Posted by AJ on .
Find the equation of the parabola whose axis is horizontal, vertex on the yaxis and which passes through (2,4) and (8,2).

Analytic Goemetry 
Reiny,
the general formula for a parabola with known vertex and horizontal axis is
x = a(yk)^2 + h, where (h,k) is the vertex
in your case the vertex , being on the yaxis , can be called (0,k) , and the equation would be
x = a(y  k)^2 + 0
for the point (2,4)
2 = a(4k)^2 (#1)
for the point (8,2)
8 = a(2k)^2 (#2)
divide #2 by #1
4 = (2k)^2 / (4k)^2
4(4k)^2 = (2k)^2
64  32k + 4k^2 = 4 + 4k + k^2
3k^2 36 + 60 = 0
k^2  12 + 20 = 0
(k10)(k2) = 0
k = 10 or k = 2
if k=2, x = a(y2)^2
and using (2,4)
2 = a(42)^2
a = 1/2 > x = (1/2)(y2)^2
if k=10 , x = a(y10)^2
again using (2,4)
2 = a(410)^2
a = 1/18 > x= (1/18)(y10)^2
Notice that there are two possible equations,
btw, I checked both equations using the point(8,2), it also satisfies both equations.