Solved Problems on Enveloping Cones and Locus of Vertex

Let the coordinates of the vertex $P$ be $(\alpha,\beta,\gamma)$. The equation of the enveloping cone of the ellipsoid is written as $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}+\frac{z\gamma}{c^2}-1\right)^2.$$ Its section by the plane $z=0$ is obtained by putting $z=0$ in the above equation. The resulting curve lies in the $xy$-plane and is given to be a rectangular hyperbola. Hence, the sum of the coefficients of $x^2$ and $y^2$ must be zero. This condition yields $$\begin{aligned} &\frac{1}{a^2}\left(\frac{\alpha^2}{a^2} +\frac{\beta^2}{b^2} +\frac{\gamma^2}{c^2}-1\right) -\frac{\alpha^2}{a^4} +\frac{1}{b^2}\left(\frac{\alpha^2}{a^2} +\frac{\beta^2}{b^2} +\frac{\gamma^2}{c^2}-1\right) -\frac{\beta^2}{b^4} =0. \end{aligned}$$ On simplification, this reduces to $$\frac{1}{a^2}\left(\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}\right) +\frac{1}{b^2}\left(\frac{\alpha^2}{a^2}+\frac{\gamma^2}{c^2}\right) =\frac{1}{a^2}+\frac{1}{b^2}.$$ Further simplification gives $$c^2(\alpha^2+\beta^2)+(a^2+b^2)\gamma^2=c^2(a^2+b^2),$$ or, $$\frac{\alpha^2+\beta^2}{a^2+b^2}+\frac{\gamma^2}{c^2}=1.$$ Hence, the locus of the vertex $(\alpha,\beta,\gamma)$ is $$\frac{x^2+y^2}{a^2+b^2}+\frac{z^2}{c^2}=1.$$ \quad $\blacksquare$

Let the vertex of the cone be $P(\alpha,\beta,\gamma)$. The equation of the enveloping cone corresponding to the ellipsoid is written as $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}+\frac{z\gamma}{c^2}-1\right)^2.$$ Its section by the plane $z=0$ is obtained as $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}-1\right)^2.$$ Since the section is a parabola, the condition $$(\text{coefficient of }2xy)^2 = (\text{coefficient of }x^2)\,(\text{coefficient of }y^2)$$ must be satisfied. Hence, $$\begin{aligned} \left(-\frac{\alpha\beta}{a^2b^2}\right)^2 &= \left(\frac{\beta^2}{a^2b^2}+\frac{\gamma^2}{a^2c^2}-\frac{1}{a^2}\right) \left(\frac{\alpha^2}{a^2b^2}+\frac{\gamma^2}{b^2c^2}-\frac{1}{b^2}\right). \end{aligned}$$ On simplification, this leads to $$\begin{aligned} \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) (\gamma^2-c^2)=0. \end{aligned}$$ Since the vertex lies outside the ellipsoid, $$\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}\neq 1.$$ Therefore, $$\gamma^2-c^2=0.$$ Hence, the locus of the vertex is $$z^2=c^2 \quad \text{or} \quad z=\pm c.$$ \quad $\blacksquare$

Let the luminous point be $(\alpha,\beta,\gamma)$. This point serves as the vertex of the enveloping cone to the ellipsoid, whose truncated portion produces the shadow. The equation of the enveloping cone is $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}+\frac{z\gamma}{c^2}-1\right)^2.$$ Its section by the plane $z=0$ is given by $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right) \left(\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \left(\frac{x\alpha}{a^2}+\frac{y\beta}{b^2}-1\right)^2.$$ Since the section is a circle, the coefficient of $xy$ must vanish and the coefficients of $x^2$ and $y^2$ must be equal. Hence, $$\alpha\beta=0. \quad (1)$$ Also, $$\frac{1}{a^2}\left(\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) = \frac{1}{b^2}\left(\frac{\alpha^2}{a^2}+\frac{\gamma^2}{c^2}-1\right). \quad (2)$$ \textit{Case I:} $\alpha=0$. From (2), $$\begin{aligned} \frac{1}{a^2}\left(\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}-1\right) &= \frac{1}{b^2}\left(\frac{\gamma^2}{c^2}-1\right), \end{aligned}$$ which reduces to $$\frac{\beta^2}{b^2-a^2}+\frac{\gamma^2}{c^2}=1.$$ Hence, the locus is $$\frac{y^2}{b^2-a^2}+\frac{z^2}{c^2}=1,\quad x=0.$$ \textit{Case II:} $\beta=0$. From (2), $$\begin{aligned} \frac{1}{a^2}\left(\frac{\gamma^2}{c^2}-1\right) &= \frac{1}{b^2}\left(\frac{\alpha^2}{a^2}+\frac{\gamma^2}{c^2}-1\right), \end{aligned}$$ which gives $$\frac{\alpha^2}{a^2-b^2}+\frac{\gamma^2}{c^2}=1.$$ Hence, the locus is $$\frac{x^2}{a^2-b^2}+\frac{z^2}{c^2}=1,\quad y=0.$$ Thus, the required loci are obtained. \quad $\blacksquare$

Without loss of generality, the equation of a plane parallel to the given plane may be written as $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1,$$ where $a,b,c$ are constants. Hence, the intercepts on the coordinate axes are obtained as $$A(a,0,0), \quad B(0,b,0), \quad C(0,0,c).$$ The equations representing the circle $ABC$ are taken as $$\left. \begin{aligned} x^2+y^2+z^2-ax-by-cz &= 0,\\ \frac{x}{a}+\frac{y}{b}+\frac{z}{c} &= 1, \end{aligned} \right\} \quad (1)$$ where the first equation denotes the sphere passing through $O,A,B,C$, and the second denotes the plane containing the circle. Let a point $P(x,y,z)$ be chosen on the required cone, and let the generator through $P$ intersect the guiding circle $ABC$ at $(\alpha,\beta,\gamma)$. Since $(\alpha,\beta,\gamma)$ lies on the circle $ABC$, it satisfies $$\alpha^2+\beta^2+\gamma^2-a\alpha-b\beta-c\gamma=0 \quad (2)$$ and $$\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}=1. \quad (3)$$ The equations of the generator through $(\alpha,\beta,\gamma)$ are expressed as $$\frac{x}{\alpha}=\frac{y}{\beta}=\frac{z}{\gamma}=r,$$ or equivalently, $$x=\alpha r,\quad y=\beta r,\quad z=\gamma r. \quad (4)$$ From equations (3) and (4), the parameter $r$ is obtained as $$\begin{aligned} \frac{x}{a}+\frac{y}{b}+\frac{z}{c} &= \frac{\alpha r}{a}+\frac{\beta r}{b}+\frac{\gamma r}{c} \\ &= r\left(\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}\right) \\ &= r. \end{aligned}$$ Hence, $$r=\frac{x}{a}+\frac{y}{b}+\frac{z}{c}. \quad (5)$$ Again, using equations (2) and (4), it follows that $$\begin{aligned} \frac{x^2+y^2+z^2}{r^2} &= \frac{ax+by+cz}{r}, \end{aligned}$$ which gives $$x^2+y^2+z^2=r(ax+by+cz).$$ Substituting the value of $r$ from (5), one obtains $$\begin{aligned} x^2+y^2+z^2 &=\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)(ax+by+cz). \end{aligned}$$ On simplification, the equation reduces to $$\left(\frac{b}{c}+\frac{c}{b}\right)yz +\left(\frac{c}{a}+\frac{a}{c}\right)zx +\left(\frac{a}{b}+\frac{b}{a}\right)xy=0.$$ Thus, the circle $ABC$ is shown to lie on the above cone with vertex at the origin. \quad $\blacksquare$