1. 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
(1)
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
(2)
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:
(3)
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)
(4)
(5)
and orthogonal to eachother
(6)
Furthermore vectors 
and 
may be chosen such that
(7)
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
(8)
with
(9)
and
(10)
In case 
Equation (8) can be written as a conic in translational form
(11)
in the variables 
and 
with
(12)
For 
and 
the line of intersection is of the form
(13)
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
:
(14)
where 
is the distance of plane (2) from the origin. Substituting 
into (10) one obtains employing (4), (5), (6) and (7)
(15)
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
(16)
and the Grassmann expansion theorem for the double cross product
(17)
2. Construction of Vectors 
and 
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:
3. A Quadratic Equation
Theorem 1: Let 
be the unit normal vector of the plane and let vectors 
and 
satisfy
, 
, 
and condition (7). Putting
(18)
and 
are solutions of the following quadratic Equation:
(19)
Proof: Utilising Corollary 1 one obtains:
Applying diagonality condition (7) and the identity of Lagrange (16) leads to:
(20)
For the cross products 
one obtains:
(21)
with the diagonal matrices
(22)
According to Grassmann’s expansion theorem for the double cross product (17)
(23)
follows, since 
and
. Applying (20), (21), (23) one obtains:
(24)
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.
4. A Formular for d
Theorem 2: Under the assumptions of Theorem 1 with 
and 
the expression for 
in (15) is given by:
(25)
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:
(26)
With Corollary 2 and the diagonal matrices
(27)
one obtains:
(28)
and it follows by substituting (28) into (26)
(29)
Introducing expressions
(30)
one obtains from (29) using (18) and (30)
(31)
Combining both Equations (31) for 
and 
leads to
(32)
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
(33)
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
(34)
Squaring both sides of (34) and substituting the expressions from (31) leads to:
Substitution of (33) results in Equation
or
(35)
Substitution of (35) in (32) leads to:
(36)
Because of (24)
(37)
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.
5. The Center of the Conic
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):
(38)
The center 
of the conic in 
is given by:
(39)
Theorem 3: Let the assumptions of Theorem 1 be fulfilled with 
and
. For the center 
of the conic of intersection in 
holds:
(40)
Proof: With diagonal matrices 
from (27) and 
from (22) utilising
and (37) one obtains a representation of 
equivalent to (40):
(41)
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:
(42)
The distance of the plane from the origin is given by:
(43)
According to (25) 
can be written as:
(44)
Substituting (18) into (12) the expressions of 
and 
are given by
(45)
where
, satisfying 
and
, are solutions of Equation (19) after substituting vector 
from (42):
(46)
With Theorem 3 one obtains by substituting 
from (42) and 
from (43) the formular for the center 
of the conic given by:
(47)
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.
6. Parabola as Curve of Intersection
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:
(48)
With the decomposition
(49)
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
(50)
With the abbreviations
(51)
the straigth line (50) can be equivalently rewritten [3]
(52)
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
(53)
results.
Constructing a vector
, fulfilling
(54)
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:
(55)
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
(56)
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
(57)
with
Instead of (49) the alternative decomposition of (48)
(58)
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
(59)
With the abbreviations
(60)
the straigth line (59) can be equivalently rewritten [3]
(61)
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.
7. Conclusion
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.