Problem:
Solution:
It is the intersection of a circle with a hyperbola, so there are only four points, there is no point 'drawing' the points.
To solve the intersection points, note that $ x^2 + y^2 = 4 $ and $ x^2y^2 = 1 $.
That gives
$ \begin{eqnarray*} x^2(4 - x^2) &=& 1 \\ x^4 - 4x^2 + 1 &=& 0 \\ x^2 &=& 2 \pm \sqrt{3} \\ x &=& \pm \sqrt{2 \pm \sqrt{3}} \\ \end{eqnarray*} $
We could have solve $ y $ by back substituting, but here we can simply use a symmetric argument to conclude the four points must be:
$ (\sqrt{2 + \sqrt{3}}, \sqrt{2 - \sqrt{3}}) $
$ (\sqrt{2 - \sqrt{3}}, \sqrt{2 + \sqrt{3}}) $
$ (-\sqrt{2 + \sqrt{3}}, -\sqrt{2 - \sqrt{3}}) $
$ (-\sqrt{2 - \sqrt{3}}, -\sqrt{2 + \sqrt{3}}) $
Solution:
It is the intersection of a circle with a hyperbola, so there are only four points, there is no point 'drawing' the points.
To solve the intersection points, note that $ x^2 + y^2 = 4 $ and $ x^2y^2 = 1 $.
That gives
$ \begin{eqnarray*} x^2(4 - x^2) &=& 1 \\ x^4 - 4x^2 + 1 &=& 0 \\ x^2 &=& 2 \pm \sqrt{3} \\ x &=& \pm \sqrt{2 \pm \sqrt{3}} \\ \end{eqnarray*} $
We could have solve $ y $ by back substituting, but here we can simply use a symmetric argument to conclude the four points must be:
$ (\sqrt{2 + \sqrt{3}}, \sqrt{2 - \sqrt{3}}) $
$ (\sqrt{2 - \sqrt{3}}, \sqrt{2 + \sqrt{3}}) $
$ (-\sqrt{2 + \sqrt{3}}, -\sqrt{2 - \sqrt{3}}) $
$ (-\sqrt{2 - \sqrt{3}}, -\sqrt{2 + \sqrt{3}}) $
Hi Andrew, I was recently working through this chapter myself. I did not understand why they said this was a "special case" of lemma 2, do you perhaps know why? I do not see what is special about it. Im not sure how I'll know if you respond to this, so if you wish you can email me at Penderm2@students.wwu.edu, or I'll keep an eye on here to see if you respond.
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