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I was studying conics and came around the topic of eliminating cross-product terms when rotating coordinates of a quadratic curve of the form $$A x^2 + B x y + C y^2 + D x + E y + F = 0$$ where $$\begin{align*} A x^2 &= A\left(\cos(\alpha) x' - \sin(\alpha) y'\right)^2\\ &= A\left(\cos^2(\alpha) x'^2 -2\cos(\alpha)\sin(\alpha) x' y' + \sin^2(\alpha) y'^2\right)\\ B x y &= B\left(\cos(\alpha) x' - \sin(\alpha) y'\right)(\sin(\alpha) x' + \cos(\alpha) y') \\ &= B\left(\sin(\alpha)\cos(\alpha)\left(x'^2 - y'^2\right) + \left(\cos^2(\alpha) - \sin^2(\alpha)\right) x' y'\right) \\ C y^2 &= C\left(\sin(\alpha) x' + \cos(\alpha) y'\right)^2 \\ &= C\left(\sin^2(\alpha) x'^2 + 2\sin(\alpha)\cos(\alpha) x' y' + \cos^2(\alpha) y'^2\right) \\ D x &= D\left(\cos(\alpha) x' - \sin(\alpha) y'\right) \\ E y &= E\left(\sin(\alpha) x' + \cos(\alpha) y'\right) \end{align*}$$

I am stuck when they converted from the above equation to $$A' x'^2 + B' x' y' + C' y'^2 + D' x' + E' y' + F' = 0$$

I want to know how they found the value of $ A', B', C'$ etc. as shown below: $$\begin{align*} A' &= A\cos^2\theta + B\cos\theta\sin\theta+ C\sin^2\theta \\ B' &= B\left(\cos^2\theta - \sin^2\theta\right) + 2\left(C - A\right)\sin\theta\cos\theta\\ C' &= A\sin^2\theta - B\sin\theta\cos\theta + C\cos^2\theta \\ D' &= D\cos\theta + E\sin\theta \\ E' &= -D\sin\theta + E\cos\theta \\ F' &= F \end{align*}$$

I tried using double angle formulas but wasn't able to derive the value in the image and I need your help. Any of $A', B', C'$ would do. I will do the rest myself. Just need direction.

I believe that I'm missing some information to provide so please let me know.

an4s
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Syed Ahmed Jamil
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  • Please use MathJax to format your mathematical expressions. You can find a quick reference here. Also, try to make your questions self-contained. Instead of linking to images on external sites that might go stale (and appear to be inaccessible without logging into Google Drive), take the time to include important context in the body of your question. You’re much more likely to get an answer that way instead of getting the question closed. – amd Mar 17 '18 at 17:44

1 Answers1

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Just expand and collect the expressions right above. For example, focusing on the $\,\color{red}{{x'}^2}\,$ terms:

$$ \begin{align} A x^2 + B x y + C y^2 + D x + E y + F &= \color{red}{A}\left(\color{red}{\cos^2(\alpha) x'^2} -2\cos(\alpha)\sin(\alpha) x' y' + \sin^2(\alpha) y'^2\right) \\ &\;\; + \color{red}{B}\left(\color{red}{\sin(\alpha)\cos(\alpha)}\left(\color{red}{x'^2} - y'^2\right) + \left(\cos^2(\alpha) - \sin^2(\alpha)\right) x' y'\right) \\ &\;\; + \color{red}{C}\left(\color{red}{\sin^2(\alpha) x'^2} + 2\sin(\alpha)\cos(\alpha) x' y' + \cos^2(\alpha) y'^2\right) \\ &\;\;+ \ldots \\ &= \underbrace{\color{red}{\left(A \cos^2(\alpha) + B \sin(\alpha)\cos(\alpha) + C \sin^2(\alpha)\right)}}_{\color{red}{\Large{A'}}}\color{red}{{x'}^2} + \ldots \end{align} $$

dxiv
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