Question

What is the vertex form of the equation of the parabola with a focus at (6,5) and a directrix of `y=19`?

Answer 1

Vertex form of the equation of the parabola is
`y=-1/28(x-6)^2+12`

Explanation:

Here the directrix is a horizontal line `y=19`.

Since this line is perpendicular to the axis of symmetry, this is a regular parabola, where the `x` part is squared.

Now the distance of a point on parabola from focus at `(6,5)` is always equal to its distance from the directrix.

Let this point be `(x,y)`.

Its distance from focus is `sqrt((x-6)^2+(y-5)^2)` and from directrix will be `|y-19|`

Hence, `(x-6)^2+(y-5)^2=(y-19)^2`

or `x^2-12x+36+y^2-10y+25=y^2-38y+361`

or `x^2-12x+28y-300=0`

As vertex form of equation is `y=a(x-h)^2+k`

the above is `28y=-x^2+12x+300` or `28y=-(x^2-12x+36-336)`

or `28y=-(x-6)^2+336` or

`y=-1/28(x-6)^2+336/28` or

`y=-1/28(x-6)^2+12`

graph{-1/28(x-6)^2+12 [-12.83, 27.17, -2.08, 17.92]}