Simplify this term—best approach?

[ElectronicTeaCup]

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?

Spoiler: Answer

$x^{\prime 2}(5 / 2-\sqrt{13} / 2)+y^{\prime 2}(5 / 2+\sqrt{13} / 2)-4=0$

 

[fresh_42]

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]

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$$\left(x \sqrt{\left(\frac{3+\sqrt{13}}{2 \sqrt{13}}\right)}-y\sqrt{\left(\frac{3-\sqrt{13}}{2 \sqrt{13}}\right)}\right)~?$$

 

[ElectronicTeaCup]

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]

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

 

[epenguin]

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... ?