Rotation of axes to eliminate the xy component

[pathos]

Hello,

I am a parent with a science background, but have not been able to help my son with his precalc problem of axes rotation. The basic problem is to discuss the equation- 34x^2-24xy+41y^2-25=0. Basically, I understand that the idea is to eliminate the x and y components and express the equation in terms of x' and y'. We have used the half angle formulas to determine sin theta and cos theta. As we plug the values in and try to solve the equation, we are always left with an x'y' component, which does not cancel out. I assume that we may be missing some basic algebraic manipulation, but I can't see where. Our working of this problem has filled a whole sheet of paper, and hence I'm not sure if it is best to reproduce it here. Please let me know if posting the work in total would be appropriate or necessary.

Thank you for your help,

 

[soroban]

Hello, pathos!

I sympathize . . . it took me three tries to get it.
You probably made a simple error in trig or in algebra . . . I did both!

Discuss the equation: $\displaystyle \,34x^2\,-\,24xy\,+\,41y^2\,-\,25\:=\:0$

Here are the formulas that I used . . .

Given the second=degree equation: $\displaystyle \,Ax^2\,+\,Bxy\,+\,Cy^2\,+\,Dx\,+\,Ey\,+\,F\;=\;0$

$\displaystyle \;\;$the angle of rotation $\displaystyle \theta$ is given by: $\displaystyle \:\tan2\theta\;=\;\frac{B}{A\,-\,C}\;\;$ [1]

$\displaystyle \;\;$and the change of coordinates* is: $\displaystyle \:\begin{array}{cc}x'\:=\:x\cdot\cos\theta\,-\,y\cdot\sin\theta\\ y'\:=\:x\cdot\sin\theta\,+\,y\cdot\cos\theta\end{array}\;\;$ [2]

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From [1], we have: $\displaystyle \tan2\theta\:=\:\frac{-24}{34\,-\,41}\:=\:\frac{24}{7}$

We have a 7-24-25 right triangle: $\displaystyle \,\sin2\theta\,=\,\frac{24}{25},\;\cos2\theta\,=\,\frac{7}{25}$

Then: \(\displaystyle \:\begin{array}{cc}\sin^2\theta\:=\:\frac{1\,-\,\cos2\theta}{2}\:=\:\frac{1\,-\,\frac{7}{25}}{2}\:=\:\frac{9}{25}\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{3}{5} \\


\cos^2\theta\:=\:\frac{1\,+\,\cos2\theta}{2}\:=\:\frac{1\,+\,\frac{7}{25}}{2}\:=\:\frac{16}{25}\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{4}{5}\end{array}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

From [2], we have: $\displaystyle \:\begin{array}{cc}x'\:=\;\frac{4}{5}x\,-\,\frac{3}{5}y \\ y'\:=\:\frac{3}{5}x\,+\,\frac{4}{5}y\end{array}$

Substitute: \(\displaystyle \:34\left(\frac{4}{5}x\,-\,\frac{3}{5}y\right)^2\,-\,24\left(\frac{4}{5}x\,-\,\frac{3}{5}y\right)\left(\frac{3}{5}x\,+\,\frac{4}{5}{y\right)\,+\,41\left(\frac{3}{5}x\,+\,\frac{4}{5}y\right)^2\,-\,25\;=\;0\)

$\displaystyle \;\;\frac{34}{25}(16x^2\,-\,24xy\,+\,9y^2) - \frac{24}{25}(12x^2\,+\,7xy\,-\,12y^2)\,+\,\frac{41}{25}(9x^2\,+\,24xy\,+\,26y^2)\,-\,25\;=\;0$

$\displaystyle \;\;\frac{1}{25}\left(544x^2\,-\,816xy\,+\,306y^2\,-\,288x^2\,-\,168xy\,+\,288y^2\,+\,369x^2\,+\,984xy\,+\,656y^2\right)\,-\,25\;=\;0$

\(\displaystyle \;\;\;\;\frac{1}{25}\left(625x^2\,+\,\)0xy$\displaystyle \,+\,1250y^2)\,-\,25\;=\;0\;\;$ . . .YAY!


We have: $\displaystyle \,25x^2\,+\,50y^2\;=\;25\;\;\Rightarrow\;\;x^2\,+\,\frac{y^2}{\frac{1}{2}}\:=\:1$

This is an ellipse** with semimajor axis $\displaystyle a\,=\,1,\;$ semiminor axis $\displaystyle b\,=\,\frac{1}{\sqrt{2}}$
$\displaystyle \;\;$ rotated through an angle of: $\displaystyle \,\sin^{-1}\left(\frac{3}{5}\right)\:\approx\:36.87^o$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* There is a set of formulas that lets us go directly to the new equation.

$\displaystyle \;\;A'\;=\;A\cdot\cos^2\theta\,+\,B\cdot\sin\theta\cdot\cos\theta\,+\,C\cdot\sin^2\theta$

$\displaystyle \;\;B'\;=\;B\cdot\cos2\theta\,+\,(C-A)\sin2\theta$

$\displaystyle \;\;C'\;=\;A\cdot\sin^2\theta\,-\,B\cdot\sin\theta\cdot\cos\theta\,+\,C\cdot\cos^2\theta$

$\displaystyle \;\;D'\;=\;D\cdot\cos\theta\,+\,E\cdot\sin\theta$

$\displaystyle \;\;E'\;=\;-D\cdot\sin\theta\,+\,E\cdot\cos\theta$

$\displaystyle \;\;F'\;=\;F$


I used these to check my work, but I wanted to do it "the long way".

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** By the way, these problems have a "discriminant", too: $\displaystyle \;\Delta\;=\;B^2\,-\,4AC$

$\displaystyle \Delta\,=\,0:\text{ parabola}$
$\displaystyle \Delta\,<\,0:\text{ ellipse}$
$\displaystyle \Delta\,>\,0:\text{ hyperbola}$

 

[pathos]

Soroban,

Thanks very much, yes, clearly we made our mistake in the algebraic manipulations during the substitution of x' for x. As frustrating as this problem was, I did enjoy refreshing my aquaintance with conics and trig. I remember in the "old days", having to bump and grind through problems such as this, and if there was a stubling block, having to wait for the prof to help out when possible. This online resource is an incredible asset to folks current taking math.

Best regards,

Pathos

 

[bike]

Thank you for the post, literally made on the day I was born. It has now helped me in my math class 15 years later!

Hello,

I am a parent with a science background, but have not been able to help my son with his precalc problem of axes rotation. The basic problem is to discuss the equation- 34x^2-24xy+41y^2-25=0. Basically, I understand that the idea is to eliminate the x and y components and express the equation in terms of x' and y'. We have used the half angle formulas to determine sin theta and cos theta. As we plug the values in and try to solve the equation, we are always left with an x'y' component, which does not cancel out. I assume that we may be missing some basic algebraic manipulation, but I can't see where. Our working of this problem has filled a whole sheet of paper, and hence I'm not sure if it is best to reproduce it here. Please let me know if posting the work in total would be appropriate or necessary.

Thank you for your help,

Hello, pathos!

I sympathize . . . it took me three tries to get it.
You probably made a simple error in trig or in algebra . . . I did both!


Here are the formulas that I used . . .

Given the second=degree equation: $\displaystyle \,Ax^2\,+\,Bxy\,+\,Cy^2\,+\,Dx\,+\,Ey\,+\,F\;=\;0$

$\displaystyle \;\;$the angle of rotation $\displaystyle \theta$ is given by: $\displaystyle \:\tan2\theta\;=\;\frac{B}{A\,-\,C}\;\;$ [1]

$\displaystyle \;\;$and the change of coordinates* is: $\displaystyle \:\begin{array}{cc}x'\:=\:x\cdot\cos\theta\,-\,y\cdot\sin\theta\\ y'\:=\:x\cdot\sin\theta\,+\,y\cdot\cos\theta\end{array}\;\;$ [2]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

From [1], we have: $\displaystyle \tan2\theta\:=\:\frac{-24}{34\,-\,41}\:=\:\frac{24}{7}$

We have a 7-24-25 right triangle: $\displaystyle \,\sin2\theta\,=\,\frac{24}{25},\;\cos2\theta\,=\,\frac{7}{25}$

Then: \(\displaystyle \:\begin{array}{cc}\sin^2\theta\:=\:\frac{1\,-\,\cos2\theta}{2}\:=\:\frac{1\,-\,\frac{7}{25}}{2}\:=\:\frac{9}{25}\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{3}{5} \\


\cos^2\theta\:=\:\frac{1\,+\,\cos2\theta}{2}\:=\:\frac{1\,+\,\frac{7}{25}}{2}\:=\:\frac{16}{25}\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{4}{5}\end{array}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

From [2], we have: $\displaystyle \:\begin{array}{cc}x'\:=\;\frac{4}{5}x\,-\,\frac{3}{5}y \\ y'\:=\:\frac{3}{5}x\,+\,\frac{4}{5}y\end{array}$

Substitute: \(\displaystyle \:34\left(\frac{4}{5}x\,-\,\frac{3}{5}y\right)^2\,-\,24\left(\frac{4}{5}x\,-\,\frac{3}{5}y\right)\left(\frac{3}{5}x\,+\,\frac{4}{5}{y\right)\,+\,41\left(\frac{3}{5}x\,+\,\frac{4}{5}y\right)^2\,-\,25\;=\;0\)

$\displaystyle \;\;\frac{34}{25}(16x^2\,-\,24xy\,+\,9y^2) - \frac{24}{25}(12x^2\,+\,7xy\,-\,12y^2)\,+\,\frac{41}{25}(9x^2\,+\,24xy\,+\,26y^2)\,-\,25\;=\;0$

$\displaystyle \;\;\frac{1}{25}\left(544x^2\,-\,816xy\,+\,306y^2\,-\,288x^2\,-\,168xy\,+\,288y^2\,+\,369x^2\,+\,984xy\,+\,656y^2\right)\,-\,25\;=\;0$

\(\displaystyle \;\;\;\;\frac{1}{25}\left(625x^2\,+\,\)0xy$\displaystyle \,+\,1250y^2)\,-\,25\;=\;0\;\;$ . . .YAY!


We have: $\displaystyle \,25x^2\,+\,50y^2\;=\;25\;\;\Rightarrow\;\;x^2\,+\,\frac{y^2}{\frac{1}{2}}\:=\:1$

This is an ellipse** with semimajor axis $\displaystyle a\,=\,1,\;$ semiminor axis $\displaystyle b\,=\,\frac{1}{\sqrt{2}}$
$\displaystyle \;\;$ rotated through an angle of: $\displaystyle \,\sin^{-1}\left(\frac{3}{5}\right)\:\approx\:36.87^o$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* There is a set of formulas that lets us go directly to the new equation.

$\displaystyle \;\;A'\;=\;A\cdot\cos^2\theta\,+\,B\cdot\sin\theta\cdot\cos\theta\,+\,C\cdot\sin^2\theta$

$\displaystyle \;\;B'\;=\;B\cdot\cos2\theta\,+\,(C-A)\sin2\theta$

$\displaystyle \;\;C'\;=\;A\cdot\sin^2\theta\,-\,B\cdot\sin\theta\cdot\cos\theta\,+\,C\cdot\cos^2\theta$

$\displaystyle \;\;D'\;=\;D\cdot\cos\theta\,+\,E\cdot\sin\theta$

$\displaystyle \;\;E'\;=\;-D\cdot\sin\theta\,+\,E\cdot\cos\theta$

$\displaystyle \;\;F'\;=\;F$


I used these to check my work, but I wanted to do it "the long way".

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** By the way, these problems have a "discriminant", too: $\displaystyle \;\Delta\;=\;B^2\,-\,4AC$

$\displaystyle \Delta\,=\,0:\text{ parabola}$
$\displaystyle \Delta\,<\,0:\text{ ellipse}$
$\displaystyle \Delta\,>\,0:\text{ hyperbola}$