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

which contains an interior point or a boundary point of hyperboloid (1). A plane spanned by vectors, and containing the point is described in parametric form by

Inserting the components of into the Equation of hyperboloid (1) leads to the line of intersection as a quadratic form in the variables and. Let the scalar product in for two vectors and be denoted by

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 nonnegative. Since need not be a scalar product in, the 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.

Let and be unit vectors orthogonal to the unit normal vector of plane (2)

and orthogonal to eachother

Furthermore vectors and may be chosen such that

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

with

and

In case Equation (8) can be written as a conic in translational form

in the variables and with

For and the line of intersection is of the form

with

If holds, (13) represents a parabola in the variables and. 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

and the Grassmann expansion theorem for the double cross product

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

Then Equations (5) and (6) are fulfilled: is a unit vector, as can be shown by the identity of Lagrange (16), utilising, and:

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, which imply conditions (4), (5) and (6). The expression

becomes zero, when choosing such that

holds.

Corollary 1: For the unit vectors and orthogonal to each other and the following statement holds:

This statement follows by substituting the definition of and utilising, and. For one obtains for instance:

Theorem 1: Let be the unit normal vector of the plane and let vectors and satisfy, , and condition (7). Putting

and are solutions of the following quadratic Equation:

Proof: Utilising Corollary 1 one obtains:

Applying diagonality condition (7) and the identity of Lagrange (16) leads to:

For the cross products one obtains:

with the diagonal matrices

According to Grassmann’s expansion theorem for the double cross product (17)

follows, since and. Applying (20), (21), (23) one obtains:

Corollary 2: Under the assumptions of Theorem 1 the following three pairs of Equations are valid:

The first pair of Equations was verified in the proof of Theorem 1. The second and the third pair of Equations follow analogously.

Theorem 2: Under the assumptions of Theorem 1 with and the expression for in (15) is given by:

where is taken from (14).

Proof: The verification of (25) consists of three steps.

Step 1: Applying the identity of Lagrange (16) the following statements hold:

With Corollary 2 and the diagonal matrices

one obtains:

and it follows by substituting (28) into (26)

Introducing expressions

one obtains from (29) using (18) and (30)

Combining both Equations (31) for and leads to

Step 2: Analogously to the verification of (24) the application of the identity of Lagrange (16) yields:

With the diagonal matrices

for the cross products holds:

Therefore one obtains

or

In contrast to the verification of (24), where diagonality condition (7) holds, the analogous expression in (33) need not be zero.

Step 3: Applying the identity of Lagrange (16) again leads to

Substituting the involved cross products according to Corollary 2 and considering diagonality condition (7) one obtains

or

Squaring both sides of (34) and substituting the expressions from (31) leads to:

Substitution of (33) results in Equation

or

Substitution of (35) in (32) leads to:

Because of (24)

holds and with (15) one finally obtains relation (25)

Corollary 3: Under the assumptions of Theorem 1 and in case of a hyperboloid of one sheet assuming for, in case of a hyperboloid of two sheets assuming for and, the intersection of hyperboloid (1) and a plane with unit normal vector and distance from the origin is an ellipse, the area of which is given by:

In this formula corresponds to a hyperboloid of one sheet, to a hyperboloid of two sheets.

Proof: With for both sides of Equation (37) are positive. Thus according to (25) is negative for, and zero for. In case of a hyperboloid of one sheet the numerator of for in (12) is positive. In case of a hyperboloid of two sheets the numerator of for in (12) is positive for. Substituting for according to (18), for are positive. In both of these cases therefore the curve of intersection (11) is an ellipse with the semi axes

The area of the ellipse is given by:

By applying (25) and (37) one obtains the formula in Corollary 3.

Remark 1: In the special case that the plane of intersection of the hyperboloid is parallel to the x-y-plane, i.e. the normal vector with and furthermore, can be chosen satisfying (4), (5), (6), (7) and, , the formula for the area of the ellipse of intersection reduces to:

The same result is obtained from (1) putting and calculating the area of an ellipse with the semi axes

As stated above in case of a hyperboloid of two sheets

has to be assumed.

Remark 2: Assuming for i = 1, 2 and in case of a hyperboloid of two sheets would also result in positive for according to (18) and (12). However for two vectors and in the conditions for and cannot be fulfilled simultaneously.

for would imply

and thus

Because of

holds. Substituting this Equation into the above inequality gives

Deleting equal terms on both sides of the inequality finally results in

which is impossible for vectors and with real components.

Corollary 4: Under the assumptions of Theorem 1 and assuming and the intersection of hyperboloid (1) and a plane with unit normal vector and distance from the origin is for a hyperbola and for a pair of straight lines.

Proof: With and both sides of Equation (37) are negative. Thus according to (25) is positive or zero. In case holds for a hyperboloid of one sheet with the semi axes

the line of intersection is a hyperbola of the form

In case holds for a hyperboloid of one sheet with the semi axes

the line of intersection is a hyperbola of the form

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

with the axes interchanged, as in the previous case.

In case of according to (8), after substituting and from (18), the line of intersection is a pair of straight lines of the form

or

Remark 3: For and the roles of the variables and have to be interchanged.

Substituting according to (14) in formulars (9) for the coordinates of the center of the conic in the plane spanned by and one obtains using (7):

The center of the conic in is given by:

Theorem 3: Let the assumptions of Theorem 1 be fulfilled with and. For the center of the conic of intersection in holds:

Proof: With diagonal matrices from (27) and from (22) utilising

and (37) one obtains a representation of equivalent to (40):

It is sufficient to show that for the difference

holds. Thus the coefficients in the expansion of in with respect to the orthonormal basis are zero, i.e., is the zero vector.

Applying representation (39) and (24) one obtains:

Furthermore one obtains:

and by interchanging the roles of and:

Both previous expressions are zero; this follows by applying diagonality condition (7), the identity of Lagrange (16) and Corollary 2:

Interchanging the roles of and leads to:

Corollary 5: 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 and, given in the proof of Corollary 3, and the apexes

Proof: Clearly and 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 Equation (15).

Corollary 6: Under the same assumptions as in Corollary 4 the line of intersection of hyperboloid (1) and a plane is in case of a hyperbola with the semi axes and 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 hyperbola.

Proof: The asymptotes of the hyperbola are given by

with

or

The point of intersection of the asymptotes fulfills the following linear system

As this homogeneous linear system for the unknowns and has a nonzero determinant, it can only have the trivial solution, which implies

Corollary 7:

Proof: This can be verified, as in the proof of Corollary 5, using in the form (39) and employing condition (7) and Equation (15).

Because of Corollary 7

holds, if and only if is an interior point of a hyperboloid of one sheet,

holds, if and only if is an interior point of a hyperboloid of two sheets,

holds, if and only if is an exterior point of a hyperboloid of one sheet,

holds, if and only if is an exterior point of a hyperboloid of two sheets.

In case of one obtains from (25)

.

The center (40) of the conic of intersection therefore becomes a tangent contact point

of hyperboloid and plane, where the -sign corresponds to a hyperboloid of one sheet and the -sign to a hyperboloid of two sheets.

Example: Determine the line of intersection of hyperboloid (1) and a plane, having the normal vector and containing the point, situated in the interior or on the boundary of (1):

The unit normal vector of the plane has the form:

The distance of the plane from the origin is given by:

According to (25) can be written as:

Substituting (18) into (12) the expressions of and are given by

where, satisfying and, are solutions of Equation (19) after substituting vector from (42):

With Theorem 3 one obtains by substituting from (42) and from (43) the formular for the center of the conic given by:

In the special case of a plane containing the origin, i.e. is the zero vector, it follows by (43), (44) and (47) that, and is the zero vector also. Furthermore the expressions of and in (45) reduce to

As described in Corollary 3 for a hyperboloid of one sheet and for one obtains for. Then the line of intersection is an ellipse. As stated in Corollary 4 for a hyperboloid of one sheet and, one obtains,. For a hyperboloid of two sheets and, one obtains,. In both of these cases the line of intersection is a hyperbola.

In a second special case with. the above formulas (43), (44) and (47) reduce to:

and

Because of in (14) holds and (38) reduces to

where and are solutions of the quadratic Equation (46) and vectors and have to be determined as described above in Section 2. As stated in Corollaries 3 and 4, if for are both positive, an ellipse as curve of intersection is obtained, and if for are of different sign, a hyperbola as curve of intersection results.

A parabola (13) as curve of intersection is obtained in case of and. A hyperboloid of one sheet, given in (1), may be factorized in the following form:

With the decomposition

for any value of these Equations represent a straight line, as the intersection of two planes in. This straight line lies on (48) because, if the members of (49) are multiplied together, (48) results. Rearranging (49) one obtains

With the abbreviations

the straigth line (50) can be equivalently rewritten [3]

with a point on (50) and.

Putting

holds, because

Choosing a vector on the surface of a hyperboloid of one sheet, as given in (1), for instance

results.

Constructing a vector, fulfilling

a plane spanned by vectors and is obtained, containing the straight line (52). The two linear Equations in (54) for the components of can be rewritten:

Solving for s₁ and s₂ under the assumptions and gives:

Dividing by one obtains for

and thus the following normalized vector:

fulfilling (54) and giving

In case, this signifies rotational symmetry of the hyperboloid with regard to the z-axis, the coefficient matrix of (55) is singular. The condition for solvability of (55) is

As this can be reduced to

Both sides of (56) are equal to only for. A solution vector may then be chosen as

fulfilling (54). This leads to

For according to (51) results. Then the linear system (55) is solvable for arbitrary and

. Choosing, as above

holds.

Using vector given in (53)

is obtained. Thus parabola (13) has the form

with

Instead of (49) the alternative decomposition of (48)

for any value of may be considered; (58) also describes a straight line as intersection of two planes in. This straight line as well lies on (48) because, if the members of (58) are multiplied together, (48) results. Rearranging (58) one obtains

With the abbreviations

the straigth line (59) can be equivalently rewritten [3]

with a point on (59) and.

As previously with the terms now with the terms vectors and can be defined satisfying

Choosing a vector as in (53), in the end a parabola of the form (57) is obtained.

Mathematica programs modelling the cases described in Corollaries 3 and 4 and in Section 6 may be obtained from the author upon request.

The intention of this paper is to look at cases which are not treated in mathematical textbooks where the plane intersecting a hyperboloid of one sheet or of two sheets is not necessarily parallel to the coordinate planes and thus produces all kinds of conics: ellipses, hyperbolas and parabolas.