Projection of the Semi-Axes of the Ellipse of Intersection

It is well known that the line of intersection of an ellipsoid and a plane is an ellipse (see for instance [1]). In this note the semi-axes of the ellipse of intersection will be projected from 3d space onto a 2d plane. It is shown that the projected semi-axes agree with results of a method used by Bektas [2] and also with results obtained by Schrantz [3].


Introduction
Let an ellipsoid be given with the three positive semi-axes 1 a , 2 a , 3 a and a plane with the unit normal vector ( ) , , , n n n = n which contains an interior point ( ) , , q q q = q of the ellipsoid.A plane spanned by vectors ( ) , , r r r = r , ( ) , , s s s = s and containing the point q is described in parametric form by ( ) with , , .t u x x x = + + = x q r s x (2)   Inserting the components of x into the equation of the ellipsoid (1) leads to the line of intersection as a quadratic form in the variables t and u.Let the scalar product in 3 R for two vectors ( ) ( ) With the diagonal matrix the line of intersection has the form: ( ) .
r s r s s s q r q s q q (3) As q is an interior point of the ellipsoid the right-hand side of Equation ( 3) is positive.
Let r and s be unit vectors orthogonal to the unit normal vector n of the plane ( ) ( ) If vectors r and s have the additional property ( ) the 2 2 × matrix in (3) has diagonal form.If condition (7) does not hold for vectors r and s , it can be fulfilled, as shown in [1], with vectors r and s  obtained by a transformation of the form cos sin , sin cos with an angle ω according to ( ) ( ) ( ) Relations ( 4), ( 5) and (6) hold for the transformed vectors r and s  instead of r and s .If plane (2) is written instead of vectors r and s with the transformed vectors r and s  the 2 2 × matrix in (3) has diagonal form because of condition (7): .
Then the line of intersection reduces to an ellipse in translational form ( ) ( ) with the center ( ) ) and  , , and the semi-axes ( ) ( ) ) Because of ( ) , and , the semi-axes A, B given in (12) can be rewritten as In [1] it is shown that 1 β and 2 β according to (14) are solutions of the following quadratic equation Furthermore it is proven in [1] that d according to (13) satisfies

Projection of the Ellipse of Intersection onto a 2-d Plane
The curve of intersection in 3d space can be described by θ ∈ and vectors r and s  obtained after a suitable rotation (8) starting from initial vectors r and s (see for instance [1]).Without loss of generality the plane of projection of the ellipse (18) shall be the x x − plane.The angle between the plane of intersection (2) containing the ellipse (18) and the plane of projection is denoted by Ω .The same angle is to be found between the unit normal n of the plane of intersection (2) and the 3 x -direction, normal to the plane of projection.Denoting the unit vector in 3 x -direction by 3 e the definition of the scalar product (see for instance [4]) yields ( ) where cos 0 Ω > holds for Let us assume that the plane of intersection (2) is not perpendicular to the plane of projection, the 1 2 x x − plane.This means that π 0 2 ≤ Ω < is valid and The ellipse of intersection (18) projected from 3d space onto the 1 2 x x − plane has the following form: are not orthogonal because their orthogonality in 3d space implies which need not be zero.In order to calculate the lenghts of the semi-axes A and B projected from 3d space onto the 1 2 x x − plane the following linear system deduced from (20) with the abbreviations 1 1 x x m ′ = − is treated: The determinant of the linear system (21), ( ) AB r s r s −     , is different from zero.This can be shown by noting that 1 2 2 1 r s r s −     is the third component of the vector × r s   .At first this vector is not affected by rotation (8):

s r s r s
This result was obtained by applying the rules for the cross product in 3 R .
Furthermore one obtains employing the Grassman expansion theorem (see for instance [4]): .Thus one ends up with which is positive because of (19) for angles Ω with π 0 2 ≤ Ω < .Solving the linear system (21) leads to ( ) ( ) ) Expanding the squares on the left side and using the denotations ( ) L as a real symmetric matrix can be diagonalized and thus is similar to the diagonal matrix of its eigenvalues ( ) with a nonsingular transformation matrix S , being orthogonal, i.e.  ) ( ) The eigenvalues ( ) λ are positive because L is positive definite; this is true since the terms 11 l and for the second term, the determinant of L , holds because of ( 22): This is an ellipse projected from 3d space (18) onto the 1 2 x x − plane with the semi-axes ( ) ( ) , .

Calculation of Semi-Axes According to a Method Used by Bektas
Let the ellipsoid (1) be given and a plane in the form The unit normal vector of the plane is: ( ) The distance between the plane and the origin is given by The plane written in Hessian normal form then reads: Without loss of generality 3 0 A ≠ shall be assumed.Then 3 0 n ≠ holds: ( ) x and substituting into equation (1) gives: In the sequel the determinant of the following matrix will be needed: In order to get rid of the linear terms 1 x and 2 x in (33) the following translation can be performed: + with parameters h and k to be determined later.After substitution into (33) one obtains: , .
The linear system (37) has M as matrix of coefficients, the determinant of which is given in (35).It is nonzero because of the assumption 3 0 n ≠ .Solving the linear system (37) yields: Substituting the terms (34) into (38) gives the result: With the terms h and k from (39) the constant term in (36) turns out to be, together with (17): ( ) ( ) M as a real symmetric matrix can be diagonalized and thus is similar to the diagonal matrix of its eigenvalues ( ) with a nonsingular transformation matrix T , being orthogonal, i.e.

T
T T − = , the inverse of T is equal to the transpose of T .Putting ( ) ( ) The eigenvalues ( ) This is an ellipse in the 1 2 x x − plane with the semi-axes ( ) ( )

Calculation of Projected Semi-Axes According to Schrantz
In [3] the ellipse [ ) G as a real symmetric matrix can be diagonalized and thus is similar to the diagonal matrix of its eigenvalues ( ) with a nonsingular transformation matrix R , being orthogonal, i.e.
the inverse of R is equal to the transpose of R .Putting ( ) ( ) the quadratic equation ( 48) in ( ) This is an ellipse in the 1 2 x x − plane with the semi-axes ( ) ( )

Some Auxiliary Means
Let H stand for the following 2 2 × matrix: and be a place holder for the matrices M and G used above.The semi-axes L A , L B projected onto the 1 2 x x − plane, given in (28), are compared with the semi-axes H A , H B .It will be shown that the two polynomials , have the same coefficients and thus have the same zeros: ( .
In the first step L L H H A B A B = will be proven.In the second step will be shown.This is sufficient, since by adding 2 2

L L H H A B A B =
to both sides of (55) one obtains: + since the semi-axes are positive.
( ) λ are the zeros of the characteristic polynomial of L .This can be expressed in two ways: .
In continuation of (64), because and s  fulfilling ( 4) and (5), the following relations hold: β and 2 β are solutions of (16).Combining (64), ( 68), ( 69) and (67) one obtains: ( ) To simplify the term in round brackets of (70) the following relations are used: according to (14).The term in round brackets of (70) thus becomes: because r and s  have been chosen in such a way that condition (7) is fulfilled.As already described in the beginning of Section 4 the ellipse (43) is projected from the original plane E onto the plane E′ .Both planes are forming an angle Ω with π 0 2 ≤ Ω ≤ .Without loss of generality the intersection of E and E′ , E E′  , shall be the 1 x -axis of the coordinate system in plane E′ .

Comparison of the Semi-Axes AL, BL with AG, BG
The original plane E thus contains the following three points: ( ) 1, 0, 0 − , ( ) 1, 0, 0 , ( ) 0, cos ,sin Ω Ω and can therefore be described by the following equation: The unit normal vector n of plane (80) given by ( 31) is ( ) In order to describe a unit vector r in the plane E the equations (4) must hold: ( ) ( ) If the unit vector r is forming the angle α with the 1 x -axis and 1 e is designating a unit vector in 1 x -direction according to the definition of the scalar product (see for instance [4]) holds ( ) of S is equal to the transpose of S .Putting

(
− are positive.For 11 l this is clear; Thus the quadratic equation (36) reduces to: