Consider the family of planes in three dimensions that intersect the positive x, y, and z axes at point A, B, and C, respectively, such that the tetrahedron whose vertices are these three points and the origin, O, has a fixed volume V . Find the equation of the envelope surface defined by this family of planes.
I can't even imagine the planes, hence can't determine the family of planes. Maybe someone can show me the picture of planes and the equation of the family of planes?
The equation of the plane so defined is $\dfrac{x}{A}+\dfrac{y}{B}+\dfrac{z}{C}=1$ Here is a drawing of it:
It shows the four points taken as vertices of a tetrahedron. From elementary formulas its volume is $V=ABC/6$. As stated in the problem's wording, this volume has to be constant for all the planes into the family, so one among $A,B,C$ is determined by the other two. Let it be $C$: $C=\dfrac{6V}{AB}$
We have three equations to find the equation of the envelope (by eliminating $A$ and $B$)
$$\left\{ \begin{matrix} \dfrac{x}{A}+\dfrac{y}{B}+\dfrac{ABz}{6V}=1 \\ \dfrac{\partial}{\partial A}\left(\dfrac{x}{A}+\dfrac{y}{B}+\dfrac{ABz}{6V}-1\right)=0 \\ \dfrac{\partial}{\partial B}\left(\dfrac{x}{A}+\dfrac{y}{B}+\dfrac{ABz}{6V}-1\right)=0 \\ \end{matrix} \right.$$