Simplify this term—best approach?

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##

fresh_42
Mentor
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.

kuruman
Homework Helper
Gold Member
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$$\left(x \sqrt{\left(\frac{3+\sqrt{13}}{2 \sqrt{13}}\right)}-y\sqrt{\left(\frac{3-\sqrt{13}}{2 \sqrt{13}}\right)}\right)~?$$

archaic
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.

fresh_42
Mentor
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:
ElectronicTeaCup
epenguin
Homework Helper
Gold Member
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... ?