How do you find the focus, vertex, and directrix of #y^2=-4x#?

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Answer 1 · 2016-11-26T04:25:41

Its vertex is #(0, 0)#
Its focus is #(-1, 0)#
Its directrix is #x=1#

given -

#y^2=-4x#

It is in the form

#y^2=-4ax#

If it is so, then -

Its vertex is #(0, 0)#
Its focus is #(-a, 0)#
Its directrix is #x=a#

Apply this in the given equation

For better understanding the given equation can be written as

#y^2=4*(-1)*x#

Then-

Its vertex is #(0, 0)#
Its focus is #(-1, 0)#
Its directrix is #x=1#