On the Intersection Equation of a Hyperboloid and a Plane

In this note, the ideas employed in [1] to treat the problem of an ellipsoid intersected by a plane are applied to the analogous problem of a hyperboloid being intersected by a plane. The curves of intersection resulting in this case are not only ellipses but rather all types of conics: ellipses, hyperbolas and parabolas. In text books of mathematics usually only cases are treated, where the planes of intersection are parallel to the coordinate planes. Here the general case is illustrated with intersecting planes which are not necessarily parallel to the coordinate planes.


Introduction
The problem of a hyperboloid being intersected by a plane is described in Section 1.The means to treat the problem are provided in Sections 2, 3 and 4. In the end of Section 4 first results can be formulated in Corollaries 3 and 4. Further results concerning the center of the conic of intersection are given in Section 5. Finally in Section 6 the case of a parabola as intersecting curve is treated.
Let a hyperboloid be given with the three positive semi axes a, b, c where on the right hand side of (1) corresponds to a hyperboloid of one sheet, on the right hand side of (1) to a hyperboloid of two sheets.Let furthermore a plane be given with the unit normal vector , , , n n n  n which contains an interior point or a boundary point of hyperboloid (1).A plane spanned by , , q q q  q r r  r   Inserting the components of x into the Equation of hyperboloid (1) leads to the line of intersection as a quadratic form in the variables and u .Let the scalar product in for two vectors and be denoted by . w With the diagonal matrices the line of intersection has the form: .
As is an interior point or a boundary point of hyperboloid (1) the right-hand side of Equation ( 3) is non- need not be a scalar product in , the 3 R  matrix in Equation ( 3) is in general no Gram matrix.If the matrix in (3) is positive definite, then the line of intersection is an ellipse.
and orthogonal to eachother   1 1 2 2 3 3 , 0 r s r s r s     r s (6) Furthermore vectors and may be chosen such that r s   holds.This will be shown in the next Section.Condition (7) ensures that the matrix in (3) has diagonal form.


In case  and the line of intersection reduces to In case Equation (8) can be written as a conic in translational form in the variables and u with t For and , , and , ., , If holds, (13) represents a parabola in the variables and u .This will be discussed further in Section 6.

 
In order to show that the expression in (10) is independent of the choice of this vector may be decomposed orthogonally with respect to : where  is the distance of plane (2) from the origin.
Substituting into (10) one obtains employing (4), ( 5), (6) and ( 7) The following rules of computation for the cross product in ( [2], p.147) will be applied repeatedly later on.For vectors of the identity of Lagrange holds 3

R
, , , and the Grassmann expansion theorem for the double cross product

Construction of Vectors and r s
Let be a unit vector orthogonal to the unit normal vector of the plane, so that Equations (4) hold.A suitable vector is obtained as a cross product r n s .  s n r Then Equations ( 5) and ( 6) are fulfilled: is a unit vector, as can be shown by the identity of Lagrange (16), utilising

      s s n r n r n n r r n r
Furthermore one obtains according to the rules applying to the spar product: In case Equation (7) is not fulfilled for the initially chosen vectors and , i.e. , the following transformation may be performed with The transformed vectors and satisfy the following conditions: , and ( ,  n r) 0      s n r , which imply conditions (4), (5) and (6).The expression ,

A Quadratic Equation
Theorem 1: Let be the unit normal vector of the plane and let vectors and satisfy , , and condition (7).Putting  and 2  are solutions of the following quadratic Equation: Proof: Utilising Corollary 1 one obtains: For the cross products Applying diagonality condition (7) and the identity of Lagrange (16) leads to: r s s r with the diagonal matrices According to Grassmann's expansion theorem for the double cross product (17) follows, since The second and the third pair of Equations follow analogously.

A Formular fo
Theorem 2: Under the as where  is taken from (14).Proo The verification of ( 2Step 1: Applying the identity of Lagrange (16) the llowing statements hold: and it follows by substituting (28) into (26) Introducing expressions one obtains from (29) using ( 18) and (30) Combining both Equations (31) for 1 0 Step usly to the verification of (24) the application of the identity of Lagrange (16) yields: In contrast to the verification of ( 24), where diagonalthe analog ssion , , Substituting the involved cross products accor Corollary 2 and considering diagonality condition ob ding to (7) one tains Squaring both sides of (34) and substituting the expressions from (31) leads to: Substitution of (35) in (32) leads to: .
holds and with (15) one finally obtains relation (25 with .
In this formula The area of the ellipse is given by: pplying (25) and (37) one ob the formula in Corollary 3.  Remark 1: In the special case that the plane of inter-By a tains section of the hyperboloid is parallel to the x-y-plane, i.e. the normal vector with and furthermore osen satisfying (4 ), (5), (6), (7 0,1,

F ab
The same result is obtained from (1) putting x q  and calculating the llipse w es area of an e ith the semi ax As stated above in case of a hyperboloid of two sheets with the axes interchanged.Since is positive or zero, is fulfilled, so that for a hyperboloid of two sheets with the semi axes the line of intersection is a hyperbola of the form B with the axes interchanged, as in the prev us case.
In case of 0 according to (8), after substi-

 and 2
 from (18), the line of intersection is a pair of straight lines of the form


The center m of the conic in 3 R is given , , , .

D D t D D u
by: Theorem 3: Let the assu ptions of Theorem 1 be . For the center of the conic of lds: Proof: With diagonal matrices from ( 27) and an ob ent to (40 d (37) one tains a representation of m equival ): It is sufficient to show tha or the diff  Furthermore one obtains: and by interchanging the ro s of r and s : Both previous expressions are zero; this f ws by applying diagonality condition (7), the identity of Lagrange (16) and Corollary 2: Interchanging the roles of and leads to: s are points of the plane cutting the hyperboloid.In order to show that they are belonging to the ellipse of intersection, it has to be verified that they are situated on hyperboloid (1) i.e. the following equalities hold: This can be verified using in the form (39) and employing condition (7) and Eq ion (15).
Corollary 6: Under the same assumptions as in Coroll ry 4 the line of intersection of hyperboloid (1) and a plane is in ca a hyperbola with the semi axes

 
A and B given in the proof of Corollary 4. The center of the hyperbola given in ( 9) is equal to the point of intersection of the asymptotes of the hyperbol Proof: The asymptotes of the hyperbola a a. re given by t u fulfills the following linear system As this homogeneous linear system for the unknowns and has a nonzero dete inant, it can ve th lution, which implies . q r q s orollary 7: Proof: This can be verified, as in the proof of Corollary 5, using in the form (39) and e ploying condition (7) and Equation (15).holds, if and only if is an interior point of a hyperboloid of one sheet, a n b n c n The center (40) of the conic of intersection therefore becomes a tangent contact point   or on the boundary of (1 The unit normal vector of the plane has the form: The distance of the plane from , , .i j k (42) s given by: the origin i (43)


According to (25) d can be written as: 0.
Open Access AM P. P. KLEIN 47 With Theorem 3 one obtains by substituting n from (42) and from (43) the formular for the conic given by: In the special case of a plane conta nd ons ining the origin, i.e. q tha therm is the zero vector, it follows by ( 43


. In both of these cases the line of intersection is a hyperbola.
In a second special case with . the ove formulas (43), (44) and (47) reduce to:

Parabola as Curve of Intersection
A parabola (13) as curve of intersection is obtai case of and , may be factor .A rbol n in ( ized llowing form: hype in the fo oid of one sheet, give omposi With the dec tion   these Equations represent a straight line, as the intersection of two planes in 3 R .This str ght lin on (48) because, if the members of (49) are multiplied together, (48) results.earg (49) one obtains R e lies ai R rangin the straigth line (50) can be equivalently rewritten with a point on (50) and Choosing a vector on the surface of a hyperboloid of one s (1), for instance , , q q q  q heet, as given in sin , cos , 0 a b In case , this signifies rotational symmetry of the hyperboloid with rega the z-ax matrix of (55) is singul i (55) is (56) y f Both sides of are equal to 1 onl or 3 s 0  .A solution vector s may then be chosen as   in f one sheet or of two sheets is not necessarily parallel to the coordinate planes and thus produces all kinds of conics: ellipses, hy parabolas.

2 2 
Letand be unit vectors orthogonal to the unit normal vector of plane(2)

orollary 2 :
Under the assumptions of Theorem 1 the fo r The first pair of Equations was verified in the proof of Th r d sumptions of Theorem 1 with  C llowing three pairs of Equations are valid:

2
With the diagonal matrices a b for the cross products holds: variables t and u have to be interchanged.f the Substituting according to (14) in formulars (9) for the coordinates of the center of the conic in the e span and one obtains using (7): Under the same assumptions as in Corollary 3 the line of intersection of hyperboloid (1) and a plane is an ellipse with the semi axes A and , given in the proof of Corollary 3, and the apexes B hyperboloid and plane, where the -sign corresponds to o line of intersection of hyperbo d (1) and a plane, having the normal vector an , situ interior  a hyperboloid of one sheet and the  -sign to a hyperboloid of tw sheets.
line (52).The two linear Equations in (54) for the components of can be rewritten: For k  ccording to (51) 0 l  results.Then the linear system (55) is solvable for arbitrary 1 considered; (58) also describes a straight line as intersection of two lanes p