SOLUTION: Write an equation of a parabola with a vertex at the origin. 8. focus (0, 2) 9. focus (4, 0) 10. directrix at y = -8

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Write an equation of a parabola with a vertex at the origin. 8. focus (0, 2) 9. focus (4, 0) 10. directrix at y = -8      Log On


   



Question 193913: Write an equation of a parabola with a vertex at the origin.

8. focus (0, 2)
9. focus (4, 0)
10. directrix at y = -8

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!

8. focus (0, 2)
This is a "vertical" parabola, if so, it follows:
y+=+%281%2F%284c%29%29%28x-h%29%5E2+%2B+k
.
Since, the vertex is at the origin:
(h,k) = (0,0)
'c' is the distance from the vertex to the focus
therefore
c =2
Now, we plug it in to
y+=+%281%2F%284c%29%29%28x-h%29%5E2+%2B+k
y+=+%281%2F%284%2A2%29%29%28x-0%29%5E2+%2B+0
y+=+%281%2F8%29x%5E2
.
9. focus (4, 0)
This is a "horizontal" parabola, if so, it follows:
x+=+%281%2F%284c%29%29%28y-k%29%5E2+%2B+h
.
Since, the vertex is at the origin:
(h,k) = (0,0)
'c' is the distance from the vertex to the focus
therefore
c =4
.
Now, we plug it in to
x+=+%281%2F%284c%29%29%28y-k%29%5E2+%2B+h
x+=+%281%2F%284%2A4%29%29%28y-0%29%5E2+%2B+0
x+=+%281%2F16%29y%5E2
.
10. directrix at y = -8
This is a "vertical" parabola, if so, it follows:
y+=+%281%2F%284c%29%29%28x-h%29%5E2+%2B+k
.
Since, the vertex is at the origin:
(h,k) = (0,0)
'c' is the distance from the vertex to the directrix
therefore
c =-8
Now, we plug it in to
y+=+%281%2F%284c%29%29%28x-h%29%5E2+%2B+k
y+=+%281%2F%284%28-8%29%29%29%28x-0%29%5E2+%2B+0
y+=+%28-1%2F32%29x%5E2