What is the equation for a parabola with a vertex at (5,-1) and a focus at (3,-1)?

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Answer 1 · 2017-03-25T16:03:24

#x=-1/8(y+1)^2+5#

Since the #y#-coordinates of the vertex and focus are the same, vertex is to the right of focus.

Hence, this is a regular horizontal parabola and as vertex #(5,-1)# is to the right of focus, it opens to the left.and #y# part is squared.

Therefore, equation is of the type

#(y+1)^2=-4p(x-5)#

As vertex and focus are #5-3=2# units apart, then #p=2# equation is

#(y+1)^2=-8(x-5)# or #x=-1/8(y+1)^2+5#

graph{x=-1/8(y+1)^2+5 [-21, 19, -11, 9]}