Question

How do you find the vertex, focus and directrix of `4x-y^2-2y-33=0`?

Answer 1

Vertex is `(8,-1)`; focus is `(9,-1)` and directrix is `x=7`

Explanation:

Let us rewrite `4x-y^2-2y-33=0` as

`4x=y^2+2y+33`

i.e. `4x=(y+1)^2+32`

or `x=1/4(y+1)^2+8`

Hence, vertex is `(8,-1)` and axis of symmetry is `y+1=0`

If the equation is of the type `(y-k)^2=4p(x-h)`,

the focus is `(h+p,k)` and the directrix is `x=h-p`

Here it is of the form `(y+1)^2=4(x-8)`

Hence, `k=-1`, `h=8` and `p=1`

Hence focus is `(9,-1)` and directrix is `x=7`

Parabola appears as

graph{(4x-y^2-2y-33)((x-8)^2+(y+1)^2-0.01)((x-9)^2+(y+1)^2-0.01)(x-7)=0 [2.35, 12.35, -3.46, 1.54]}