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I am trying to describe an ellipsoid of the general form

$A\rightarrow\frac{(x-x_{o})^{2}}{a^{2}}+\frac{(y-y_{o})^{2}}{b^{2}}+\frac{(z-z_{0})^{2}}{c^{2}}=1$

Into this form(quadric form)

$A\rightarrow a_{1}x^{2}+b_{1}y^{2}+c_{1}z^{2}+2f_{1}yz+2g_{1}zx+2h_{1}xy+2p_{1}x+2q_{1}y+2r_{1}z+d_{1}=0$

I expanded the first equation, to get the following equation

$A\rightarrow(bc)^{2}x^{2}+(ac)^{2}y^{2}+(ab)^{2}z^{2}-2(bc)^{2}xx_{0}-2(ac)^{2}yy_{0}-2(ab)^{2}zz_{0}+(bc)^{2}x_{0}^{2}+(ac)^{2}y_{0}^{2}+(ab)^{2}z_{0}^{2}-a^{2}b^{2}c^{2}=0$

I was working under the assumption that both the forms should be able to be interchanged. So comparing the second and third equation, I get

$a_{1}=b^{2}c^{2}$

$b_{1}=a^{2}c^{2}$

$c_{1}=a^{2}b^{2}$

$d_{1}=(bc)^{2}x_{0}^{2}+(ac)^{2}y_{0}^{2}+(ab)^{2}z_{0}^{2}-a^{2}b^{2}c^{2}$

$p_{1}=-2(bc)^{2}x_{0}$

$q_{1}=-2(ac)^{2}y_{0}$

$r_{1}=-2(ab)^{2}z_{0}$

But I am not able to figure out how to get $f_1$,$g_1$ and $h_1$. How do I do that? Are they simply zero? When would they not be zero?

I found this link: Formula for an Ellipsoid? which asks exactly my question, but I do not see how the answer addressed the question. So I decided it was I needed a fresh question and answer.

P.S. I am not a math or physics major, and therefore have a limited exposure to complex mathematical symbols and concepts, so please try to keep it as simple and physical as possible.

Ayush Agrawal
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  • Your assumption that these formula are interchangeable is in general false. An ellipsoid is one type of quadric, so the general quadric formula can correspond to more varied surfaces than just ellipsoids. If you head to https://en.wikipedia.org/wiki/Quadric, they have some illustrations for other quadric. – N.Bach Nov 16 '17 at 10:55
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    Beyond "quadric vs ellipsoid", if you know that you have an ellipsoid, and that it is axis-aligned with the axes of your coordinate system, then the coefficients $f_1,g_1,h_1$ are zero, and the rest of your identities are verified. If you rotate your ellipsoid, then $f_1,g_1,h_1$ become non-zero, and your other identities also become wrong. – N.Bach Nov 16 '17 at 10:58

1 Answers1

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Your first equation describes an axis-aligned ellipsoid: the main axes of the ellipsoid align with the axes of the coordinate system. This leads to the mixed-coordinate coefficients $f_1,g_1,h_1$ being zero. If you want to go the other way and have non-zero mixed terms, you need to introduce a rotation into your formula. See e.g. this post of mine for a 2d example of the equation for an ellipsis which includes rotation.

MvG
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