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Find the equation for the parabola with focus at (-1,-1) and vertex at (4,-1)
The focus S(-1,-1) and the vetex A(4,-1) are on the line y =-1 which is a horizontal line parallel to the x-axis at a distance 1 unit below it
Axis horizontal means the parabola is square in y and linear in x
The focus is to the left of the vertex, which means the parabola is facing west
The equation to the parabola will be of the form
(y-k)^2 = -4a(x-h)----(1)
Where A(h,k ) = A(4,-1) and a = distance of the focus from the vertex.
The distance SA = (one unit to the y-axis plus 4 units to the vertex) = 1+4 = 5
Therefore putting h =4, k = -1 and a = 5
Therefore (1) becomes
[y-(-1)]^2 = -4X5(x-4)
(y+1)^2 = -20(x-4)
y^2+2y+20x-79 = 0
which DOES NOT tally with any of the following given answers:
Remark: Either the set of answers given is wrong or the given particulars about the parabola are wrong. Please give the correct problem