Question

How do you find the focus, vertex, and directrix of `y^2+4x-4y-8=0`?

Answer 1

The focus is at`(1,3)`, the vertex is `(2,3)` and the directrix is at `x = 3`

Explanation:

Reformulate the equation to have one variable on its own on the left hand side. In this case it should be `x` because `y` is the one that is raised to a power.

`y^2 +4x - 4y -8=0`
`4x = -y^2+4y+8`
`x=-1/4(y^2 -4y-8)`

This represents a horizontal parabola, because x is being expressed in terms of y. It is opening to the left, because of the minus sign.

We need to get this into vertex form `x= a(y-b)^2 +c` where `(b,c)` is the vertex.

`x= -1/4((y-2)^2-4 -8)`
`x = -1/4(y-2)^2 +12/4 = -1/4(y-2)^2 +3`

The vertex is `(2,3)`

Rearranging the equation as `-4(x-3) = (y-2)^2` which is in the form `4p(x-h) = (y-k)^2` where `p` is the distance between the vertex and the focus which equals the distance from the vertex to the directrix.

In this case `p=-1` so the focus is at`(1,3)` one unit to the left of the vertex.

The directrix is therefore at `x = 3`