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https://ibb.co/8DfMSvC as it is shown in the image I would be grateful if you could explain me how did we get orange highlighted equation from yellow highlighted one. Can you show me the steps of it?

$$ \begin{split} 2(x' \cos{\theta})^2 \\-4(x' \cos{\theta}y' \sin{\theta}) \\+ 2 (y' \sin{\theta})^2 \\+\sqrt{3} [ (x' \cos{\theta} - y'\sin{\theta}) (x'\sin{\theta} + y' \cos{\theta}) ]\\ + (x' \sin{\theta})^2 -2 (x'\sin{\theta} y' \cos{\theta}) + (y' \cos{\theta})^2\\ = 8 \\ \end{split} $$ $$ \Rightarrow \underbrace{-2\cos{\theta}\sin{\theta}}_{=-2\sin{2\theta}} + \sqrt{3} \underbrace{(\cos^2 2{\theta} - \sin^2{\theta})}_{=\cos{2\theta}} = 0 $$

Question asks us to get rid of x'y' through finding a perfect translation angle. I just couldn't simplfy from yellow to orange...

Thanks in advance.

Matti P.
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zkngkdnz
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    Please review my transcription of the equations. It could be that I have made a small mistake there. – Matti P. Jan 10 '20 at 07:28

1 Answers1

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It seems fairly straightforward. The substitution is: $$x=x'\cos\theta-y'\sin\theta,y=x'\sin\theta+y'\cos\theta$$ That gives the coefficient of the $x'y'$ term on the lhs as $$-4\cos\theta\sin\theta+\sqrt3(\cos^2\theta-\sin^2\theta)+2\cos\theta\sin\theta $$ The idea is to choose the constant $\theta$ to make this term vanish. So we need $$2\cos\theta\sin\theta=\sqrt3(\cos^2\theta-\sin^2\theta)$$ You then use the double angle formulae to get $$\sin2\theta=\sqrt3\cos2\theta$$ Incidentally, this gives a rotation of the coordinate axes, rather than a translation.

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More detail on the substitution. We start with $2x^2+\sqrt3 xy+y^2=8$ and we make the substitution above. The first term becomes $$2x^2=2(x'\cos\theta-y'\sin\theta)^2$$ $$=2x'^2\cos^2\theta-4x'y'\cos\theta\sin\theta+2y'^2\sin^2\theta\quad(1)$$ The second term becomes $$\sqrt3 xy=\sqrt3(x'\cos\theta-y'\sin\theta)(x'\sin\theta+y'\cos\theta)$$ $$=\sqrt3 x'^2\sin\theta\cos\theta+x'y'\sqrt3(\cos^2\theta-\sin^2\theta)-y'^2\sqrt3\sin\theta\cos\theta\quad(2)$$ The third term becomes $$y^2=(x'\sin\theta+y'\cos\theta)^2$$ $$=x'^2\sin^2\theta+2x'y'\cos\theta\sin\theta+y'^2\cos^2\theta\quad(3)$$ We are interested in the $x'y'$ terms. Adding these three terms together we get the result above.

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We start with the equation $$2x^2+\sqrt3\ xy+y^2=0$$ We make the substitutions. The three terms turn out as above at (1), (2), (3). So the equation turns into the sum of the right-hand sides of (1), (2), (3) equals 0. If we add these right-hand sides we can rearrange to get a term in $x'^2$, a term in $x'y'$ and a term in $y'^2$. We are only interested in the term in $x'y'$ because we want that term to be 0. So from each of (1), (2), (3) we pick out the $x'y'$ term and we add them. (1) gives $$-4x'y'\cos\theta\sin\theta$$ (2) gives $$x'y'\sqrt3(\cos^2\theta-\sin^2\theta)$$ and (3) gives $$2x'y'\cos\theta\sin\theta$$ Adding these gives $$-2x'y'\cos\theta\sin\theta+x'y'\sqrt3(\cos^2\theta-\sin^2\theta)$$ [Note that $-4x'y'\cos\theta\sin\theta$ and $2x'y'\cos\theta\sin\theta$ add to give $-2x'y'\cos\theta\sin\theta$.] That takes us to the fourth line of this answer.

Note that I have not worked out the final result of adding (1), (2), (3), because we don't care about the coefficients of $x'^2$ and $y'^2$. All we care about is finding the right value of $\theta$ to eliminate the $x'y'$ term.

almagest
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  • Sorry for misused term. Can you show step by step how we get to the x'y' term on the lhs as like that simple? – zkngkdnz Jan 10 '20 at 13:49
  • Sorry. Part of a line got missed out. Now fixed. Is that clear now? – almagest Jan 10 '20 at 14:13
  • I couldn't make myself understood sorry, what I can't manage to do here is: when we substitute for x and y, we get a looong equation, right? This one that is shared on my question. I want to be shown every little step of getting into that simple equation which starts with -4cosTHETAsinTHETA... From where should I start (may I use wrong term but) taking into common factors? Or is it bracketing? Can you show me step by step how did you get equation starting with -4cosTHETAsinTHETA... ? Because that's the only part that I am stuck with. Sorry for hardwork and thank you. – zkngkdnz Jan 11 '20 at 14:46
  • @zkngkdnz I have added something on that. – almagest Jan 11 '20 at 15:05
  • Look, I know. The question is so simple that you can't understand how I can't understand... Can you please show the steps of adding these three terms? One by one if possible. Like: "First we add these, and then these, and we get that and we sum it up with that. We substitute for the result. We replace the equation with bla bla because it is formula." etc.. What I'm asking for is adding up these three terms actually. That's the only part that I cannot manage to do. – zkngkdnz Jan 13 '20 at 07:19
  • @zkngkdnz I have added a few more lines, but I am still unclear what your difficulty is. – almagest Jan 13 '20 at 08:44
  • Thank you! My bad... sorry. I didn't understand "we are looking for only coefficient of x'y' not x'^2y'^2" first but finally I do. Thank you and sorry for being a pain in your neck. – zkngkdnz Jan 14 '20 at 11:44