On Loci of Points

S. F. Sibirtsev

(Presented by the scientific seminar of the Department of Descriptive Geometry and Graphics)

Introduction

In descriptive geometry, loci of points are used in the solution of many problems. The aim of the present work is to examine loci of points and to give their classification.

In addition, the paper considers the locus of points equidistant from two skew lines.

Loci of Points

A locus of points (l.p.) is a set of points that contains all points possessing some property and contains not a single point that does not possess this property.

Basic Loci of Points

1. The l.p. equidistant from one point is the surface of a sphere with center at the given point.

2. The l.p. equidistant from two points is a plane passing through the midpoint of the segment joining the given points and perpendicular to it.

3. The l.p. equidistant from three given points \(A, B, C\), not lying on one straight line, is a straight line perpendicular to the plane of triangle \(ABC\) and passing through the center of the circle circumscribed about triangle \(ABC\).

4. The l.p. equidistant from four given points not lying in one plane is the center of the sphere whose surface passes through these points.

5. The l.p. equidistant from one straight line is the surface of a right circular cylinder whose axis is the given straight line.

6. The l.p. equidistant from two parallel straight lines is a plane perpendicular to the segment defining the shortest distance between the given lines and passing through its midpoint.

7. The l.p. equidistant from two intersecting straight lines is a pair of planes perpendicular to the plane determined by the given straight lines and passing through the bisectors of the angles between them.

8. The l.p. equidistant from three parallel straight lines not lying in one plane is the axis of a cylinder whose surface passes through the given straight lines.

1. The locus of points equidistant from three lines passing through one point is the axis of a cone whose surface passes through the given lines.

(If the directions of the given lines are taken into account, then we obtain four lines corresponding to the given locus. An example is the locus of points equidistant from the coordinate axes. This locus will correspond to four axes of cones passing through the following octants: I—VII; II—VIII; III—V; IV—VI.)

1. The locus of points equidistant from one plane is a pair of planes parallel to the given plane and situated on different sides of it at the given distance.

2. The locus of points equidistant from two intersecting planes is the two bisector planes of the dihedral angles formed by these planes.

3. The locus of points equidistant from two parallel planes is a plane parallel to the given planes and situated midway between them.

4. The locus of points equidistant from three planes passing through one point is the axis of a cone inscribed in the trihedral angle formed by the given planes (if the direction of the planes is taken into account, then we obtain four lines).

5. The locus of points equidistant from three planes forming a prism (having a common improper point) is the axis of a cylinder inscribed in the given prism (if the planes are extended beyond the boundary of the prism, then we obtain four lines).

6. The locus of lines passing through a given point and inclined at an angle \(\alpha\) to a given plane is the surface of a (right circular) cone whose vertex lies at the given point, and whose generators are inclined at an angle \(\alpha\) to the given plane.

7. The locus of points from which a given segment is seen under a given angle is the surface obtained by rotating a segment, resting on the given segment and subtending the given angle (for \(\alpha = 90^\circ\) we obtain the surface of a sphere).

8. The locus of points equidistant from a circle is the surface of a torus whose axis is the given circle.

9. The locus of points equidistant from two skew lines is the surface of a hyperbolic paraboloid. The proof of this proposition is considered separately in the paper.

Classification of Loci

All the loci indicated above can be divided into three kinds. To the first kind one may assign those loci that have meaning only for a value of the argument that is unique for each separate case.

An example of such a locus is the locus of points equidistant from one point. In this case the determining argument is the distance from the given point. This locus has meaning only for a quite definite value of the distance. If the distance is regarded as varying from zero to infinity, then the given locus loses its meaning.

To loci of the first kind one may assign the following: 1, 5, 10, 15, 16, 17.

To the second kind one may assign those loci for which the argument determining them may vary within certain

limits. An example of such a locus may be the locus of points equidistant from two given points. The defining argument in this case will be the distance from the two points. It is understood that the distance in the given case may take values from a quantity equal to half the distance between the two given points to infinity. The following may be assigned to loci of this kind: 2, 3, 6, 7, 9, 11, 13, 18.

To loci of the third kind one may assign those loci which do not belong to the first two. The argument defining the given locus will be constant, and its value is completely determined by the specification of the given geometric forms.

Fig. 1

An example of such a locus may be the locus of points equidistant from four given points. The argument defining the given locus will be, in this case, the radius of the sphere. The value of the radius of the sphere is completely determined by the specification of the given points. The following may be assigned to loci of the third kind: 4, 8, 12, 14. It is easy to observe that, except for the fourth locus, the others are special cases of the corresponding loci of the second kind. For example, the locus of points equidistant from three parallel straight lines

(GML No. 8) may be regarded as a special case of the locus of points equidistant from three intersecting straight lines (GML No. 9), when the point of their intersection becomes improper.

Similarly, loci Nos. 12 and 14 may be regarded as special cases, respectively, of loci Nos. 11 and 13.

On the locus of points equidistant from two skew lines

Consider the skew lines \(MM\) and \(KK\), situated at right angles; the line \(KK\) is parallel to the plane \(V\) (Fig. 1). Let the distance between the lines be denoted by \(2a\). The locus of points equidistant from one line is the surface of a cylinder. Obviously, for a prescribed distance (greater in magnitude than \(a\)) the desired locus will be the line of intersection of the surfaces of the cylinders. Varying the magnitude of the distance from \(a\) to infinity, we obtain the required locus.

To investigate the locus thus obtained, let us examine its sections by planes. To simplify the investigation, construct two circular cylinders I and II of given radius \(R\), whose axes are, respectively, the lines \(MM_1\) and \(KK_1\).

Section by the plane \(R\), passing through the point \(C\) and parallel to the plane \(H\)

The section of cylinders I and II by the plane \(R\) gives two pairs of parallel lines, lying at the same distance from the lines \(KK_1\) and \(MM_1\). Obviously, the lines obtained will intersect on the bisectors of the angles between the lines \(KK_1\) and \(MM_1\), i.e., on the lines \(CF\) and \(CG\). Thus, the section by a plane of the locus obtained gives two straight lines, which are the bisectors of the angles between \(KK_1\) and \(MM_1\), i.e., the lines \(CF\) and \(CG\).

Section by the plane \(S\), passing through the point \(C\) and parallel to the plane \(V\)

The section of cylinder I by the plane \(S\) on the vertical projection gives a circle of radius \(R\), while from cylinder II we obtain two straight lines at distance \(R\) from \(KK_1\). In the drawing one line \(DE\) is shown. Let us denote the points of intersection of the circle and the line by \(D\) and \(E\). These points will be at the same distance from the center \((m'm_1')\) and from the line \(k'k_1'\). Obviously, when the magnitude \(R\) changes, this property is preserved for all points. Therefore the section of the locus under consideration by the plane \(S\) gives a parabola with focus at the point \((m'm_1')\) and directrix \(k'k_1'\). The branches of the parabola are directed upward.

Section by the plane \(U\), passing through the point \(C\) and parallel to the plane \(W\)

Reasoning analogously to the preceding case, on the profile plane of projection we obtain a parabola with focus at the point \((k''k_1'')\) and directrix \(m_1''m''\). The branches of this parabola will be directed downward.

Section by the plane \(P\) parallel to the plane \(H\)

The section of the two cylinders I and II by the plane \(P\) gives two pairs of parallel lines, which intersect at the points 1, 2, 3, and 4. As the...

as the value of the radius \(R\) changes, these points will describe a curve. As the radius \(R\) is increased, the distance between points 1 and 3 increases, and the points approach the straight line \(CF\). Therefore, for a radius equal to infinity we obtain on the curve one improper point. Arguing similarly for points 2 and 4, we obtain a second improper point on the curve.

Thus, the second-order curve obtained has two improper points and therefore is a hyperbola, whose asymptotes are the straight lines \(CF\) and \(CG\).

In the same way one can prove that any section by a plane parallel to the plane \(H\) and situated above the point \(C\) gives a hyperbola with asymptotes \(CF\) and \(CG\).

Section by a plane \(Q\), parallel to the plane \(H\) and situated below the point \(C\)

Arguing exactly as in the preceding case, we prove that the section by the plane \(Q\) gives a hyperbola with asymptotes \(CF\) and \(CG\), except that the axes of this hyperbola will be rotated by \(90^\circ\).

Any section by a plane parallel to the plane \(H\) and situated below the point \(C\) gives a hyperbola of the same form.

Section by a plane \(T\), parallel to the plane \(V\) and at a distance \(b\) from the point \(C\)

The section by the plane \(T\) on the vertical projection of the first cylinder gives a circle of radius \(R\) with center at the point \((m'_1m'_1)\), and from the other—two straight lines parallel to \(k'k'_1\), one of which intersects the circle

Fig. 2

at points 5 and 6. If the value of the radius \(R\) is varied, then points 5 and 6 describe a curve. We shall investigate this curve.

For convenience of analysis, let us superpose the frontal and profile projections in one drawing (Fig. 2). Let us denote the focus of the parabola by the point-

whose \(D\); we transfer the origin of coordinates to the vertex of the parabola \(C\); the profile trace of the plane \(T\) will take the position of the line \(AB\). \(FB\) is the directrix of the parabola. We shall prove that the section by the plane \(T\) gives a parabola the same as the section by the plane \(S\), only lowered downward.

Take a point \(N(xz)\) on the parabola obtained from the section by the plane \(S\). The equation of this parabola will have the form

\[ x^2 = 4az. \]

or

\[ z=\frac{x^2}{4a}. \tag{1} \]

On the second parabola take a point \(M(xz_1)\), having the same abscissa as the point \(N\).

By construction \(EF=AB=a+z_1\).

From the right triangle \(\triangle FAB\) we have

\[ FA=R=\sqrt{(a+z_1)^2+b^2}. \]

By construction

\[ DM=FA=\sqrt{(a+z_1)^2+b^2}. \]

Fig. 3

From the right triangle \(\triangle DEM\) we have:

\[ x^2=(a+z_1)^2+b^2-(z_1-a)^2, \]

\[ x^2=a^2+2az_1+z_1^2+b^2-z_1^2+2az_1-a^2, \]

\[ x^2=4az_1+b^2 \]

or

\[ z_1=\frac{x^2-b^2}{4a}. \tag{2} \]

Subtracting the second from the first, we obtain

\[ z-z_1=\frac{x^2}{4a}-\frac{x^2-b^2}{4a}=\frac{b^2}{4a}. \]

Thus, for any two points taken on the obtained curves whose abscissas are equal, the difference of the applicates is equal to the constant value \(\frac{b^2}{4a}\).

Consequently, if the first curve is a parabola, then the second will also be a parabola, only lowered by the amount \(\frac{b^2}{4a}\).

Considering the section by the plane \(T\), one may conclude that the locus under investigation can be obtained from the motion of a parabola which moves parallel to itself and whose vertex slides along another parabola.

The branches of the latter parabola are directed in the opposite direction. It is known from analytic geometry that such a surface is called a hyperbolic paraboloid.

Let us consider the case when the angle between the skew lines is not a right angle (Fig. 3).

In the drawing, \(MM_1\) is parallel to the plane \(V\), and the angle between \(MM_1\) and \(KK_1\) is \(60^\circ\). About each line describe a cylindrical surface of radius \(R\), and consider the sections of these surfaces by various planes.

Section by planes parallel to the plane \(H\)

The section by the plane of projections \(V\) of the surface of the cylinder described about the line \(KK_1\) gives an ellipse. To avoid constructing ellipses, let us apply the method of auxiliary projection onto two planes. As auxiliary planes we take two horizontal-projecting planes \(R\) and \(S\), equally inclined to the generators of the cylinders and to the depth-projecting lines.

The planes \(R\) and \(S\) intersect the cylinders in ellipses, which are projected onto the plane \(V\) as circles. In what follows, reasoning in the same way as in the first case, we shall prove that the sections by the plane \(Q\), passing through the point \(C\), give on the horizontal plane of projection two straight lines \(ab\) and \(ef\), which are the bisectors of the angles formed by the lines \(kk_1\) and \(mm_1\). Similarly, we shall prove that the section by the plane \(P\) on the horizontal projection gives a hyperbola whose asymptotes will be the straight lines \(ab\) and \(ef\).

The axes of the hyperbola obtained will in this case be rotated by \(15^\circ\) clockwise. It is evident that for any angle between the lines \(MM_1\) and \(KK_1\), the section by a plane parallel to the plane \(H\) of the locus under investigation gives a hyperbola whose asymptotes are mutually perpendicular.

Section by vertical planes

Before constructing the section by vertical planes, let us consider the loci of points obtained from the intersection of circles and ellipses.

Let us consider the locus of points obtained from the intersection of pairs of concentric circles of equal radii, whose centers are located at the points \(M\) and \(N\) (Fig. 4). It is obvious that the points of the desired locus \(E, C, D, F\) belong to the line \(AB\). The line \(AB\) is perpendicular to \(MN\) and passes through the point \(O\),

Fig. 4

the midpoint of the segment \(MN\). Let us prove this mathematically. To this end, place the origin at the point \(O\), and direct the \(OX\) axis along the line \(AB\). The equations of the circles with centers at the points \(M, N\) and radius \(R\) will have the form:

\[ x^{2} + (z - a)^{2} = R^{2}, \quad \mathrm{I} \]

\[ x^{2} + (z + a)^{2} = R^{2}. \quad \mathrm{II} \]

To obtain the desired locus, let us eliminate the radius \(R\) from the equations obtained; then we get:

\[ x^{2} + (z - a)^{2} = x^{2} + (z + a)^{2}, \]

\[ z^{2} - 2az + a^{2} = z^{2} + 2az + a^{2}, \]

\[ 4az = 0, \]

\[ z = 0. \]

We have obtained the equation of the line \(AB\).

Let us consider the locus of points obtained from the intersection of a circle with center at the point \(M\) and similar ellipses with center at the point \(N\). The minor axes of the ellipses are equal to the corresponding diameters of the circles (Fig. 5). If the ellipse is replaced by a circle with center at the point \(N\) and radius \(R\), then circles I and II intersect at the points \(A\) and \(B\). Replacing circle II by an ellipse, we obtain the intersection points of circle I and the ellipse, \(C\) and \(D\). It is obvious that the points \(C\) and \(D\) are located above the line \(AB\), and the distance between them has increased. As the radius \(R\) is increased, the points \(A\) and \(B\) will describe a straight line, while the points \(C\) and \(D\) will describe a curve. Let us prove that the curve obtained will be a parabola. By replacing the circle with an ellipse, we establish such a one-to-one correspondence between the points of the line and the curve that to each point of the line there will corresp-

there will be a single point of the curve. As the radius \(R\) increases, the distance of point \(A\) from point \(O\) increases and the ordinate of point \(C\) increases. For a radius equal to infinity, we obtain the improper point \(N\infty\) of the line \(AB\), to which there will correspond the single improper point of the curve. The curve under investigation is a curve of the second order and has one improper point; consequently, it is a parabola. The branches of the parabola are directed upward, toward the cir-

Fig. 5

cumference. Let us prove this mathematically. Place the origin of coordinates at point \(O\), and direct the axis of abscissas along the line \(AB\). The minor semiaxis of the ellipse is equal to the radius \(R\), and we denote the major semiaxis by \(R\kappa\), where \(\kappa>1\). Then the equations of the circle and the ellipse will have the form:

\[ x^2+(z-a)^2=R^2 \quad \text{equation of the circle,} \]

\[ \frac{x^2}{\kappa^2R^2}+\frac{(z+a)^2}{R^2}=1 \quad \text{equation of the ellipse.} \]

Eliminating the radius from these equations, we obtain:

\[ x^2+(z-a)^2=\frac{x^2}{\kappa^2}+(z+a)^2, \]

\[ x^2+z^2-2az+a^2=\frac{x^2}{\kappa^2}+z^2+2az+a^2, \]

\[ x^2-\frac{x^2}{\kappa^2}=4az, \]

\[ x^2\left(1-\frac{1}{\kappa^2}\right)=4az. \]

We have obtained the equation of a parabola.

For

\[ \kappa>1; \qquad 1-\frac{1}{\kappa^2}<1, \]

therefore, the branches of the parabola are flatter than those of the parabola

\[ x^2=4az. \]

Let us consider the locus of points obtained from the intersection of ellipses whose centers are at the points \(M\) and \(N\).

For this purpose we replace the upper circle by an ellipse whose minor semiaxis is equal to the radius \(R\), and whose major semiaxis is equal to \(R\kappa_1\). Then the equations of the ellipses will have the form:

\[ \frac{x^2}{R^2\kappa_1^2}=\frac{(z-a)^2}{R^2}=1 \qquad \text{the equation of the ellipse with center} \]
\[ \text{at the point } M. \]

\[ \frac{x^2}{R^2\kappa^2}+\frac{(z+a)^2}{R^2}=1 \qquad \text{the equation of the ellipse with center} \]
\[ \text{at the point } N. \]

Eliminating the radius \(R\) from the equations, we obtain:

\[ \frac{x^2}{\kappa_1^2}+(z-a)^2 = \frac{x^2}{\kappa^2}+(z+a)^2; \]

\[ \frac{x^2}{\kappa_1^2}+z^2-2az+a^2 = \frac{x^2}{\kappa^2}+z^2+2az+a^2, \]

\[ x^2\left(\frac{1}{\kappa_1^2}-\frac{1}{\kappa^2}\right)=4az \]

we have obtained the equation of a parabola.

For \(\kappa>\kappa_1\);

\[ \left(\frac{1}{\kappa_1^2}-\frac{1}{\kappa^2}\right)>0, \]

therefore the branches of the parabola are directed upward.

For \(\kappa<\kappa_1\);

\[ \left(\frac{1}{\kappa_1^2}-\frac{1}{\kappa^2}\right)<0 \]

then the branches of the parabola are directed downward.

Let us return again to Fig. 3. Rotate the axis of the hyperbola \(X_1X_1\) by \(15^\circ\) counterclockwise; then it will take a horizontal position. Let us cut the surfaces of the cylinder by a horizontal-projecting plane passing through the axis \(X_1X_1\); then on the vertical plane of projection we obtain two ellipses whose centers lie on one vertical straight line. The minor semiaxes of the ellipses are equal to the radius of the cylinder surfaces, while the major semiaxis of the lower ellipse is greater than the major semiaxis of the upper ellipse. Increasing the value of the radius of the cylinders, we obtain two families of similar ellipses. As was proved above, the locus of the points of intersection of the corresponding ellipses is a parabola whose branches are directed upward.

Similarly we shall prove that the section by a horizontal-projecting plane passing through the axis \(Y_1Y_1\) will give a parabola whose branches are directed downward.

In conclusion it may be said that in this case as well, when the angle between the skew lines is not equal to \(90^\circ\), the locus of points equidistant from them will be a hyperbolic paraboloid.

In what follows we shall prove all the propositions derived by the method of analytic geometry.

Analytic proof of the locus of points under consideration

Consider the case when the angle between the skew lines is a right angle and one of the lines \(MM_1\) is perpendicular to the plane \(V\), while the other \(KK_1\) is parallel to it (Fig. 6).

For convenience, only the vertical projections of the lines \(MM_1\) and \(KK_1\) are given in the drawing. The distance between the lines is equal to \(2a\).

The locus of points equidistant from one line is the surface of a cylinder whose axis is the given line. Let us write the equations of the cylinders whose axes are the given lines:

1. \(x^2+(z-a)^2=R^2\) — the equation of the cylinder with axis \(MM_1\).

2. \(y^2+(z+a)^2=R^2\) — the equation of the cylinder with axis \(KK_1\).

The points equidistant from the lines \(MM_1\) and \(KK_1\) at a distance \(R\) must satisfy the first and second equations. In order to obtain the locus of points equidistant from the given lines, it is necessary that the radius take all values from \(a\) to infinity. For this it is necessary to eliminate the radius from equations 1 and 2; then we obtain:

\[ x^2+z^2-2az+a^2=y^2+z^2+2az+a^2, \]

\[ x^2-y^2=4az \]

or

\[ \frac{x^2}{2a}-\frac{y^2}{2a}=2z. \]

We have obtained the equation of the surface of an equilateral hyperbolic paraboloid.

Fig. 6.

Let us derive the equation of the surface of the required locus for any angle between skew lines. For this, first let us derive the equation of the surface of a cylinder whose axis is parallel to the plane \(H\) and forms an angle \(\alpha\) with the plane \(V\).

The normal equations of a line passing through the point \(K(x_1,y_1,z_1)\) and the point \(M(x,y,z)\), with current coordinates, have the form

\[ \frac{x-x_1}{\cos\alpha}=\frac{y-y_1}{\cos\beta}=\frac{z-z_1}{\cos\gamma}. \]

Denote the point belonging to the directrix of the cylinder by \(m(x,y,z)\) (Fig. 7), and the point with current coordinates belonging to the corresponding generator by \(M(X,Y,Z)\); taking into account that \(\cos\beta=\sin\alpha\), and \(\cos\gamma=0\), the equations of the generators will have the form:

\[ \frac{X-x}{\cos\alpha}=\frac{Y-y}{\sin\alpha}=\frac{Z-z}{0}. \tag{1} \]

To obtain the equation of the directrix, take the equation of a sphere at the point \(B\)

\[ x^2+y^2+(z-a)^2=R^2 \tag{2} \]

and intersect it with the plane passing through the point \(B\) and perpendicular to the line \(AB\).

\[ x\cos\alpha+y\sin\alpha=0 \tag{3} \]

Eliminating \(x,y,z\) from the system of equations \((1,2,3)\) and replacing \(X,Y,Z\) by \(x,y,z\), we obtain the equation of the required cylinder

\[ (x\sin\alpha-y\cos\alpha)^2+(z-a)^2=R^2. \tag{4} \]

Let us derive the equation of the locus of points equidistant from two skew lines \(AB\) and \(CD\), located at

at an angle \(\alpha\). The line \(AB\) is arranged as shown in Fig. 7, while the line \(CD\) is parallel to the \(OX\) axis and the point \(D\) lies on the \(OZ\) axis below the point \(O\) by the amount \(a\).

Then the equations of the cylinders expressing the loci of points equidistant from the given straight lines will respectively be:

\[ \text{for } AB \ldots\quad (x\sin\alpha - y\cos\alpha)^2 + (z-a)^2 = R^2, \tag{1} \]

\[ \text{for } CD \ldots\quad y^2 + (z+a)^2 = R^2. \tag{2} \]

As in the case considered above, eliminate \(R\) from these equations; then we obtain

\[ x^2\sin^2\alpha - y\sin^2\alpha - 2xy\sin\alpha\cdot\cos\alpha = 4az. \tag{3} \]

We have obtained the equation of a hyperbolic paraboloid.

Fig. 7

For \(\alpha=\dfrac{\pi}{2}\), equation (3) assumes the form of the equation derived above, i.e.

\[ \frac{x^2}{2a} - \frac{y^2}{2a} = 2z. \]

Let us reduce equation (3) to canonical form. To do this, rotate the entire system about the \(OZ\) axis through an angle \(\beta\).

\[ \left. \begin{aligned} x &= x'\cos\beta - y'\sin\beta,\\ y &= x'\sin\beta + y'\cos\beta,\\ z &= z' \end{aligned} \right\} \quad \text{rotation formula} \tag{4} \]

Substituting formulas (4) into equation (3) and carrying out the necessary transformations, we obtain:

\[ \begin{aligned} &x'^2(\sin^2\alpha\cos^2\beta - \sin^2\alpha\cdot\sin^2\beta - 2\sin\alpha\cdot\cos\alpha\cdot\sin\beta\cdot\cos\beta) -\\ &\quad - y'^2(\sin^2\alpha\cdot\cos^2\beta - \sin^2\alpha\cdot\sin^2\beta - 2\sin\alpha\cdot\cos\alpha\cdot\sin\beta\cdot\cos\beta) -\\ &\quad - 2(2\sin^2\alpha\sin\beta\cos\beta + \cos^2\beta\sin\alpha\cos\alpha - \sin^2\beta\sin\alpha\cos\alpha)x'y' = 4az. \end{aligned} \tag{5} \]

In order to reduce equation (5) to canonical form, we find those values of the angle \(\beta\) for which the coefficient of \(x'y'\) becomes zero. Equating this coefficient to zero, we obtain two values for the angle \(\beta\):

\[ \beta = \frac{\alpha}{2} - \frac{\pi}{4} \quad\text{and}\quad \beta = \frac{\pi}{4} + \frac{\alpha}{2}. \tag{6} \]

For these values of the angle \(\beta\), equation (5) takes the form

\[ Ax^2 - Ay^2 = az, \tag{7} \]

where

\[ A=\sin^2\alpha\cdot\cos^2\beta - \sin^2\alpha\cdot\sin^2\beta - 2\sin\alpha\cos\alpha\cdot\sin\beta\cos\beta. \]

Equation (7) is the equation of an equilateral hyperbolic paraboloid.

For \(\alpha=\dfrac{\pi}{2}\), \(A=1\). This will be the greatest value of the coefficient.

For \(\alpha<\dfrac{\pi}{2}\), \(A<1\).

In this case the generator and directrix of the parabola will be more gently sloping. The smaller the angle \(\alpha\), the more the branches of the parabolas, the generator and the directrix, will approach the plane \(XOY\).

For \(\alpha = 0\), equation (5) will take the form: \(4az = 0\), or \(z = 0\). We have obtained the equation of the plane determined by the coordinate axes \(OX\) and \(OY\).

Having considered the locus of points equidistant from two skew lines, one may speak of loci of points equidistant from three and four skew lines. The locus of points equidistant from three skew lines will be the line obtained from the intersection of the surfaces of two corresponding hyperbolic paraboloids. This line may exist, but may also fail to exist, depending on the position of the given lines.

For four skew lines, one may speak of the existence of points equidistant from the four given lines. These points may exist, but may also fail to exist, depending on the position of the given lines in space.

LITERATURE

1. N. A. Popov. A Course in Descriptive Geometry. 1947.
2. S. M. Kolotov. Auxiliary Projection. 1956.
3. A. K. Rudaev. Collection of Problems in Descriptive Geometry. 1948.
4. N. F. Chetverukhin, V. S. Levitskii, Z. I. Pryanishnikova, A. M. Tevlin, G. I. Fedorov. A Course in Descriptive Geometry. 1956.
5. O. N. Tsuberbiller. Problems and Exercises in Analytic Geometry. 1934.
6. N. A. Khomenko. Surfaces and Lines in Space. 1959.

Corrections and Noted Misprints