Question 30464: (M) Question:
Find an equation for the parabola with focus at (-1,1) and vertex at (4,1)
(A) x^2-8x+20y+28=0
(B) x^2-8x+20y+12=0
(C) y^2-2y+20x+81=0
(D) y^2+2y+20x-79=0
Thanks!
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! 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 above 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+1= -20x+80
y^2-2y+20x+1-80 =0
y^2-2y+20x-79 = 0
Answer:y^2-2y+20x-79 = 0
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which DOES NOT tally with any of the following given answers:
(A) x^2-8x+20y+28=0
(B) x^2-8x+20y+12=0
(C) y^2-2y+20x+81=0
(D) y^2+2y+20x-79=0
Remark: Either the set of answers given is wrong or the given particulars about the parabola are wrong. Please give the correct problem
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