What is the equation for a parabola with Focus at (4,0), directrix is y=-4?

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Answer 1 · 2018-03-04T00:17:46

#y = 1/8(x-4)^2-2#

Given: parabola with focus #(4,0)#, directrix: #y=-4#

General equation: #y = a(x-h)^2 +k#, where vertex:#(h, k)#,

focus:#(h, k+p)#, directrix:#" "y = k - p#, #" "a = 1/(4p)#

From the focus: #h = 4, " " k+p = 0#

From the directrix: #k-p = -4#

Using elimination:
#k + p = 0#
#ul(k-p=-4)#
#2k = -4; " " k = -2#

#-2 + p = 0; " " p = 2#

vertex: #(h, k) = (4, -2)#

The vertical distance from the vertex to the directrix: #p = 2#

#y = a(x-4)^2-2#

#a = 1/(4p) = 1/(4*2) = 1/8#

Parabola equation: #" "y = 1/8(x-4)^2-2#