Question

How do you find the vertex, focus and directrix of `x^2-2x+8y+9=0`?

Answer 1

The vertex is `(1,-1)`, the focus is `(1,-3)`, and the directrix is `y=1`.

Explanation:

`x^2-2x+8y+9=0`

Complete the square to put the equation in the vertex form of a parabola.

`x^2-2xcolor(white)(aaa)=-8y-9`

Divide the coefficient of the `x` term by 2 and square it. Then add the square to both sides of the equation.

`-2/2 =-1 => (-1)^2=1`

`x^2-2x+1=-8y-9+1`

Factor the left side. Note that the constant term in the factors is the quotient of the coefficient of the `x` term divided by 2.

`(x-1)(x-1)=-8y-8`

Rewrite the left side as the square of a binomial, and factor out a GCF from the right side.

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

Rearrange to match the vertex form of a parabola that opens up or down: `4p(y-k)=(x-h)^2` where `(h,k)` is the vertex and `p` is used to find the focus and directrix, as shown below.

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

The vertex `(h,k)=(1,-1)`

Because the `4p=-8` is negative, the parabola opens down.

`p=-2`

The focus is "inside" the parabola, `p` units away from the vertex. Because the parabola opens down, the focus is BELOW the vertex, and its coordinates are `(1,-3)`

The directrix is a line "outside" the parabola, and wil be a horizontal line two units ABOVE the vertex. The equation of the directrix is `y=1`.