Simplify this term—best approach?

  • #1
Homework Statement:
$$
\left(x \sqrt{\left(\frac{3+\sqrt{13}}{2 \sqrt{13}}\right)}-y \sqrt{\frac{3-\sqrt{13}}{2 \sqrt{13}}}\right)^{2}-2\left(x \sqrt{\left(\frac{3+\sqrt{13}}{2 \sqrt{13}}\right)-y\left(\frac{3-\sqrt{13}}{2 \sqrt{13}}\right)}\right)\left(x \sqrt{\frac{3-\sqrt{13}}{2 \sqrt{13}}}-y \sqrt{\frac{3+\sqrt{3}}{2 \sqrt{13}}}\right)+4\left(x \sqrt{\frac{3-\sqrt{13}}{2 \sqrt{13}}}-y \sqrt{\frac{3+\sqrt{13}}{2 \sqrt{13}}}\right)^{2}-4=0
$$
Relevant Equations:
Substituting in second order polynomial to rotate it to remove xy values. But this question is just to ask for a technique for simplifying.
I'm not sure how to simplify this without spending a lot of time on it. Is there a pattern that I need to weed out?
##x^{\prime 2}(5 / 2-\sqrt{13} / 2)+y^{\prime 2}(5 / 2+\sqrt{13} / 2)-4=0##
 

Answers and Replies

  • #2
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You can write the quadratic form as ##(x.y)A(x,y)^\tau## with a ##2\times 2## matrix ##A## and then normalize ##A##, i.e. determine the eigenvectores, find an eigenbase, and use those vectors as new ##A##.

Look up: 'standard (normal) form of conic sections (quadratic equations)' or similar.
 
  • #3
kuruman
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Is the first factor in the second term as you have it
$$\left(x \sqrt{\left(\frac{3+\sqrt{13}}{2 \sqrt{13}}\right)-y\left(\frac{3-\sqrt{13}}{2 \sqrt{13}}\right)}\right)$$or did you mean to write$$\left(x \sqrt{\left(\frac{3+\sqrt{13}}{2 \sqrt{13}}\right)}-y\sqrt{\left(\frac{3-\sqrt{13}}{2 \sqrt{13}}\right)}\right)~?$$
 
  • #4
Is the first factor in the second term as you have it
$$\left(x \sqrt{\left(\frac{3+\sqrt{13}}{2 \sqrt{13}}\right)-y\left(\frac{3-\sqrt{13}}{2 \sqrt{13}}\right)}\right)$$or did you mean to write$$\left(x \sqrt{\left(\frac{3+\sqrt{13}}{2 \sqrt{13}}\right)}-y\sqrt{\left(\frac{3-\sqrt{13}}{2 \sqrt{13}}\right)}\right)~?$$
Oh yes, I meant the latter. Sorry for the confusion that it may have lead to.
 
  • #5
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I assume that ##\sqrt{3}## is an error, too. And I would definitely check all signs before carrying on: this is essential!

What you can do is bring into a form ##8\sqrt{13}=z_1x^2+2z_2xy+z_3y^2## with ##z_j\in \mathbb{C}## and transform it into a normal form where the axis of this ellipse in ##\mathbb{C}^2## are parallel to the coordinate axis.

The eigenvalues I calculated where awful if we were looking for an eigenbasis, which indicates that
  • there has been a sign error somewhere
  • the requested result is of the form shown above
  • the result is indeed complicated
  • I made a mistake
 
Last edited:
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  • #6
epenguin
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You do not seem to have written out your full formula. It might ir it might not be simpler if you do. What is there after the -2(x... ?
 

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