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I have an oblate spheroid (E): (x/a)^2+(y/a)^2+(z/b)^2 = 1 and a plane (P): ux+vy+wz+d=0 / d=0 (passing by origin)

I have plugged z = - (ux+vy)/w from (P) into (E), then I became this equation:

[(wb)^2+(au)^2]x^2 + [(wb)^2+(av)^2]y^2 + (2*uva^2)xy -(wab)^2 = 0

That is the equation of conics (here in this case, an ellipse, bcz intersection of an plane passing by origin with an oblate spheroid is always an ellipse except when it is parallel to plane xy is a circle)

General equation of conics:

Ax^2 + Bxy + Cy^2 + F = 0 [Dx and Ey vanish bcz the ellipse centered always at O(0,0,0)]

I have rotated and everything done, the I got this form:

A'x^2 + C'y^2 - F' = 0 with A' = a^2*[u^2 + v^2] + (bw)^2 C' = (bw)^2 F' = (abw)^2

When I continue to find the semi-axis major a' and semi-axis minor b' of this ellipse, I get:

a' = a and b' = (abw)/sqrt[a^2*(u^2 + v^2) + (b*w)^2]

My problem, is: the semi-axis major a' shouldn't equal the semi-major of the oblate spheroid a at each case!!! What is wrong?!!

Thanks for any help, regards!

1 Answers 1

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The equation you got is the projection of the ellipse (intersection between spheroid and plane) on the $(x,y)$ plane. As your plane cuts the "equator" of the spheroid along a diameter, then yes: the semi-axis major $a'$ of the projection is equal to the semi-major of the oblate spheroid. And the same is true for the intersection, of course.

You can see below what happens: the plane $ACF$ through the origin (light gray) cuts the $(x,y)$ plane (dark gray) along line $AC$. But of course $AC$ is a diameter of the spheroid equator (black) and is the major axis of the intersection ellipse (blue). Line $EF$ lies on the plane but has no special meaning: it is there just to help you visualize the situation.

enter image description here

To convince you that $AC$ is indeed the major axis, here's the same diagram seen from "above".

enter image description here

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    @Arentino thanks, you are right but what I look for, to calculate a' in general for each ellipse as an intersection and of course also b', passing by Origin O(0,0,0) and not only in this case when it's at equator – 2017-02-03
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    If the plane passes by the origin then its intersection with the $(x,y)$ plane is a line through the origin, and the intersection of the spheroid with the $(x,y)$ plane is a circle (which I called "equator" for short) centered at the origin and of radius $a$. That line always cuts the equator along a diameter, which is also the major axis of the intersection. – 2017-02-03
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    I don't know if my question is well asked and/or answered. But there is infinty planes passing with center O(0,0,0) but they all are not parallel or perpendiculat to plane xy, at this case, there is semi-axis major and minor – 2017-02-03
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    Would be of help to make you see a diagram to convince you? Or should I suggest a simpler algebric approach? – 2017-02-03
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    Could you see this? http://www.gnu.org/software/3dldf/graphics/elpsd_17.png – 2017-02-03
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    This plane passes by the origin O(0,0,0) and it's an ellipse (bcz it's intersection with an spheroid, except it would be a circle when it passes at equator), and when it would be only rotation about x-axis by an angle alpha, then yes also a' = a but this plane has two angles of freedom, first around x-axis then about y-axis by another angle – 2017-02-03
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    Neither of the semi-axis a' or b' coincide along x-axis – 2017-02-03
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    As the spheroid is symmetric for rotations around the $z$ axis, the direction of the plane doesn't matter. I've just added a couple of pictures, I hope they can be of help. – 2017-02-03
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    By the way: I made the above diagrams with GeoGebra, a free software which is very powerful: you could give it a try. – 2017-02-03
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    Thanks you very much, sry for time... and I thank you also for pictures. So a'=a always b= as mentioned above – 2017-02-03
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    Remember that your $b'$ is the semiaxis of the projected ellipse, not of the intersection. To find the semi-minor axis $b''$ of the intersection, you must add the $z$ part: $b''=\sqrt{b'^2+(u^2+v^2)b^2/w^2}$. – 2017-02-03
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    So for all planes that they pass by origin O(0,0,0) and have N=(u,v,w) as normal vector, when they intersect an oblate spheroid, we get an ellipse with a' = a and b' = b*sqrt[1+(u^2+v^2)/w^2] = (b/w)*sqrt(u^2+v^2+w^2), right? – 2017-02-03
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    No: the first term under square root is $b'^2= (abw)^2/[a^2*(u^2 + v^2) + (b*w)^2]$. – 2017-02-03