How do you find the standard form of the equation of the parabola with a focus at (0, -6) and a directrix at y = 6?

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Answer 1 · 2018-03-07T12:13:04

The equation of parabola is #y=-1/24x^2 #.

Focus is at #(0,-6) #and directrix is #y=6#. Vertex is at midway

between focus and directrix. Therefore vertex is at #(0,(-6+6)/2)#

or at #(0,0)# . The vertex form of equation of parabola is

#y=a(x-h)^2+k ; (h.k) ;# being vertex. # h=0 and k =0#

So the equation of parabola is #y=a(x-0)^2+0 or y =ax^2 #.

Distance of vertex from directrix is #d=6-0=6#, we know

# d = 1/(4|a|) :. 6 = 1/(4|a|) or |a|= 1/(4*6)=1/24#. Here the directrix

is above the vertex , so parabola opens downward and #a# is

negative #:a=-1/24#. Hence the equation of parabola is

#y=-1/24x^2 #.
graph{-1/24 x^2 [-160, 160, -80, 80]} [Ans]