The distance between the two intersections of \(x=y^4\) and \(x+y^2=1\) is \(\sqrt{u+v\sqrt5}\). Find the ordered pair, (u,v).
x = y^4
x + y^2 = 1 sub the first into the second and we have that
y^4 + y^2 = 1 we can complete the square on y thusly
y^4 + y^2 + 1/4 = 1 + 1/4 factor and simplify
(y^2 + 1/2)^2 = 5/4 take both roots
y^2 + 1/2 = ± sqrt (5) / 2
y^2 = ±sqrt (5) / 2 - 1/2
y^2 = [ -1 ± sqrt (5) ] / 2 take both roots again
y = ± sqrt [ [ -1 ± sqrt (5) ] / 2 ]
Let y = a , -a
So ......x = y^4 = a^4
So we have the points (a^4, a ) and ( a^4, - a)
So....the distance between these points =
sqrt [ (a^4 - a^4) + ( a - -a)^2 ] = sqrt [ (2a)^2] = 2a
So...we have, 2a =
2 √ [ ( -1 ± √5) / 2 ] but we cannot take a root of a negative so the distance must be
2 √ [ ( -1 + √5) / 2 ] =
√ [ 4 ( -1 + √5) / 2 ] =
√ [ - 2 + 2√5 ]
So (u, v) = (-2, 2)