Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4

• May 28th 2011, 01:34 PM
Lizziepaquette
Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4
In essence I'm trying to take a horizontal ellipse and put it at an angle, rotating it around the origin. It's for my conics project, I'm making the elliptical orbits of the planets.

So I have the a, b, and c terms of the ellipse, (as in the distance from vertice to center and such) which are 28, 21, and $7\sqrt{7}$ respectively. When plugged into the equation they give:

${x}^{2} / 784 +{y}^{2} /441 =1$

which turns into

$441 {x}^{2} + 784{y}^{2 } - 345,744 = 0$

from staring at my math book i've assumed that these x's and y's are actually x prime and y prime in the grand scheme of how we normally look at rotations.

so actually

$441{x'}^{2 } + 784{y'}^{2 } - 345,744 = 0$

with this is mind I tried to apply the transformation equations knowing that I was trying to acheive a rotation of $\pi / 4$

$x = x' \cos \pi /4 - y' \sin \pi /4$

$x = (x' - y')/\sqrt{2}$

$\sqrt{2}x = x' - y'$

$\sqrt{2}y = x' + y'$

But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable and create a Bxy term.

So please, its due Wednesday... help?
• May 29th 2011, 12:20 AM
Opalg
Quote:

Originally Posted by Lizziepaquette
I was trying to achieve a rotation of $\pi / 4$

$x = x' \cos \pi /4 - y' \sin \pi /4$

$x = (x' - y')/\sqrt{2}$

$\sqrt{2}x = x' - y'$

$\sqrt{2}y = x' + y'$

But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable

If you get from $(x',y')$ to $(x,y)$ by rotating through $\pi/4$, then you can get back from $(x,y)$ to $(x',y')$ by rotating through $-\pi/4$. Therefore if $x = x' \cos \pi /4 - y' \sin \pi /4$ then $x' = x \cos(- \pi /4) - y \sin(- \pi /4) = \tfrac1{\sqrt2}x +\tfrac1{\sqrt2}y.$
• May 29th 2011, 06:47 AM
Lizziepaquette
Hmm let me plug it in
• May 29th 2011, 06:48 AM
Lizziepaquette
I did the math out with that concept and my end result is

$597.5{x}^{2} -373xy + 597.5{y}^{2} -345,744 = 0$

I haven't done the forward math to check but i graphed it and it looks about right.

Thanks (Smile)
The text passage describes a student's attempt to rotate an ellipse by an angle of $\pi/4$ and their struggle with substituting the transformed coordinates into the original equation to find the $Bxy$ term. The student receives help from another user, which leads to a solution.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4

• May 28th 2011, 01:34 PM
Lizziepaquette
Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4
In essence I'm trying to take a horizontal ellipse and put it at an angle, rotating it around the origin. It's for my conics project, I'm making the elliptical orbits of the planets.

So I have the a, b, and c terms of the ellipse, (as in the distance from vertice to center and such) which are 28, 21, and $7\sqrt{7}$ respectively. When plugged into the equation they give:

${x}^{2} / 784 +{y}^{2} /441 =1$

which turns into

$441 {x}^{2} + 784{y}^{2 } - 345,744 = 0$

from staring at my math book i've assumed that these x's and y's are actually x prime and y prime in the grand scheme of how we normally look at rotations.

so actually

$441{x'}^{2 } + 784{y'}^{2 } - 345,744 = 0$

with this is mind I tried to apply the transformation equations knowing that I was trying to acheive a rotation of $\pi / 4$

$x = x' \cos \pi /4 - y' \sin \pi /4$

$x = (x' - y')/\sqrt{2}$

$\sqrt{2}x = x' - y'$

$\sqrt{2}y = x' + y'$

But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable and create a Bxy term.

So please, its due Wednesday... help?
• May 29th 2011, 12:20 AM
Opalg
Quote:

Originally Posted by Lizziepaquette
I was trying to achieve a rotation of $\pi / 4$

$x = x' \cos \pi /4 - y' \sin \pi /4$

$x = (x' - y')/\sqrt{2}$

$\sqrt{2}x = x' - y'$

$\sqrt{2}y = x' + y'$

But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable

If you get from $(x',y')$ to $(x,y)$ by rotating through $\pi/4$, then you can get back from $(x,y)$ to $(x',y')$ by rotating through $-\pi/4$. Therefore if $x = x' \cos \pi /4 - y' \sin \pi /4$ then $x' = x \cos(- \pi /4) - y \sin(- \pi /4) = \tfrac1{\sqrt2}x +\tfrac1{\sqrt2}y.$
• May 29th 2011, 06:47 AM
Lizziepaquette
Hmm let me plug it in
• May 29th 2011, 06:48 AM
Lizziepaquette
I did the math out with that concept and my end result is

$597.5{x}^{2} -373xy + 597.5{y}^{2} -345,744 = 0$

I haven't done the forward math to check but i graphed it and it looks about right.

Thanks (Smile)
user
The text passage describes a student's attempt to rotate an ellipse by an angle of $\pi/4$ and their struggle with substituting the transformed coordinates into the original equation to find the $Bxy$ term. The student receives help from another user, which leads to a solution.

TYPE: SINGLE_QUESTION_WITH_ANSWER