Finding equation of an ellipsoid Consider I have an ellipsoid (let say an egg) lies in a general form in 3D space. Suppose, I have the equations of two projected views of this egg (e.g. one projected view on x-y plane and another one on y-z plane and naturally, each projected view would be an ellipse with an equation like: $ax^2+bxy+cy^2+dx+ey+f=0$).
Please be advised that since the egg has been placed in space in a general form, none of the axes of the ellipsoid has necessarily the same length as the corresponding axis in its projected view.
When I suppose the ellipsoid center to be at the Origin, its equation would be: $$
\begin{pmatrix}x & y & z\end{pmatrix}
\begin{pmatrix}
\alpha_1 & \beta_3 & \beta_2\\
\beta_3 & \alpha_2 & \beta_1\\
\beta_2 & \beta_1  & \alpha_3
\end{pmatrix}
\begin{pmatrix}x\\y\\z\end{pmatrix} = 1\tag{1}$$
If I write equations of two projected ellipses on xy and yz planes, after simplifying I can make a set of 6 Eqs. with 6 unknowns. However, I think that only 5 of them are independent and so they are not enough to derive the equation of ellipsoid and I need one more equation. What is your opinion?
(LHS values are known)
$$
\begin{cases}
X_1= \beta_1 - \frac{\beta_2\beta_3}{\alpha_1}\\
X_2= \alpha_2 - \frac{\beta_3^2}{\alpha_1}\\
X_3= \alpha_3 - \frac{\beta_2^2}{\alpha_1}\\
Z_1= \alpha_1 - \frac{\beta_2^2}{\alpha_3}\\
Z_2= \alpha_2 - \frac{\beta_1^2}{\alpha_3}\\
Z_3= \beta_3 - \frac{\beta_1\beta_2}{\alpha_3}\\
\end{cases}\tag{2}
$$
 A: The following example is a supplement to achille hui's definitive answer.
Consider the family of ellipsoids
$${\cal E}_\lambda:\quad x^2-2\lambda xy+y^2+(1-\lambda^2)z^2=1-\lambda^2\qquad(-1<\lambda<1)\ .\tag{1}$$
Note that ${\cal E}_0$ is just the unit sphere.
We now project ${\cal E}_\lambda$ along the $y$-axis onto the $(x,z)$-plane. The points on ${\cal E}_\lambda$ giving rise to a boundary point of ${\cal Y}_\lambda:=\pi_y({\cal E}_\lambda)$ are the points where
$${\partial\over\partial y}\bigl(x^2-2\lambda xy+y^2+(1-\lambda^2)z^2\bigr)=2y-2\lambda x=0\ .\tag{2}$$
The points on  ${\cal E}_\lambda$ satisfying $(2)$ are the points $(x,\lambda x,z)$ that also satisfy $(1)$, which amounts to the condition $$x^2+z^2=1\ .$$
It follows that ${\cal Y}_\lambda$ is the unit disk in the $(x,z)$-plane, irrespective of the value of $\lambda$.
By symmetry it follows that the projected ellipsoid ${\cal X}_\lambda:=\pi_x({\cal E}_\lambda)$ is the unit disk in the $(y,z)$-plane, irrespective of the value of $\lambda$.
