How to reduce a conic to it's canonical form

I'm having trouble with this :

5x^2 + 4xy + 8y^2 -32 x - 56y + 80 = 0

What i know :

square form is : 5x^2 + 4xy + 8y^2 so the matix is A = 5 2

2 8

from the characteristic eq P(λ) = 0 ; |5-λ 2|

. | 2 8-λ|

so i reach λ^2 - 13λ +36 = 0

delta = 25

λ1,2 = 9 , 4

using the formula sgn(λ1-λ2)=sgn(a12) to get detR=+1 (if a12 is pos then sgn(a12) = +1 if neg = -1)

so λ1 = 9 and λ2=4

for λ1 x2=2x1 so v1(1,2)

for λ2 x1=-2x2 so v2(-2,1)

e1 = v1 / || v1 || = (1/sqrt5 , 2/sqrt5)

e2 = v2 / || v2 || = (-2/sqrt5, 1/sqrt5)

so R = (1/sqrt5 -2/sqrt5)

(2/sqrt5 1/sqrt5)

now the rotation (x) = R(x`) => x = 1/sqrt5 ( x`-2y` ) and y = 1/sqrt5(2x`+y`)

. (y) (y`)

My problem starts here i don't know how to replace x and y in the eq of the conic. I have this problem solved the next part would be to reach this :

9x`^2 - 144/sqrt5 *x` +4y`^2 +8/sqrt5 * y` +80 = 0 , i just can't get to that idk what i'm doing wrong can someone explain to me this next step in detail.

Re: How to reduce a conic to it's canonical form

You've done all of the hard work. You have $\displaystyle x=\frac{1}{\sqrt{5}}(x'- 2y')$ and [tex]y= \frac{1}{\sqrt{5}}(2x'+ 5y'). Now put those into "$\displaystyle x^2$", "$\displaystyle xy$", and $\displaystyle y^2$. Of course we will have, in each $\displaystyle \frac{1}{\sqrt{5}}\frac{1}{\sqrt{5}}= \frac{1}{5}$ so we really just need to do $\displaystyle 5x^2=5(\frac{1}{5}) (x'- 2y')^2= x'^2- 4x'y'+ 4 y'^2$, $\displaystyle xy=4(\frac{1}{5} (x'- 2y')(2x'+ y')= \frac{4}{5}(2x'^2- 3x'y'- 2y'^2)$, and $\displaystyle 8y^2= \frac{8}{5}(2x'+ y')^2= \frac{8}{5}(4x'^2+ 4x'y'+ y'^2)$. Put each of those into the equation, and "combine like terms".

Also, $\displaystyle -32x= -\frac{32}{\sqrt{5}}(x'+ 2y')$, $\displaystyle -56y= -\frac{56}{\sqrt{5}}(2x'- y')$.

Re: How to reduce a conic to it's canonical form

so funny for me 1/sqrt 5 * 1/sqrt 5 = 1/ 2sqrt 5...

thnx for the help

Re: How to reduce a conic to it's canonical form

I have one more but this one i just can't find it's vectors v1 or v2

7x^2 - 8xy - y^2 - 2x - 4y -1 = 0

so i reached at λ1 = 3-4*sqrt2 and λ2 = 3+4*sqrt2; i dunno if this is right or not please check my math out.

and for this one idk what to do cose delta = 0 so λ1=λ2 and when i make the systems idk how to continue

x^2 - 4xy + y^2 + 6x - 12y +8 = 0

P(λ) = 0 => λ^2 - 2λ + 1 = 0

Re: How to reduce a conic to it's canonical form

First, it's not a good idea to post a new question in the same thread. People who have already responded may not look at it again.

Yes, we can write the quadratic portion as $\displaystyle \begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}7 & -4 \\ -4 & -1\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}$. That has eigenvalue equation $\displaystyle \left|\begin{array}{cc}7-\lambda & -4 \\ -4 & -1-\lambda\end{array}\right|$$\displaystyle = (7- \lamba)(-1- \lambda)- 16= \lambda^2- 6\lambda- 23= 0$ and that has roots $\displaystyle 3+ 4\sqrt{2}$ and $\displaystyle 3- 4\sqrt{2}$. An eigenvector corresponding to eigenvalue $\displaystyle 3- 4\sqrt{2}$ must satisfy $\displaystyle 7x- 4y= (3+ 4\sqrt{2})x$ which is the same as $\displaystyle 4y= (4- 4\sqrt{2})x$ so is of the form $\displaystyle \left<x, (1-\sqrt{2})x\right>$. An eigenvector corresponding to eigenvalue $\displaystyle 3+ 4\sqrt{2}$ must satisfy $\displaystyle 7x- 4y= (3- 4\sqrt{2})x$ which is the same as $\displaystyle 4y= (4+ 4\sqrt{2})x$ so is of the form $\displaystyle \left<x, (1+\sqrt{2})x\right>$.

Let $\displaystyle x'= y- (1-\sqrt{2})x$ and $\displaystyle y'= y- (1+\sqrt{2})x$ so that the axes are rotated to be parallel to those eigenvectors. It really isn't necessary to determine the angle but if you want to know it, look at $\displaystyle \theta= tan^{-1}(1-\sqrt{2})$ and $\displaystyle tan^{-1}(1+ \sqrt{2})$. Those angles are complemetary so you can rotate through either angle reversing x' and y'.

For $\displaystyle x^2- 4xy+ y^2+ 6x- 12y+ 8= 0$, the coefficient matrix is$\displaystyle \begin{bmatrix}1 & -2 \\ -2 & 1\end{bmatrix}$ which has characteristic equation $\displaystyle \left|\begin{array}{cc}1-\lambda & -2 \\ -2 & 1-\lambda\end{array}\right|= (1-\lambda)^2-4= \lambda^2- 2\lambda- 3= (\lambda- 3)(\lambda+ 1)= 0$ so that $\displaystyle \lambda= 3$ and $\displaystyle \lambda= -1$. They are NOT the same. You appear to have forgotten the "4xy" part.

(Of course, it **is** possible that the two eigenvalues are the same. But then you would still have two independent eigenvectors corresponding to that eigenvalue. The conic would be a circle, if the common eigenvalue is positive, or an equilateral hyperbola, if negative.)

Re: How to reduce a conic to it's canonical form

Can you please continue the first ex to the end i'm kinda lost it ; It would help me a lot to see it to the end ; for the second yep ... another silly mistake.