PLEASE HELP WITH MY TRIG HW: Parabolas *BEST ANSWER AWARDED*?

Let the equation be

(y-2)^2=4a(x+5)

The focus is (1,2)=>

a=sqr[(1+5)^2+(2-2)^2]=>

a=6

=>

(y-2)^2=24(x+5)

or

y^2-24x-4y-116=0

is the required equation.

First determine the orientation of the parabola.

The vertex and focus are horizontally aligned (they have the same y-coordinate), so the parabola is horizontal.

The focus lies to the right of the vertex, so the parabola opens to the right.

Vertex-form equation for a right-opening parabola:

 x = a(y-k)² + h

with

 a > 0

 vertex (h,k)

 focal length p = 1/|4a|

 focus (h+p,k)

Apply your data and solve for h, k, and a.

vertex (h,k) = (-5,2), so h = -5 and k = 2

focus (h+p,k) = (1,2)

h+p = 1

-5+p = 1

p = 6

a = 1/|4p| = 1/24

The equation becomes

x = (1/24)(y-2)² - 5

axis of symmetry is parallel to the x-axis, it has an equation of (y - k)^2 = 4p (x - h), where the focus is (h + p, k)

focus (1,2) vertex (-5,2) ... h = -5, k = 2

. . . h+p = 1 .. p = 6

just fill in the equation ...

(y - k)^2 = 4p (x - h)

(y - 2)^2 = 24(x + 5)

Focus and vertex lie on a horizontal line so the parabola will be horizontal, taking form

x-h = a(y-k)^2

Where (h,k) are coordinates of vertex.

The y term is squared as opposed to an "everyday" vertical parabola where the x term is squared.

Here a = 1/(4p) where p=6 is distance between focus and vertex.

Focus is to the right of the vertex so parabola opens right and a>0.

If the parabola opened left then we would have a = -1/(4p).

Put together

x+5 = (1/24)(y-2)^2

A parabola is basically a curve