How do you find the vertex, focus, and directrix of the parabola #(x+3)+(y-2)^2=0#?

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Answer 1 · 2018-07-01T12:54:28

Vertex is at #(-3,2)#, directrix is # x = - 2.75# and
focus is at # (-3.25,2)#

# (x+3)+(y-2)^2=0 or (y-2)^2=-(x+3) # or

#(y-2)^2=-4*0.25(x+3) # The equation of horizontal

parabola opening left is

#(y-k)^2 = -4p(x-h) ; h=-3 ,k=2 , p = 0.25 ; h ,k#

being vertex .Therefore vertex is at #(-3,2)#

Distance between focus and vertex is #p=0.25#

Here the directrix is at right of the vertex. Therefore the

equation of directrix is #x=(-3+0.25) or x = - 2.75#

Vertex is at midway between focus and directrix. Therefore,

focus is at #(-3-0.25),2 or (-3.25,2)#

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