RATIO OF THE AXES OF AN ELLIPSE OBTAINED BY INTERSECTING A CONE WITH A PLANE

B. A. MASHUKOV

(Presented by the scientific seminar of the departments of descriptive geometry of the Tomsk higher educational institutions)

In Fig. 1 two cones are given, with vertices \(S\) and \(S_1\) and a common base.

From the point \(F\), cutting planes \(Q_{10}, Q_{20}, Q_{30}\ldots Q_n\) are drawn through every \(10^\circ\). The angle measured upward from the base is taken as positive, and downward as negative.

The plane \(Q_{60}\) gives the ratio of the axes of the ellipse

\[ \frac{b}{a}=\frac{0}{a}=0. \]

The plane

\[ Q_0-\frac{b}{a}=\frac{d}{d}=1. \]

The plane

\[ Q_{60}-\frac{b}{a}=\frac{\infty}{\infty}=\infty, \]

and thus it is possible to obtain an ellipse with the ratio of the axes from zero to infinity.

The author sets as his goal to find the dependence between the cutting plane and the ratio of the axes of the ellipse.

Let us consider a cone with vertex \(S\). The cutting planes intersect the cone in ellipses, and the major axis of each ellipse is obtained in its true magnitude, while the minor axis is projected to the points \(O_1, O_2, O_3\ldots O_n\) and at that point divides the major axis in half.

In order to find the points \(O_2, O_3\), etc., it is sufficient to divide the generator \(FS\) in half (point \(R\)) and connect the point \(R\) with the point \(O\), and to mark the points of intersection of the cutting planes \(Q_{10}, Q_{20}\), etc., with the plane \(T\). But the plane \(T\) is parallel to the generator of the cone; consequently, it will intersect the cone in a parabola.

On the horizontal projection of the cone, the magnitudes of the minor axes of the ellipses have been found for each cutting plane \(Q_{10}, Q_{20}\), etc.

If the ends of the minor axes are connected, we obtain a parabola. The magnitude of the major and minor axes of the ellipse can be determined at once for any cutting plane in the following way. We draw a cutting plane from the point \(F\) and mark the point \(O_n\) on the trace of the plane \(T\). The magnitude of the major axis is obtained in true magnitude on the trace of the cutting plane \(Q_n\) (for example, \(Q_{y20}\) for an angle of \(20^\circ\)), and the minor axis is obtained as follows: we drop a perpendicular from the point to the horizontal

projection and mark two points \(K\) and \(K_1\) between the branches of the parabola. This will be the length of the minor axis of the ellipse in full size.

The cone with vertex \(S_1\) is intersected by cutting planes \(Q_1, Q_2, \ldots, Q_n\), and the minor axis of the ellipse also lies in the plane that cuts the cone along a parabola. From Fig. 1 it is clear that no matter how many cones with a common base we take, the vertices of the parabolas will be projected to one point and the parabolas will have one common horizontal projection. The vertices of the parabolas lie in a plane that cuts the cone along a hyperbola (see Fig. 1).

This parabola can be constructed not by points, as was done in Fig. 1, but from the equation of the parabola \(y^2 = 2px\).

Let us find the value of \(p\) as a function of the diameter \(d\) of the base of the cone. \(p = \dfrac{y^2}{2x}\). Expressing, at point 1 (see Fig. 1, horizontal projection), \(x\) and \(y\) in terms of the diameter of the circumference of the base of the cone, we shall have

\[ y = \frac{d}{2}, \quad \text{and} \quad x = \frac{d}{4}, \quad \text{then} \quad p = \frac{d^2}{4} : \frac{2d}{4} = \frac{d}{2}; \]

substituting \(\dfrac{d}{2}\) into the formula of the parabola, we obtain \(y^2 = 2 \dfrac{d}{2} x = dx\), and \(y = \sqrt{dx}\), where

\[ y = \frac{b}{2} \]

is one half of the minor axis of the ellipse.

Fig. 1.

In Fig. 2 two projections of two cones are given, and on the horizontal projections of the cones the parabola is constructed from the equation \(y = \sqrt{dx}\). For a cone with the angle of inclination of its generators to the plane \(H\), \(\angle \alpha = 70^\circ\) (Fig. 2), a graph has been constructed (Fig. 3), in which the dependence is given between the angle of inclination of the cutting plane and the quantity \(\dfrac{a}{b}\) (the ratio of the axes of the ellipse).

Fig. 2.

Vertical axis: Ratio \(b/a\)

Horizontal axis: \(\alpha\) in degrees

Note:
1. Diameter of the base of the cone
\[ b = 100\ \text{mm} \]
2. Angle at the vertex of the cone
\[ \beta = 40^\circ \]
3. Angle of inclination of the generatrices to the plane of the base
\[ \lambda = 70^\circ \]

Fig. 3.

From this graph (Fig. 3) one can determine the angle of inclination of the cutting plane so that the ratio of the axes \(\frac{b}{a}\) is prescribed in advance; for example, \(\frac{b}{a}=\frac{1}{2}\). We draw a straight line from \(\frac{b}{a}\) to its intersection with the curve, drop a perpendicular to the abscissa axis, and note that for the given ellipse the cutting plane must be set at an angle of inclination to the plane \(H\) of \(\angle \alpha \approx 52^\circ\).