conic sections

**heroic** · February 22nd 2008, 10:46 AM

the following equation is a hyperbola,
5x^2 + 38x - 8xy - 22y - y^2 + 47 = 0
and i need to know how to translate and rotate it so that it is in the standard form, AX^2 + BY^2 = 1
does anyone have any ideas??? any help would be great.

**TKHunny** · February 22nd 2008, 11:35 AM

You should have at least one example in your book. The most important part would be: $tan(2\theta)\;=\;\frac{B}{A-C}$ . After that, you're down to arithmetic.

**heroic** · February 22nd 2008, 01:25 PM

i don't think i need any trig equations, my lecturer said to use a diagonalise function, using eigenvales and vectors. but i dont know how to 'translate' the function so that it can be rotated using eigenvalues. etc. cheers anyway. could u possibly suggest anything else.

**mr fantastic** · February 22nd 2008, 02:07 PM

Originally Posted by heroic

i don't think i need any trig equations, my lecturer said to use a diagonalise function, using eigenvales and vectors. but i dont know how to 'translate' the function so that it can be rotated using eigenvalues. etc. cheers anyway. could u possibly suggest anything else.

Try having a read of this first ......

**TKHunny** · February 22nd 2008, 02:47 PM

Originally Posted by heroic

i don't think i need any trig equations, my lecturer said to use a diagonalise function, using eigenvales and vectors. but i dont know how to 'translate' the function so that it can be rotated using eigenvalues. etc. cheers anyway. could u possibly suggest anything else.

There, then, is a nice lesson in providing the actual problem statement. I suppose I could have guessed in the calculus section that you would not be using elementary analytic geometry methods, but please go read your first post and point to where that is obvious. The rule is this, the more accurate the problem statement, the more useful information concerning your personal efforts, the better the response.

Let's see what you get.

**heroic** · February 23rd 2008, 02:26 AM

sorry for not being more clearer, i hope my second post didn't sound nasty because that wasn't intended. anyway thanks for that link very useful however it doesn't consider the 38x and 22y term could you possibly help me there. i hope im being clear. lol

**mr fantastic** · February 23rd 2008, 03:15 AM

Originally Posted by heroic

sorry for not being more clearer, i hope my second post didn't sound nasty because that wasn't intended. anyway thanks for that link very useful however it doesn't consider the 38x and 22y term could you possibly help me there. i hope im being clear. lol

Let $X = \left( \begin{array}{c} x \\ y \end{array} \right) \,$ , $\, M = \left( \begin{array}{cc} 5 & -4 \\ -4 & -1 \end{array} \right)\,$ and $\, B = \left( \begin{array}{c} 38 \\ -22 \end{array} \right) \,$ .

Then the conic $\, 5x^2 + 38x - 8xy - 22y - y^2 + 47 = 0 \,$ can be expressed in matrix form as $X^T M X + B^T X + 47 = 0$ .

Let $Y = X - C \,$ where $\, C \,$ is a column matrix to be determined. Then $\, X = Y + C \,$ and so:

$(Y + C)^T M (Y + C) + B^T (Y + C) + 47 = 0$

$\Rightarrow Y^T M Y + C^T M Y + Y^T M C + C^T M C + B^T Y + B^T C + 47 = 0$

$\Rightarrow Y^T M Y + Y^T (2M C + B) + \beta = 0$

(since $\, M\,$ is real symmetric) where $\beta = C^T M C + B^T C + 47$ .

Since $\, M \,$ is invertible there exists a unique matrix $\, C \,$ such that $\, 2 M C + B = 0\,$ . With this choice of $\, C \,$ you have:

$Y^T M Y = - \beta$ .

The substitution $\, Y = X - C\,$ corresponds to a translation of the origin to the point with coordinate matrix $\, C \,$ .

All the dirty calculational details are left to you ......

**heroic** · February 23rd 2008, 03:20 AM

thanks for that, really appreciate it. i'll have a crack at it now, and let you know how i go on. thanks again

**heroic** · February 23rd 2008, 03:54 AM

i get c as -3 and 1. however i'm having trouble with beta, where BtransposeC. i get 68 + [270 -130] + 47 but obviously i can't add that up, could you help plz.

**mr fantastic** · February 23rd 2008, 04:15 AM

Originally Posted by heroic

i get c as -3 and 1. Mr F says: Correct.

however i'm having trouble with beta, where BtransposeC. i get 68 + [270 -130] + 47 but obviously i can't add that up, could you help plz.

$\beta = C^T M C + B^T C + 47 = 68 + (38 \times -3 + -22 \times 1) + 47 = 68 - 136 + 47 = -21$ .

The 68 is good. But I don't know where the [270 - 130] has come from

So $ Y^T M Y = 21 $ .

**heroic** · February 23rd 2008, 04:26 AM

now that i have to rotate it using the eigenvalues and eigenvectors i get the 2x2 matrix [-3 0, 0 7] therefore i get the standard form equation -3x + 7y = 21
therefore it would be -1/7x + 1/3y = 1
could you possibly confirm if this is correct plz.

**mr fantastic** · February 23rd 2008, 04:50 AM

Originally Posted by heroic

now that i have to rotate it using the eigenvalues and eigenvectors i get the 2x2 matrix [-3 0, 0 7] therefore i get the standard form equation -3x + 7y = 21
therefore it would be -1/7x + 1/3y = 1
could you possibly confirm if this is correct plz.

You forgot the 'squared' in each term

.....

Otherwise, I get the gist and it looks good.

One thing ....

I'd write it as $-3(x^{'})^2 + 7 (y^{'})^2 = 21$ etc. where $\, x' \,$ and $\, y' \,$ are the rotated coordinates.

And you understand the relationship between the unrotated coordinates and the rotated ones ..... $\, Y = PX^{'} \,$ where P is as defined in the link I gave you and $X^{'} = \left( \begin{array}{c} x^{'} \\ y^{'} \end {array} \right)$ .

For practice, I'd advise checking using the technique TKHunny initially suggested .....

(Caveat emptor: My arithmetic is as bad as anyone's ..... but I'm pretty sure everything (that is, C, $\beta$ etc.) is OK. And, since I'm lazy, I checked the eigenvalues using my TI-89 so I'm certain they're OK!)

**heroic** · February 23rd 2008, 05:22 AM

haha i'm lazy too, i got a matrix function on my calculator.

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