How do you find the equation of a parabola with vertex at the origin and focus (0, -3/2)?

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Answer 1 · 2018-03-15T11:56:16

The equation of parabola is #y=-1/6x^2 #

Focus is at #(0,-3/2) #and vertex is at #0,0#. Vertex is at midway

between focus and directrix. Therefore ,directrix is #y=3/2#.

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=ax^2 #

Here directrix is above the vertex , so parabola opens downward

and #a# is negative. Distance of vertex from directrix is

#d= 3/2-0=3/2#, we know # d = 1/(4|a|):. 3/2 = 1/(4|a|)# or

# |a|= 2/(3*4)=1/6 :. a= -1/6#.

So the equation of parabola is #y=-1/6x^2 #

graph{-1/6 x^2 [-10, 10, -5, 5]} [Ans]