Posted by **AJ** on Wednesday, March 23, 2011 at 7:09am.

Find the equation of the parabola whose axis is horizontal, vertex on the y-axis and which passes through (2,4) and (8,-2).

- Analytic Goemetry -
**Reiny**, Wednesday, March 23, 2011 at 9:05am
the general formula for a parabola with known vertex and horizontal axis is

x = a(y-k)^2 + h, where (h,k) is the vertex

in your case the vertex , being on the y-axis , can be called (0,k) , and the equation would be

x = a(y - k)^2 + 0

for the point (2,4)

2 = a(4-k)^2 (#1)

for the point (8,-2)

8 = a(-2-k)^2 (#2)

divide #2 by #1

4 = (-2-k)^2 / (4-k)^2

4(4-k)^2 = (-2-k)^2

64 - 32k + 4k^2 = 4 + 4k + k^2

3k^2 -36 + 60 = 0

k^2 - 12 + 20 = 0

(k-10)(k-2) = 0

k = 10 or k = 2

if k=2, x = a(y-2)^2

and using (2,4)

2 = a(4-2)^2

a = 1/2 ---------> x = (1/2)(y-2)^2

if k=10 , x = a(y-10)^2

again using (2,4)

2 = a(4-10)^2

a = 1/18 ----------> x= (1/18)(y-10)^2

Notice that there are two possible equations,

btw, I checked both equations using the point(8,-2), it also satisfies both equations.

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