Year: 1950
Topic: Geometry Analysis

Problem:
A plane varies so that it includes a cone of constant volume equal to $\pi a^3 / 3$ with the surface the equation of which in rectangular coordinates is $2xy = z$. Find the equation of the envelope of the various positions of this plane. \newline State the result so that it applies to a general cone (that is, conic surface) of the second order.

Solution:
Make the change of coordinates \[ x = \frac{1}{2} \sqrt{2} (u + v), \quad y = \frac{1}{2} \sqrt{2} (u - v). \] Then the $uv$-axes are orthogonal and rotated by $\pi/4$ radians from the $xy$-axes. The equation of the given surface in the new coordinates is \[ z^2 + v^2 = uz, \] which is a right circular cone with the $u$-axis. \newline Next we find the volume of the conical region cut off from the solid cone by a plane. Because of rotational symmetry we need only consider planes of the form $v = mu + b$. In order that the plane should cut off a bounded region it is necessary that $|m| < 1$. The region cut off is then a cone with an elliptical base. We will find the altitude of that cone and the area of its base. \newline The altitude is the distance from the origin to the plane, namely \[ \frac{|b|}{\sqrt{1 + m^2}}. \] The area of the base is $\sqrt{1 + m^2}$ times the area of the ellipse obtained by projecting it orthogonally onto the $vz$-plane. To find the equation of the projected ellipse we eliminate $u$ between the equations of the cone and the plane, getting \[ z^2 + v^2 = (mv + b)^2. \] Collecting the $v$ terms and completing the square we get \[ z^2 + (1 - m^2) \left( v - \frac{mb}{1 - m^2} \right)^2 = b^2 \frac{1}{1 - m^2}. \] [Note that this is indeed an ellipse because $|m| < 1$.] The area of this ellipse is \[ A = \pi b^2 \frac{1}{(1 - m^2)}. \] The volume of the conical region is therefore \[ \frac{1}{3} \text{ base } \times \text{ altitude } = \frac{1}{3} \left( \sqrt{1 + m^2} \pi b^2 \frac{1}{(1 - m^2)} \right) \left( \frac{|b|}{\sqrt{1 + m^2}} \right) = \frac{1}{3} \pi |b| \frac{1}{(1 - m^2)^2}. \] The problem restricts consideration to those planes which cut off a volume $\pi a^3 / 3$, that is, to planes for which \[ |b| = a \sqrt{1 - m^2}, \] where $m$ is the tangent of the angle between the plane and the $uv$-plane. \newline We are to find the envelope $E$ of all such planes. Clearly $E$ must share the rotational symmetry of the entire configuration, so it suffices to find the intersection of $E$ with the $uv$-plane. \newline Consider the plane $P$ tangent to $E$ at a point of $I$. Reflection in the $uv$-plane preserves $P$, since it preserves $E$ and fixes each point of $I$. Therefore $P$ is perpendicular to the $uv$-plane, so it has an equation of the form $v = mu + b$ where, as we have seen, $|b| = a \sqrt{1 - m^2}$. The line in which $P$ meets the $uv$-plane is clearly tangent to $I$. Hence $I$ is the envelope of all lines in the $uv$-plane having equations of the form \[ (1) \quad v = mu \pm a \sqrt{1 - m^2}. \] The envelope problem is thus reduced to two dimensions. \newline To find the envelope, we first take the positive sign in (1), and eliminate $m$ between the equation \[ (2) \quad v = mu + \sqrt{1 - m^2} \] and the equation obtained by differentiating (2) with respect to $m$, namely \[ (3) \quad 0 = v - \frac{m}{\sqrt{1 - m^2}} u + a \sqrt{1 - m^2}. \] From (3) we find \[ m^2 = \frac{v^2}{v^2 + a^2}, \quad \text{so } 1 - m^2 = \frac{a^2}{v^2 + a^2}. \] Equation (2) now becomes \[ u = \sqrt{1 - m^2} \left( \frac{m}{\sqrt{1 - m^2}} v + a \right) = \sqrt{1 - m^2} \sqrt{1 - m^2} \left( \frac{v^2}{a^2} + 1 \right), \] giving finally \[ (4) \quad u^2 = v^2 + a^2 \] as the equation of the envelope $I$. The same equation results if we start with the negative sign. \newline [Since the algebraic work could introduce extraneous points not in the locus, we should check to see that the given family of lines is exactly the family of tangent lines to the hyperbola (4). This examination reveals that the tangents to the upper branch of the hyperbola involve the plus sign in (1), while the tangents to the lower branch involve the minus sign. Thus the lower branch is indeed extraneous to the problem of finding the envelope of the family of lines (2).] \newline The three-dimensional envelope $E$ is obtained by revolving $I$ around the $u$ axis; hence its equation is \[ u^2 = v^2 + z^2 + a^2. \] Transforming back to the original coordinates its equation is \[ \boxed{2xy = z^2 + a^2}. \]