How do you find the focus, vertex, and directrix of #x^2=32y#?

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Answer 1 · 2016-02-05T14:24:43

Focus #(0, 8)#
Vertex #(0, 0)#
Directrix: #y=-8#

standard form:

#(x-h)^2=4p(y-k)#

From the given:

#x^2=32y#

#(x-0)^2=4*8(y-0)#

Therefore,# p=8#

Vertex at #(0, 0)#
Focus at #(0, 8)#
Directrix at #y=-8#

kindly check on the graph.

graph{(32y-x^2)(y+8)=0 [-30, 30, -15, 15]}

Have a nice day !! from the Philippines...