My Math Forum Simplest form of curve equation

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 April 15th, 2018, 09:22 AM #1 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 Simplest form of curve equation I'm kind of confused with this excercise: In 2d space and in respect to an orthonormal system Oxy we are given the curve with equation: $\displaystyle x^2-10\sqrt{3}xy+11y^2+64=0$ Find an orthonormal system for which the equation takes the simplest form. What is that form? I'm confused as to what "simplest form" means.. Is it the canonical form? And how do I find the new orthonormal system?
 April 15th, 2018, 11:51 AM #2 Senior Member   Joined: Oct 2009 Posts: 573 Thanks: 179 Yes, the canonical form.
 April 15th, 2018, 01:06 PM #3 Global Moderator   Joined: May 2007 Posts: 6,613 Thanks: 617 The equation looks like an ellipse. Rotate it so there is no xy term.
 April 15th, 2018, 08:25 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,731 Thanks: 1809 With axes rotated by pi/6, one gets $x^2 - 4y^2 = 16$, the equation of a hyperbola. (Or one can swap the axes to get $y^2 - 4x^2 = 16$.) Last edited by skipjack; April 16th, 2018 at 06:13 AM.
 April 16th, 2018, 03:04 AM #5 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 I rotated the axis by pi/6 but I found indeed a hyperbola but with this equation instead: $\displaystyle \frac{3x^2}{32} -\frac{y^2}{4}=1$ Here is what I did: $\displaystyle x=x'\cos(\frac{\pi}{6})-y'*\sin(\frac{\pi}{6})=\frac{\sqrt3}{2}x'-\frac{1}{2}y'$ $\displaystyle y=x'\sin(\frac{\pi}{6})+y'\cos(\frac{\pi}{6})= \frac{1}{2}x'+\frac{\sqrt3}{2}y'$ I then replace the $\displaystyle x,y$ into the equation which gave me $\displaystyle \left ( \frac{\sqrt{3}}{2}x'-\frac{1}{2}y' \right )^2-10\sqrt3\left (\frac{\sqrt3}{2}x'-\frac{1}{2}y' \right )\left (\frac{1}{2}x'+\frac{\sqrt{3}}{2}y' \right )+11\left (\frac{1}{2}x'+\frac{\sqrt{3}}{2}y' \right )^2$ And after some calculation, I ended with the final equation: $\displaystyle \frac{3x'^2}{32} -\frac{y'^2}{4}=1$ Did I miss something? Last edited by skipjack; April 16th, 2018 at 05:54 AM.
 April 16th, 2018, 05:06 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Since this is in the Linear Algebra sub-forum, write the equation as $\displaystyle \begin{bmatrix} x & y \end{bmatrix}\begin{bmatrix}1 & -5\sqrt{3} \\ -5\sqrt{3} & 11 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}+ 64= 0$. I have divided the "$\displaystyle -10\sqrt{3}$" terms into the two anti-diagonal terms so that it is a symmetric matrix and has two real eigenvalues. That matrix has characteristic equation is $\displaystyle \left|\begin{array}{cc}1- \lambda & -5\sqrt{3} \\ -5\sqrt{3} & 11- \lambda\end{array}\right|= \lambda^2- 12\lambda- 64= (\lambda- 16)(\lambda- 4)= 0$ so the eigenvalues are -4 and 16. To find the eigenvectors corresponding to eigenvalue -4, find x and y such that $\displaystyle \begin{bmatrix}1 & -5\sqrt{3} \\ -5\sqrt{3} & 11 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}x- 5\sqrt{3}y \\ -5\sqrt{3}x+ 11y\end{bmatrix}= -4\begin{bmatrix}x \\ y \end{bmatrix}$. The gives the two equations $\displaystyle x- 5\sqrt{3}y= -4x$ and $\displaystyle -5\sqrt{3}x+ 11y= -4y$. Since -4 is an eigenvalue, those two equations are equivalent and reduce to just one equation. For any given y we must have $\displaystyle 5x= 5\sqrt{3}y$ or $\displaystyle x= \sqrt{3}y$. So one eigenvector corresponding to eigenvalue -4 is $\displaystyle \begin{bmatrix}\sqrt{3} \\ 1 \end{bmatrix}$. That has length 2 so a unit eigenvector is $\displaystyle \begin{bmatrix}\frac{\sqrt{3}}{2} \\ \frac{1}{2}\end{bmatrix}$. Similarly, an eigenvector, $\displaystyle \begin{bmatrix}x \\ y \end{bmatrix}$, corresponding to eigenvalue 16 must satisfy $\displaystyle x- 5\sqrt{3}y= 16x$. For any y we must have $\displaystyle 15x= -5\sqrt{3}y$ or $\displaystyle x= -\frac{\sqrt{3}}{3}a$. So one eigenvector corresponding to eigenvalue 16 is $\displaystyle \begin{bmatrix}-\frac{\sqrt{3}}{3} \\ 1\end{bmatrix}$. That has length $\displaystyle \frac{2}{\sqrt{3}}$ so a unit eigenvector is $\displaystyle \begin{bmatrix}-\frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{bmatrix}$. Now construct the matrix, P, having those eigenvectors as columns: $\displaystyle P= \begin{bmatrix}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}$. Its inverse matrix is $\displaystyle P^{-1}= \begin{bmatrix}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}$, the transpose of P. P All of that gives $\displaystyle PAP^{-1}= \begin{bmatrix}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}\begin{bmatrix}1 & -5\sqrt{3} \\ -5\sqrt{3} & 11 \end{bmatrix}\begin{bmatrix}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}= \begin{bmatrix}-4 & 0 \\ 0 & 16\end{bmatrix}$. Writing "D" for that diagonal matrix, we can write the original equation, $\displaystyle X^TAX+ 65= 0$ as $\displaystyle X^TP^{-1}PAP^{-1}PX+ 64= (PX)^TD(PX)+ 64= 0$. With $\displaystyle X= \begin{bmatrix}x \\ y \end{bmatrix}$, $\displaystyle PX= \begin{bmatrix}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}\frac{x\sqrt{3}}{2}- \frac{y}{2} \\ \frac{x}{2}+ \frac{y\sqrt{3}}{2}\end{bmatrix}$ so that $\displaystyle (PX)^TD(PX)+ 64= 0$ becomes $\displaystyle \begin{bmatrix}\frac{x\sqrt{3}}{2}- \frac{y}{2} & \frac{x}{2}+ \frac{y\sqrt{3}}{2}\end{bmatrix}\begin{bmatrix}-4 & 0 \\ 0 & 11\end{bmatrix}\begin{bmatrix}\frac{x\sqrt{3}}{2}- \frac{y}{2} \\ \frac{x}{2}+ \frac{y\sqrt{3}}{2}\end{bmatrix}+ 64= 0$. Finally, if we take $\displaystyle x'= \frac{x\sqrt{3}}{2}- \frac{y}{2}$ and $\displaystyle y'= \frac{x}{2}+ \frac{y\sqrt{3}}{2}$ that will be written as $\displaystyle -4x'^2+ 16y'^2+ 64= 0$ which, dividing both sides by 64 gives $\displaystyle \frac{y'^2}{4}- \frac{x'^2}{16}= 1$, the equation of a hyperbola.
 April 16th, 2018, 06:16 AM #7 Global Moderator   Joined: Dec 2006 Posts: 19,731 Thanks: 1809 You introduced a sign error in your final step.

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