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# Proof of eccentricity formula of a general conic

https://en.wikipedia.org/wiki/Eccentricity_(mathematics) ...This wikipedia page has two formulas to calculate eccentricity of a conic.One deals with having the equation of general conic and find eccentricity and other deals with the angle the plane should be cut.I request proof of these results.I couldn't find it anywhere.It would be really helpful if you suggest me a book or link to the proofs.Please reply i am eager to know the proofs.

Note by Ritik Agrawal
1 month, 3 weeks ago

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Both parts can be handled by shifting and rotating the coordinate system being used.

If we start with the equation $$Ax^2 + 2Bxy + Cxy^2 + Dx + Ey + F = 0$$, we can find constants $$\alpha,\beta,\gamma$$ such that this equation becomes $A(x+\alpha)^2 + 2B(x+\alpha)(y+\beta) + C(y+\beta)^2 \; = \; \gamma$ provided that $$AC \neq B^2$$. This is just shifting the origin of the coordinate system, so we might as well consider equations of the form $Ax^2 + 2Bxy + Cy^2 \; = \; \gamma$ We can now rotate the coordinate axes, and find a new coordinate system $$0XY$$ such that the equation becomes $\lambda X^2 + \mu Y^2 \; = \; \gamma$ where $$\lambda,\mu$$ are the eigenvalues of the matrix $\left(\begin{array}{cc} A & B \\ B & C \end{array} \right)$ It is easy to obtain the eccentricity of the ellipse from $$\lambda,\mu,\gamma$$. I have not checked the details on Wikipedia, but presume that you will obtain the answer given if you follow the calculations through.

For the other, the equation of the cone is $z^2 \; =\; \tan^2\alpha(x^2 + y^2)$ and we could assume that the equation of the intersecting plane is $x \sin\beta - z\cos\beta \; = \; c$ If we introduce the orthogonal coordinate system $$0XYZ$$ defined by $X \;=\; x\sin\beta - z\cos\beta \hspace{1cm} Y \; = \; y \hspace{1cm} Z \; = \; x\cos\beta + z\sin\beta$ then we can express the equation of the cone in terms of $$X,Y,Z$$ and intersect it with the plane $$X=c$$ to obtain an equation for the ellipse in terms of $$Y,Z$$, from which it should be easy to determine the eccentricity. Again, I leave you the details.

- 1 month, 3 weeks ago

What about hyperbola?When i get two roots solving the eigenvalue of matrix,how do i know which one is coefficient of x^2(mue)? @Mark Hennings

- 1 month, 3 weeks ago

If you have a hyperbola, one of the eigenvalues will be positive and one negative. You will then be able to manipulate the equation into the form $\frac{X^2}{a^2} - \frac{Y^2}{b^2} \; = \; 1$ and hence obtain the eccentricity.

- 1 month, 3 weeks ago