What would be the equation that represents a parabola that has a focus of (0, 0) and a directrix of y = 2?

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Answer 1 · 2018-04-27T12:46:18

The equation of parabola is #y=-1/4 x^2+1 #

Focus is at #(0,0) #and directrix is #y=2#. Vertex is at midway

between focus and directrix. Therefore vertex is at #(0,1)#

The vertex form of equation of parabola is

#y=a(x-h)^2+k ; (h.k) ;# being vertex. # h=0 and k = 1#

So the equation of parabola is #y=a(x-0)^2+1 #. Distance of

vertex from directrix is #d= 2-1=1#, we know # d = 1/(4|a|)#

#:. 1 = 1/(4|a|) or |a|=1/4# Here the directrix is above

the vertex , so parabola opens downward and #a# is negative.

#:.a= -1/4# The equation of parabola is #y=-1/4 x^2+1 #

graph{-1/4 x^2 +1 [-10, 10, -5, 5]} [Ans]