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Find the eccentricity of the conic $$(3x + 9)^2 + (3y – 12)^2 – (2x – y)^2 = 6y – 12x + 9$$

For this type of problem where the axis is not parallel to $x$-axis or $y$-axis, how do I factorize it so that I can get the equation in $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$ or $\frac{X^2}{a^2}-\frac{Y^2}{b^2}=1$ form and easily find the value of $a$ or $b$.

Blue
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Samar Imam Zaidi
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  • Do you absolutely need rotating the conic, or you just want to compute the eccentricity? In the latter case you can use a simple formula. – Intelligenti pauca Sep 02 '19 at 13:42
  • Anyway, principal axes are the eigenvectors of the matrix associated to the quadratic part of the equation, see here: https://math.stackexchange.com/questions/2490486/why-do-the-eigenvectors-of-the-quadratic-form-defining-a-conic-section-give-you – Intelligenti pauca Sep 02 '19 at 13:47
  • https://en.wikipedia.org/wiki/Conic_section#Eccentricity_in_terms_of_coefficients – amd Sep 02 '19 at 20:10

1 Answers1

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Suppose you are given the general second degree equation $C\equiv ax^2+2hxy+by^2+2gx+2fy+c=0$

  1. If you want eliminate the $xy$ term , rotate the coordinate axes by $$\displaystyle\theta=\frac{1}{2}\tan^{-1}\left(\frac{2h}{a-b}\right)$$ By doing so, the respective axes of the conic will become parallel to the coordinate axes.

  2. If you want to eliminate the first degree terms, shift the origin to the center of conic which you can find by solving the two equations $$\frac{\partial(C)}{\partial x}=0\ \text{and}\ \frac{\partial(C)}{\partial y}=0.$$

Now, in your question we are concerned only with the eccentricity, so just rotating axes will do. Rotate the axes by $$\theta=\frac{1}{2}\tan^{-1}\left(-\frac{4}{3}\right)=\tan^{-1}\left(-\frac{1}{2}\right)$$ and write the equation in general form by completing square (which will still be messy). Figure for reference.

$$\begin{align}4x^2+9y^2+42\sqrt 5 x-18\sqrt 5y+216&=0\\ \frac{\left(x+\frac{21\sqrt 5}{4}\right)^2}{\left(\frac{39}{4}\right)^2}+\frac{(y-\sqrt 5)^2}{\left(\frac{13}{2}\right)^2}&=1\end{align}$$

SarGe
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