x^2 = 20(y+1)
None of the other equations represent a parabola with the given focus and vertex.
x^ = 20(y+1)
x^ = 40(y+6)
20y = x^2 -20
40y = (x+6)^2
40y = x^2 - 240
None of the other equations represent a parabola with the given focus and vertex.
The general form of the equation of a parabola with a vertical axis of symmetry is given by:
(x-h)^2 = 4p(y-k)
where (h,k) is the vertex and p is the distance between the focus and the vertex.
Given that the vertex is (0, -6), h = 0 and k = -6.
So, the equation becomes:
(x-0)^2 = 4p(y+6)
x^2 = 4p(y+6)
In this case, since the focus is (0, 4), it is 10 units away from the vertex. Therefore, p = 10.
Substitute p = 10 into the equation:
x^2 = 40(y+6)
Therefore, the equation x^2 = 40(y + 6) also represents a parabola with the focus (0, 4) and the vertex (0, -6).
So, the equations that represent a parabola with the focus (0, 4) and the vertex (0, -6) are:
1. x^2 = 20(y+1)
2. x^2 = 40(y+6)
2. x^2 = 40(y+6) works ok
The correct equation that represents a parabola with the focus (0, 4) and the vertex (0, -6) is:
2. x^2 = 40(y+6)
Thank you for pointing out the mistake.