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I want to find the equation of the plane passing the intersection of two intersecting ellipsoids. the intersection of two ellipsoids is always a ellipse. I need to find the equation of the planar surface containing this ellipse.

Does have anyone any idea in this matter?

thanks in advance for your any response

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    Put them in a equation and Try to cancel out the squared terms to attain a plane equation – 2017-02-04
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    It would have been better if you gave an example of the equations of elipsoids. – 2017-02-04
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    would not be that easy. there are three independent squared variables (x, y and z) in each ellipsoid's equation. You cal at most cancel out one of them and there still would be two squared variables. – 2017-02-05
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    The intersection of two ellipsoids is definitely not an ellipse in general. – 2017-03-09

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Only two similar and translated ellipsoids guarantee planar ellipse section:

For $\lambda > 0$,

\begin{align*} \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} &= 1 \tag{1} \\ \frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}+\frac{(z-\gamma)^2}{c^2} &= \lambda^2 \tag{2} \\ \end{align*}

$(1)-(2)$,

$$\frac{2\alpha x}{a^2}+\frac{2\beta y}{b^2}+\frac{2\gamma z}{c^2}= 1-\lambda^2+\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}$$

which is the radical plane of the two ellipsoids.

The centre of section is:

$$ \frac {1-\lambda^2+\dfrac{\alpha^2}{a^2}+\dfrac{\beta^2}{b^2}+\dfrac{\gamma^2}{c^2}}{2\left( \dfrac{\alpha^2}{a^2}+\dfrac{\beta^2}{b^2}+\dfrac{\gamma^2}{c^2} \right)} \begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} $$

Note briefly:

No intersections if $\, \dfrac{\alpha^2}{a^2}+\dfrac{\beta^2}{b^2}+\dfrac{\gamma^2}{c^2}>(1+\lambda)^2$

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    Your two ellipsoids are not only similar; they are also oriented identically. – 2017-03-09
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    Yes, thanks for your reminding. – 2017-03-09