What is the equation of the parabola with a vertex at (0,0) and a focus at (0,-3)?

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Answer 1 · 2017-07-21T06:23:36

#y=-1/12x^2#

As vertex is #(0,0)# and focus is #(0,-3)#, the equation is of the form

#y=-ax^2# and directrix is #y=3#

As every point on parabola is equidistant from focus and directrix, and the equation would be

#x^2+(y+3)^2=(y-3)^2#

or #x^2+y^2+6y+9=y^2-6y+9#

or #12y=-x^2#

or #y=-1/12x^2#

graph{(12y+x^2)(x^2+y^2-0.02)(x^2+(y+3)^2-0.01)(y-3)=0 [-10, 10, -5, 5]}

Answer 2 · 2017-07-21T06:29:57

#y=-x^2/12#

Given -
Vertex #=(0,0)#
Focus #=(0,-3)#

This parabola opens down. Look at the graph -
enter image source here

So the equation of the parabola in such cases is

#x^2=-4ay#

Where # a = 3#
It is the distance from the vertex to focus.

Then -

#x^2=-4 xx 3 xx y#
#x^2=-12y#
#-12y=x^2#
#y=-x^2/12#