Question

How do you find the vertex, focus and directrix of `(x + 0.5)^2 = 4(y - 1)`?

Answer 1

Vertex is at `(-0.5 , 1)` , Directrix is `y=0` and
Focus is at (-0.5,2)#

Explanation:

`4(y-1)= (x+0.5)^2 or y-1= 1/4(x+0.5)^2 or y= 1/4(x+0.5)^2 +1`

Comparing with standard vertex form equation `y = a(x-h)^2 +k ; (h,k)` being vertex , we get here `h = -0.5, k= 1 , a =1/4` since `a>0`, the parabola opens upwards .

Vertex is at `(-0.5 , 1)`. We know vertex is at equidistance from focus and directrix `d= 1/(4 |a|) = 1/(4*1/4)= 1`

Directrix is behind the vertex, `:.y=1-1=0`

Focus is above the vertex, at `(-0.5, (1+1)) i.e (-0.5,2)`
graph{(x+0.5)^2 =4(y-1) [-10, 10, -5, 5]} [Ans]