Question

What is the vertex form of the equation of the parabola with a focus at (11,28) and a directrix of `y=21`?

Answer 1

The equation of parabola in vertex form is `y=1/14(x-11)^2+24.5`

Explanation:

The Vertex is equuidistant from focus(11,28) and directrix (y=21). So vertex is at `11,(21+7/2)=(11,24.5)`
The equation of parabola in vertex form is `y=a(x-11)^2+24.5`. The distance of vertex from directrix is `d=24.5-21=3.5` We know, `d=1/(4|a|) or a=1/(4*3.5)=1/14`.Since Parabola opens up, 'a' is +ive.
Hence the equation of parabola in vertex form is `y=1/14(x-11)^2+24.5` graph{1/14(x-11)^2+24.5 [-160, 160, -80, 80]}[Ans]