I need to learn how to find the properties (foci, vertex, etc) of a conic given its equation, I know

mistergoneo7 2022-07-02 Answered
I need to learn how to find the properties (foci, vertex, etc) of a conic given its equation, I know how to do it when the coeficient of $xy=0$ , but I have troubles in the other case. Can someone recommend me a book, website o document where I can learn how to find the properties of a conic section given its equation please?
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Answers (1)

Gornil2
Answered 2022-07-03 Author has 20 answers
Step 1
Given the equation of the conic in the form
$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$
The term containing xy can be eliminated using a rotation. Define the change of variables as follows
$x={x}^{\prime }\mathrm{cos}\theta -{y}^{\prime }\mathrm{sin}\theta$
$y={x}^{\prime }\mathrm{sin}\theta +{y}^{\prime }\mathrm{cos}\theta$
A point (x, y) will have coordinates $\left({x}^{\prime },{y}^{\prime }\right)$ in a coordinate frame that is rotated by $\theta$ counter-clockwise with respect to the original frame.
Plug these in,
$\begin{array}{rl}& A\left({x}^{\prime 2}{\mathrm{cos}}^{2}\theta +{y}^{\prime 2}{\mathrm{sin}}^{2}\theta -{x}^{\prime }{y}^{\prime }\mathrm{sin}2\theta \right)\\ & +B\left(\frac{1}{2}\left({x}^{\prime 2}-{y}^{\prime 2}\right)\mathrm{sin}2\theta +{x}^{\prime }{y}^{\prime }\mathrm{cos}2\theta \right)\\ & +C\left({x}^{\prime 2}{\mathrm{sin}}^{2}\theta +{y}^{\prime 2}{\mathrm{cos}}^{2}\theta +{x}^{\prime }{y}^{\prime }\mathrm{sin}2\theta \right)\\ & +D\left({x}^{\prime }\mathrm{cos}\theta -{y}^{\prime }\mathrm{sin}\theta \right)+E\left({x}^{\prime }\mathrm{sin}\theta +{y}^{\prime }\mathrm{cos}\theta \right)+F=0\end{array}$
Combining like terms,
$\begin{array}{rl}& {x}^{\prime 2}\left(A{\mathrm{cos}}^{2}\theta +C{\mathrm{sin}}^{2}\theta +\frac{1}{2}B\mathrm{sin}2\theta \right)\\ & +{y}^{\prime 2}\left(A{\mathrm{sin}}^{2}\theta +C{\mathrm{cos}}^{2}\theta -\frac{1}{2}\mathrm{sin}2\theta \right)\\ & +{x}^{\prime }{y}^{\prime }\left(\left(C-A\right)\mathrm{sin}2\theta +B\mathrm{cos}2\theta \right)\\ & +D\left({x}^{\prime }\mathrm{cos}\theta -{y}^{\prime }\mathrm{sin}\theta \right)+E\left({x}^{\prime }\mathrm{sin}\theta +{y}^{\prime }\mathrm{cos}\theta \right)+F=0\end{array}$
Now we want to eliminate the ${x}^{\prime }{y}^{\prime }$ term, so we have to choose $\theta$ such that
$\left(C-A\right)\mathrm{sin}2\theta +B\mathrm{cos}2\theta =0$
that is,
$\mathrm{tan}2\theta =\frac{B}{A-C}$
Now the equation is
${A}^{\prime }{x}^{\prime 2}+{C}^{\prime }{y}^{\prime 2}+{D}^{\prime }{x}^{\prime }+{E}^{\prime }{y}^{\prime }+{F}^{\prime }=0$
where
${A}^{\prime }=A{\mathrm{cos}}^{2}\theta +C{\mathrm{sin}}^{2}\theta +\frac{1}{2}B\mathrm{sin}2\theta$
${C}^{\prime }=A{\mathrm{sin}}^{2}\theta +C{\mathrm{cos}}^{2}\theta -\frac{1}{2}B\mathrm{sin}2\theta$
${D}^{\prime }=D\mathrm{cos}\theta +E\mathrm{sin}\theta$
${E}^{\prime }=-D\mathrm{sin}\theta +E\mathrm{cos}\theta$
${F}^{\prime }=F$
Since rotation preserves lengths, you can find the axes, foci, vertices, and then rotate everything by the angle $\theta$ to obtain them in the original coordinate frame.
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