Question

What is the equation of a parabola with a focus at (-2, 6) and a vertex at (-2, 9)? What if the focus and vertex are switched?

Answer 1

The equation is `y=-1/12(x+2)^2+9`. The other equation is `y=1/12(x+2)*2+6`

Explanation:

The focus is `F=(-2,6)` and the vertex is `V=(-2,9)`

Therefore, the directrix is `y=12` as the vertex is the midpoint from the focus and the directrix

`(y+6)/2=9`

`=>`, `y+6=18`

`=>`, `y=12`

Any point `(x,y)` on the parabola is equidistant from the focus and the directrix

`y-12=sqrt((x+2)^2+(y-6)^2)`

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

`y^2-24y+144=(x+2)^2+y^2-12y+36`

`12y=-(x+2)^2+108`

`y=-1/12(x+2)^2+9`

graph{(y+1/12(x+2)^2-9)(y-12)=0 [-32.47, 32.45, -16.23, 16.25]}

The second case is

The focus is `F=(-2,9)` and the vertex is `V=(-2,6)`

Therefore, the directrix is `y=3` as the vertex is the midpoint from the focus and the directrix

`(y+9)/2=6`

`=>`, `y+9=12`

`=>`, `y=3`

`y-3=sqrt((x+2)^2+(y-9)^2)`

`(y-3)^2=(x+2)^2+(y-9)^2`

`y^2-6y+9=(x+2)^2+y^2-18y+81`

`12y=(x+2)^2+72`

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

graph{(y-1/12(x+2)^2-6)(y-3)=0 [-32.47, 32.45, -16.23, 16.25]}