+0  
 
+1
2
1
avatar+86 

The graphs of \(x=y^4\) and \(x+y^2=1\) intersect at two points. The distance between these points is of the form \(\sqrt{u+v\sqrt5}\), where \(u\) and \(v\) are integers. Find the ordered pair \((u,v)\).

 May 13, 2024
 #1
avatar+9665 
+1

\(\begin{cases} x = y^4\\ x + y^2 = 1 \end{cases}\)

Substituting,

\(y^4 + y^2 - 1= 0\\ y^2 = \dfrac{-1 \pm \sqrt 5}2\text{(reject negative root)}\\ y^2 = \dfrac{\sqrt 5 - 1}2\\ y = \pm\sqrt{\dfrac{\sqrt 5 - 1}2}\)

 

Since \(x + y^2 = 1\),

\(x = 1 - \dfrac{\sqrt 5 - 1}2 = \dfrac{3 -\sqrt 5}2\)

 

So the intersection points are \(\left(\dfrac{3 - \sqrt 5}2, \pm\sqrt{\dfrac{\sqrt 5 - 1}2}\right)\).

 

The distance is \(2 \sqrt {\dfrac{\sqrt 5 - 1}2} = \sqrt{2(\sqrt 5 - 1)} = \sqrt{-2 + 2 \sqrt 5}\).

 

Then \((u,v) = (-2, 2)\).

 May 13, 2024

0 Online Users