Question

What is the vertex form of the equation of the parabola with a focus at (55,45) and a directrix of `y=47`?

Answer 1

`y=-1/4(x-55)^2+46`

Explanation:

Parabola is locus of a point which moves so that its distance from a given point calld focus and a given line ccalled directrix is always constant.

Let the point be `(x,y)`. Here focus is `(55,45)` and distance from focus is `sqrt((x-55)^2+(y-45)^2)`. Its distance from directrix `y=47` i.e. `y-47=0` is `|y-47|`.

Hence equaion of parabola is

`(x-55)^2+(y-45)^2)=|y-47|^2`

or `(x-55)^2+y^2-90y+2025=y^2-94y+2209`

or `(x-55)^2-184=-4y`

i.e. `y=-1/4(x-55)^2+46`

Hence parabola's equation is `y=-1/4(x-55)^2+46` and vertex is `(55,46)`.

graph{((x-55)^2-184+4y)(y-47)((x-55)^2+(y-45)^2-0.1)=0 [35, 74.99, 35.02, 55.02]}