Solved Problems on Cones with Guiding Curves and Sections

Let the vertex of the cone be denoted by $(\alpha,\beta,\gamma)$. A general point $P(x,y,z)$ is taken on the cone. Let the generator through $P$ meet the guiding ellipse at $(x_1,y_1,0)$. Since this point lies on the ellipse, it satisfies $$\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1. \quad (1)$$ The equations of the generator joining $(\alpha,\beta,\gamma)$ and $(x_1,y_1,0)$ are written as $$\frac{x-\alpha}{x_1-\alpha} =\frac{y-\beta}{y_1-\beta} =\frac{z-\gamma}{-\gamma}.$$ From these relations, one obtains $$x_1=\alpha-\gamma\frac{x-\alpha}{z-\gamma},\qquad y_1=\beta-\gamma\frac{y-\beta}{z-\gamma}.$$ Substituting these expressions in equation (1), the equation of the cone is obtained as $$\frac{1}{a^2}\left(\frac{\alpha z-\gamma x}{z-\gamma}\right)^2 +\frac{1}{b^2}\left(\frac{\beta z-\gamma y}{z-\gamma}\right)^2=1.$$ The section of this cone by the plane $x=0$ is therefore $$\frac{\alpha^2 z^2}{a^2} +\frac{(\beta z-\gamma y)^2}{b^2} =(z-\gamma)^2,\quad x=0.$$ This represents a rectangular hyperbola in the $yz$-plane. Hence, the sum of the coefficients of $y^2$ and $z^2$ must vanish. Therefore, $$\begin{aligned} \frac{\gamma^2}{b^2} +\frac{\alpha^2}{a^2} +\frac{\beta^2}{b^2} -1 &= 0. \end{aligned}$$ Thus, $$\frac{\alpha^2}{a^2}+\frac{\beta^2+\gamma^2}{b^2}=1.$$ Hence, the locus of the vertex $(\alpha,\beta,\gamma)$ is $$\frac{x^2}{a^2}+\frac{y^2+z^2}{b^2}=1.$$ \quad $\blacksquare$

Let the vertex of the cone be denoted by $(\alpha,\beta,\gamma)$. A general point $P(x,y,z)$ is taken on the cone. Let the generator through $P$ meet the guiding ellipse at $(x_1,y_1,0)$. Since this point lies on the ellipse, it satisfies $$\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1. \quad (1)$$ The equations of the generator joining $(\alpha,\beta,\gamma)$ and $(x_1,y_1,0)$ are written as $$\frac{x-\alpha}{x_1-\alpha} =\frac{y-\beta}{y_1-\beta} =\frac{z-\gamma}{-\gamma}.$$ From these relations, one obtains $$x_1=\alpha-\gamma\frac{x-\alpha}{z-\gamma},\qquad y_1=\beta-\gamma\frac{y-\beta}{z-\gamma}.$$ Substituting these expressions in equation (1), the equation of the cone is obtained as $$\frac{1}{a^2}\left(\frac{\alpha z-\gamma x}{z-\gamma}\right)^2 +\frac{1}{b^2}\left(\frac{\beta z-\gamma y}{z-\gamma}\right)^2=1.$$ The section of this cone by the plane $x=0$ is therefore $$\frac{\alpha^2 z^2}{a^2} +\frac{(\beta z-\gamma y)^2}{b^2} =(z-\gamma)^2,\quad x=0.$$ This represents a rectangular hyperbola in the $yz$-plane. Hence, the sum of the coefficients of $y^2$ and $z^2$ must vanish. Therefore, $$\begin{aligned} \frac{\gamma^2}{b^2} +\frac{\alpha^2}{a^2} +\frac{\beta^2}{b^2} -1 &= 0. \end{aligned}$$ Thus, $$\frac{\alpha^2}{a^2}+\frac{\beta^2+\gamma^2}{b^2}=1.$$ Hence, the locus of the vertex $(\alpha,\beta,\gamma)$ is $$\frac{x^2}{a^2}+\frac{y^2+z^2}{b^2}=1.$$ \quad $\blacksquare$