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: \(\displaystyle \:\begin{array}{cc}\sin^2\theta\:=\:\frac{1\,-\,\cos2\theta}{2}\:=\:\frac{1\,-\,\frac{7}{25}}{2}\:=\:\frac{9}{25}\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{3}{5} \\
\cos^2\theta\:=\:\frac{1\,+\,\cos2\theta}{2}\:=\:\frac{1\,+\,\frac{7}{25}}{2}\:=\:\frac{16}{25}\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{4}{5}\end{array}\)
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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: \(\displaystyle \,\sin^{-1}\left(\frac{3}{5}\right)\:\approx\:36.87^o\)
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* 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".
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** By the way, these problems have a "discriminant", too: \(\displaystyle \;\Delta\;=\;B^2\,-\,4AC\)
\(\displaystyle \Delta\,=\,0:\text{ parabola}\)
\(\displaystyle \Delta\,<\,0:\text{ ellipse}\)
\(\displaystyle \Delta\,>\,0:\text{ hyperbola}\)
