How do you find the vertex, focus, and directrix of the parabola #(y+1/2)^2=2(x-5)#?

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Answer 1 · 2018-05-16T12:57:45

Vertex is at # (5,-0.5)#, focus is at # (5.5,-0.5)# and
directrix is # x=4.5#

#(y+1/2)^2=2 (x-5) or(y+0.5)^2=4* 0.5 (x-5) #

The equation of right opened parabola is

#(y-k)^2=4 a(x-h) ; (h,k) ;# being vertex and focus is at

#(h+a,k) :. h =5 , k= -0.5 , a= 0.5#

Vertex is at # (5,-0.5)# , Focus is at # (5.5,-0.5)#,

Vertex is at midway between focus and directrix.

Therefore directrix is # x= h-a or x = 5-0.5 or x=4.5#

graph{(y+1/2)^2=2(x-5) [-10, 10, -5, 5]} [Ans]