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

3 Answers
Jun 27, 2017

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)

Jun 27, 2017

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

Jun 27, 2017

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

Explanation:

Given -

Vertex (-2, 9)
Focus (-2, 6)
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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=3Distance 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