What is the focus of the parabola #(x − 5)^2 = −4(y + 2)#?

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Answer 1 · 2017-08-07T12:58:19

Focus is at # (5 , -3) #

# (x-5)^2 = -4 ( y+2 ) or y+2 = -1/4(x-5)^2 # or

#y = -1/4(x-5)^2 -2 # . Comparing with standard vertex form

of equation # y= a (x-h)^2 +k ; (h,k)# being vertex , we find here

#a= -1/4 , h= 5 , k = -2# So vertex is at # (5, -2)# . Distance of

directrix from vertex is #d= 1/(4|a|) = 1/(4*1/4)=1#.

since #a# is negative , the parabola opens downward . Focus is

below the vertex and at a distance of #d=1 #unit from vertex.

So focus is at # (5 , -3) #

graph{(x-5)^2=-4(y+2) [-10, 10, -5, 5]} [Ans]