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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).

 Feb 15, 2019
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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)

 

 

cool cool cool

 Feb 15, 2019

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