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).
+128408 x = y^4 (1) and x + y^2 = 1 (2)
Sub (1) into (2) and we have that
y^4 + y^2 = 1
Let y^2 = a and we have
a^2 + a = 1 complete the square on a
a^2 + a + 1/4 = 1 + 1/4 factor the left....simplify the right
(a + 1/2)^2 = 5/4 take both roots
a + 1/2 = ±√5 / 2 subtract 1/2 from both sides
a = ( - 1 ±√5 ) / 2
Since y^2 must be positive, then a must be ( √ 5 - 1 ) / 2 = y^2 (3)
Using x + y^2 = 1
x = 1 - y^2
x = 1 - [ ( √ 5 - 1 ) / 2 ] = ( 3 - √ 5 ) / 2
These curves intersect at the same x values
So...the distance between the y values is the distance between the intersection points
So.....using (3)
y = √ [ ( √ 5 - 1 ) / 2 ] or y = - √ [ ( √ 5 - 1 ) / 2 ]
So... the distance between these points is just
√ [ ( √ 5 - 1 ) / 2 ] - [ - √ [ ( √ 5 - 1 ) / 2 ] ] =
2 √ [ ( √ 5 - 1 ) / 2 ] =
√ [ 4 ( √5 - 1 ) / 2 ] =
√ [ 2√ 5 - 2 ] =
√ [ - 2 + 2√ 5 ] ⇒ (u, v) = (-2, 2)
