Question

What is the vertex form of the equation of the parabola with a focus at (1,-9) and a directrix of `y=0`?

Answer 1

`y = -1/18(x - 1)^2 - 9/2`

Explanation:

Because the directrix is a horizontal line, `y = 0`, we know that the vertex form of the equation of the parabola is:

`y = 1/(4f)(x - h)^2 + k" [1]"`

where `(h,k)` is the vertex and `f` is the signed vertical distance from the focus to the vertex.

The x coordinate of the vertex is the same as the x coordinate of the focus, `h = 1`.

Substitute into equation [1]:

`y = 1/(4f)(x - 1)^2 + k" [2]"`

The y coordinate of the vertex is the midpoint between the y coordinate of the focus and the y coordinates of the directrix:

`k = (0+ (-9))/2 = -9/2`

Substitute into equation [2]:

`y = 1/(4f)(x - 1)^2 - 9/2" [3]"`

The value of `f` is the y coordinate of the vertex subtracted from the y coordinate of the focus:

`f = -9 - -9/2`

`f = -9/2`

Substitute into equation [3]:

`y = 1/(4(-9/2))(x - 1)^2 - 9/2`

`y = -1/18(x - 1)^2 - 9/2" [4]"`

Equation [4] is the solution.