Question

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

Answer 1

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

Explanation:

Generic Equation is
y - k = 1/4p ( x - h)^2
p is distance vertex to focus = 3
(h,k) = vertex location = (-2, 9)

Answer 2

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

Explanation:

When talking about the focus and vertex of a parabola, the easiest way to write the equation is in vertex form. Luckily, you already have most of your information.

`y=a(x+2)^2+9`

However, we do not have the value of `a`.

`a=1/(4c)`

`c` is the distance between the focus and the vertex.

`c=-3`

We know this because the only difference between the two coordinates is the `y` part. The reason it is negative is because the vertex is above the focus; this means that the parabola opens downwards.

`1/(4c)`

`1/((4)(-3))`

`1/-12`

`-1/12`

Now that you have your value for `a`, you can plug this in and finalize your equation.

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

Answer 3

`y=-x^2/12-x/3+26/3`

Explanation:

Given -

Vertex `(-2, 9)`
Focus `(-2, 6)`

The focus of the parabola lies below the vertex. Hence, it opens down.

The formula for downward opening parabola having origin as its vertex is -

`x^2=-4ay`

The vertex of the given parabola is not at the vertex. it is in the 2nd quarter.

The formula is -

`(x-h)^2=-4xxaxx(y-k)`

`h=-2` x-coordinate of the vertex
`k=9` y-coordinate of the vertex
`a=3`Distance between vertex and focus
Substitute the values in the formula
`(x+2)^2=-4xx3xx(y-9)`
`x^2+4x+4=-12y+108`
`-12y+108=x^2+4x+4`
`-12y=x^2+4x+4-108`
`y=-x^2/12-4/12x+108/12`
`y=-x^2/12-x/3+26/3`