+0

help

0
196
1

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).

Jan 16, 2019

#1
+1

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)   Jan 17, 2019