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.
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.