)^2 = 4p(x - h)\)[/tex].
- Substituting [tex]\(k = 0\)[/tex], [tex]\(h = 0\)[/tex], and [tex]\(p = 3\)[/tex], we get:
[tex]\[ y^2 = 4p(x) = 4(3)x = 12x. \][/tex]
- Since the parabola opens to the left (as the focus is on the left side of the vertex), [tex]\(p\)[/tex] should be negative. Thus, [tex]\(4p = 4(-3) = -12\)[/tex].
5. Write the final equation:
- Substituting in the negative value for [tex]\(p\)[/tex], the equation becomes:
[tex]\[ y^2 = -12x. \][/tex]
Hence, the correct equation for the parabola is:
[tex]\[ \boxed{y^2 = -12x} \][/tex]
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