Question

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

Answer 1

`y=-1/16(x-1)^2+5`

Explanation:

Parabola is the locus of a point which moves so that its distance from a point called focus and a line called directrix is always same.

Hence a point, say `(x,y)` on the desired parabola will be equidistant from focus `(1,-9)` and directrix `y=-1` or `y+1=0`.

As the distance from `(1,-9)` is `sqrt((x-1)^2+(y+9)^2)` and from `y+1` is `|y+1|`, we have

`(x-1)^2+(y+9)^2=(y+1)^2`

or `x^2-2x+1+y^2+18y+81=y^2+2y+1`

or `x^2-2x+16y+81=0`

or `16y=-1(x^2-2x+1-1)-81`

or `16y=--(x^2-2x+1)+1-81`

or `y=-1/16(x-1)^2+5`

Hence, vertex is `(1,-5)` and axis of symmetry is `x=1`

graph{(y+1/16(x-1)^2+5)(y+1)(x-1)((x-1)^2+(y+9)^2-0.04)=0 [-20.08, 19.92, -17.04, 2.96]}