How do you identify the focus, directrix, and axis of symmetry of the parabola and graph the equation #x^2= 40y#?

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Answer 1 · 2017-03-19T12:58:48

The focus is #F=(0,p/2)=(0,10)#
The directrix is #y=-10#
The axis of symmetry is #x=0#

We compare this equation #x^2=40y# to

#x^2=2py#

#2p=40#

#p=20#

The vertex is #V=(0,0)#

The focus is #F=(0,p/2)=(0,10)#

The directrix is #y=-p/2#

#y=-10#

The axis of symmetry is #x=0#

graph{(x^2-40y)(y+10)((x-0)^2+(y-10)^2-0.1)=0 [-28.9, 28.83, -12.91, 15.96]}