Question

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

Answer 1

Vertex is at `(-4,2)`, focus is at `(-4.5,2)` and
directrix is `x=-3.5`

Explanation:

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

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

Comparing with the equation of horizontal

parabola opening left is `(y-k)^2 = -4p(x-h)`

We get , `h=-4 ,k=2, p=1/2 :.` vertex is at

`(h,k) or (-4,2)` , focus is at `(-4-1/2),2or (-4.5,2)`

and directrix is `x= (-4+1 /2) or x = -3.5`

graph{x= -1/2(y-2)^2-4 [-10, 10, -5, 5]}