Hi,
I came across this
here http://books.google.co.uk/books?id=...Ag#v=onepage&q=ax^2 +bxy+cy^2+dx+ey+f&f=false
I've tried to apply it to the equation \(\displaystyle x^2+2xy-y^2=\sqrt{3}\) but seem to be doing something wrong as the xy term is not being eliminated.
Here's my working...
From my equation I have \(\displaystyle A=1\ \ \ B=2\ \ \ C=-1\)
\(\displaystyle cot 2\theta=\dfrac{A-C}{B}\ \ \ \implies\ \ \ \ tan 2\theta=\dfrac{B}{A-C}\)
\(\displaystyle tan 2\theta=\dfrac{2}{1-(-1)}\ \ \ \ \ tan 2\theta=1\ \ \ \ \therefore\ \ \ 2\theta=\dfrac{\pi}{4}radians\ \ \ \ so\ \ \ \theta=\dfrac{\pi}{8}radians\)
using the substitutions from the above image I get
\(\displaystyle x=x'cos\left(\dfrac{\pi}{8}\right)-y'sin\left(\dfrac{\pi}{8}\right)\ \ \ \ \ \ \ \ y=x'sin\left(\dfrac{\pi}{8}\right) + y'cos\left(\dfrac{\pi}{8}\right)\)
to save space I'll let \(\displaystyle P=sin\left(\dfrac{\pi}{8}\right)=\dfrac{\sqrt{2-\sqrt{2}}}{2}\ \ \ \ and\ \ \ \ Q=cos\left(\dfrac{\pi}{8}\right)=\dfrac{\sqrt{2 + \sqrt{2}}}{2}\)
so I have \(\displaystyle x=Qx'-Py'\ \ \ \ \ and\ \ \ \ \ y=Px'+Qy'\)
substituting these values in the original equation I get
\(\displaystyle \left(Qx'-Py'\right)^2+2\left(Qx'-Py'\right)\left(Px'+Qy'\right)-\left(Px'+Qy'\right)^2=\sqrt{3}\)
\(\displaystyle \implies\ \ \ \ (Q^2+2PQ-P^2)x'^2+2(Q^2-2PQ-P^2)x'y'+(P^2-2PQ-Q^2)y'^2=\sqrt{3}\)
The important part is the central term which should equal 0 but...
\(\displaystyle \left(\dfrac{\sqrt{2 + \sqrt{2}}}{2}\right)^2-2\left(\dfrac{\sqrt{2-\sqrt{2}}}{2}\right)\left(\dfrac{\sqrt{2 + \sqrt{2}}}{2}\right)-\left(\dfrac{\sqrt{2-\sqrt{2}}}{2}\right)^2=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}}{2}\)
so the \(\displaystyle x'y'\) term has not been eliminated.
Any idea what I'm doing wrong ?
Thanks
I came across this
here http://books.google.co.uk/books?id=...Ag#v=onepage&q=ax^2 +bxy+cy^2+dx+ey+f&f=false
I've tried to apply it to the equation \(\displaystyle x^2+2xy-y^2=\sqrt{3}\) but seem to be doing something wrong as the xy term is not being eliminated.
Here's my working...
From my equation I have \(\displaystyle A=1\ \ \ B=2\ \ \ C=-1\)
\(\displaystyle cot 2\theta=\dfrac{A-C}{B}\ \ \ \implies\ \ \ \ tan 2\theta=\dfrac{B}{A-C}\)
\(\displaystyle tan 2\theta=\dfrac{2}{1-(-1)}\ \ \ \ \ tan 2\theta=1\ \ \ \ \therefore\ \ \ 2\theta=\dfrac{\pi}{4}radians\ \ \ \ so\ \ \ \theta=\dfrac{\pi}{8}radians\)
using the substitutions from the above image I get
\(\displaystyle x=x'cos\left(\dfrac{\pi}{8}\right)-y'sin\left(\dfrac{\pi}{8}\right)\ \ \ \ \ \ \ \ y=x'sin\left(\dfrac{\pi}{8}\right) + y'cos\left(\dfrac{\pi}{8}\right)\)
to save space I'll let \(\displaystyle P=sin\left(\dfrac{\pi}{8}\right)=\dfrac{\sqrt{2-\sqrt{2}}}{2}\ \ \ \ and\ \ \ \ Q=cos\left(\dfrac{\pi}{8}\right)=\dfrac{\sqrt{2 + \sqrt{2}}}{2}\)
so I have \(\displaystyle x=Qx'-Py'\ \ \ \ \ and\ \ \ \ \ y=Px'+Qy'\)
substituting these values in the original equation I get
\(\displaystyle \left(Qx'-Py'\right)^2+2\left(Qx'-Py'\right)\left(Px'+Qy'\right)-\left(Px'+Qy'\right)^2=\sqrt{3}\)
\(\displaystyle \implies\ \ \ \ (Q^2+2PQ-P^2)x'^2+2(Q^2-2PQ-P^2)x'y'+(P^2-2PQ-Q^2)y'^2=\sqrt{3}\)
The important part is the central term which should equal 0 but...
\(\displaystyle \left(\dfrac{\sqrt{2 + \sqrt{2}}}{2}\right)^2-2\left(\dfrac{\sqrt{2-\sqrt{2}}}{2}\right)\left(\dfrac{\sqrt{2 + \sqrt{2}}}{2}\right)-\left(\dfrac{\sqrt{2-\sqrt{2}}}{2}\right)^2=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt{2}}}{2}\)
so the \(\displaystyle x'y'\) term has not been eliminated.
Any idea what I'm doing wrong ?
Thanks
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