Find the focus of the parabola (x-3)^2=8(y+1)
To find the focus of the parabola given by the equation (x-3)^2 = 8(y+1), we need to first rewrite the equation in the standard form of a parabola: (x-h)^2 = 4p(y-k), where (h,k) is the vertex of the parabola and p is the distance between the vertex and the focus.
Rewriting the given equation in the standard form:
(x-3)^2 = 8(y+1)
=> (x-3)^2 = 8(y+1)
=> (x-3)^2 = 8(y+1)
=> (x-3)^2 = 8(y+1)/8
=> (x-3)^2 = (y+1)/8
=> (x-3)^2 = (1/8)(y+1)
Comparing this with the standard form (x-h)^2 = 4p(y-k), we see that h = 3, k = -1, and p = 1/8. Therefore, the vertex is V(3,-1) and p = 1/8.
The focus of the parabola is p units above the vertex along the axis of symmetry. In this case, the focus is 1/8 units above the vertex along the axis of symmetry.
Therefore, the focus of the parabola is at the point (3, -1 + 1/8) or (3, -7/8).