• Mar 2nd 2009, 11:26 AM
rebirthflame
im not really sure if this is to do with quadratic forms or not but tbh i have no idea of where to start with this question, i think it has something to do with diagonilising a matrix somewhere along the line but apart from that im clueless

by using suitable transformations express
$\displaystyle x^2+x-8+5xy-6y+2y^2 = 0$
in the form
$\displaystyle AX^2+BY^2=1$
• Mar 3rd 2009, 03:16 AM
General Equation of a Conic
Hello rebirthflame
Quote:

Originally Posted by rebirthflame
im not really sure if this is to do with quadratic forms or not but tbh i have no idea of where to start with this question, i think it has something to do with diagonilising a matrix somewhere along the line but apart from that im clueless

by using suitable transformations express
$\displaystyle x^2+x-8+5xy-6y+2y^2 = 0$
in the form
$\displaystyle AX^2+BY^2=1$

What you are being asked to do here is to transform a general equation of the second degree into a standard form for a conic. The usual method, starting with

$\displaystyle ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

is to let

$\displaystyle x = X\cos\theta - Y\sin\theta$

$\displaystyle y = X\sin\theta + Y\cos\theta$

which corresponds to a rotation of the axes through an angle $\displaystyle \theta$, in order to eliminate the term in $\displaystyle xy$. When you do this, it leads to the value of $\displaystyle \theta$ given by

$\displaystyle \tan 2\theta = \frac{2h}{a-b}$

You then complete the squares for $\displaystyle X$ and $\displaystyle Y$ to get the equation in the required form.

In the example you've been given:

$\displaystyle \tan 2\theta = -5$

But this doesn't seem to give straightforward (i.e. easy) values for $\displaystyle \sin\theta$ and $\displaystyle \cos\theta$ ($\displaystyle \sin\theta = \sqrt{\frac{1 + \sqrt{26}}{2\sqrt{26}}}$, for instance).

Are you sure you have the right numbers here?

• Mar 3rd 2009, 04:02 AM
rebirthflame
the numbers are defnetly right, ive had a word with my lecturer and he says like you that i need to transform the curve so that we get the equation in a quadratic form then he said something about completing squares and then using eigenvalues or diagonolising matrices, your method looks like it would work however i also know there is more than one way to do this, you are looking at eliminating the values of xy which wont be a quadratic form, i know what to do when i get to the quadratic form i think. its just getting there and choosing the suitable trandformation that is the problem.
• Mar 3rd 2009, 06:02 AM
HallsofIvy
$\displaystyle x^2+5xy+2y^2 = \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}1 & \frac{5}{2} \\ \frac{5}{2} & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$
We can diagonalize that matrix by finding its eigenvalues and eigenvectors. Its characteristic equation is $\displaystyle \left|\begin{array}{cc}1-\lambda & \frac{5}{2} \\ \frac{5}{2} & 2-\lambda\end{array}\right|= \lambda^2- 3\lambda+ 2- \frac{25}{4}$$\displaystyle = \lambda^2- 3\lambda- \frac{17}{4}= 0 You can find the eigenvalues by solving that equation using the quadratic formula. They are not, as already pointed out, "easy" numbers: something like, I think \displaystyle \frac{3}{2}\pm \frac{\sqrt{37}}{4}. Find the eigenvectors corresponding to that and construct the matrix P having those eigenvectors as columns. \displaystyle P^{-1}AP will be the diagonal matrix having the eigenvallues as entries. • Mar 3rd 2009, 08:35 AM rebirthflame Quote: Originally Posted by HallsofIvy your quadratic form can be written as a matrix multiplication as \displaystyle x^2+5xy+2y^2 = \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}1 & \frac{5}{2} \\ \frac{5}{2} & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} We can diagonalize that matrix by finding its eigenvalues and eigenvectors. Its characteristic equation is \displaystyle \left|\begin{array}{cc}1-\lambda & \frac{5}{2} \\ \frac{5}{2} & 2-\lambda\end{array}\right|= \lambda^2- 3\lambda+ 2- \frac{25}{4}$$\displaystyle = \lambda^2- 3\lambda- \frac{17}{4}= 0$
You can find the eigenvalues by solving that equation using the quadratic formula. They are not, as already pointed out, "easy" numbers: something like, I think $\displaystyle \frac{3}{2}\pm \frac{\sqrt{37}}{4}$. Find the eigenvectors corresponding to that and construct the matrix P having those eigenvectors as columns. $\displaystyle P^{-1}AP$ will be the diagonal matrix having the eigenvallues as entries.
i need to transform the curve to get rid of all the variables apart from the ones of the form $\displaystyle ax^2+bxy+cy^2$ this will centre my curve about the origin as far as i understand. then i can diagonalize the matrix as hallsofivy has done this rotates the curve so that it is of the form $\displaystyle AX^2+BY^2=1$
$\displaystyle x^2+x-8+5xy-6y+2y^2 = 0$ into a quadratic form.