SOLUTION: Find an equation of the parabola that has a focus at (9,15) and a vertex at (9,8) in terms of y.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find an equation of the parabola that has a focus at (9,15) and a vertex at (9,8) in terms of y.      Log On


   

Question 409132: Find an equation of the parabola that has a focus at (9,15) and a vertex at (9,8) in terms of y.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
a focus at (9,15) and
a vertex at (9,8)

Since the x-coordinates of the vertex and focus are the same, they are one of top of the other, so this is a regular vertical parabola, where the x+part is squared.
Since the vertex is below the focus, this is a right-side+up parabola and p+is positive.
Since the vertex and focus are 15 –8 = 7 units apart, then p+=+7.
And that's all I need for my equation, since they already gave me the vertex.


(x–h)^2=4p(y –k)
(x–9)^2 = 4*7(y –8)
(x–9)^2 = 28(y –8)
(x–9)^2/28= y–8

(x–9)^2/28 +8= y

y+=%28x%5E2-18x+%2B81%29%2F28+%2B8+
y+=%281%2F28%29x%5E2-0.64x+%2B2.9+%2B8+
y+=+%281%2F28%29x%5E2-0.64x+%2B10.9+


+graph%28+600%2C+600%2C+-35%2C+25%2C+-35%2C+25%2C+%281%2F28%29x%5E2-0.64x+%2B10.9%29+