Solved Problems on Common Vertex and Common Generators of Cones

Let the common vertex be $V(\alpha,\beta,\gamma)$. The cone through $y=0,\, z^2=4ax$ is $$\begin{equation} (\gamma y-\beta z)^2=4a(y-\beta)(\alpha y-\beta x). \end{equation}$$ The cone through $x=0,\, z^2=4by$ is $$\begin{equation} (\gamma x-\alpha z)^2=4b(x-\alpha)(\beta x-\alpha y). \end{equation}$$ Their sections by $z=0$ are respectively $$\begin{equation} \gamma^2y^2=4a(y-\beta)(\alpha y-\beta x), \end{equation}$$ $$\begin{equation} \gamma^2x^2=4b(x-\alpha)(\beta x-\alpha y). \end{equation}$$ Any curve passing through the common points of (3) and (4) is $$\gamma^2y^2-4a(y-\beta)(\alpha y-\beta x) +\lambda\{\gamma^2x^2-4b(x-\alpha)(\beta x-\alpha y)\}=0.$$ Since this curve is a circle, the coefficients of $x^2$ and $y^2$ must be equal and that of $xy$ must vanish. Hence, $$\gamma^2-4a\alpha=\lambda(\gamma^2-4b\beta),$$ $$4a\beta+4b\alpha\lambda=0.$$ Eliminating $\lambda$, one obtains $$\gamma^2\!\left(\frac{\beta}{b}+\frac{\alpha}{a}\right)=4(\alpha^2+\beta^2).$$ Thus, the vertex lies on $$z^2\!\left(\frac{x}{a}+\frac{y}{b}\right)=4(x^2+y^2),$$ as required. $\blacksquare$

Let the common vertex be $(\alpha,\beta,\gamma)$. Consider a point $P(x,y,z)$ on the cone whose guiding curve is $zx=a^2,\; y=0$. Suppose the generator through $P$ meets the guiding curve at $(x_1,0,z_1)$. Then $$z_1x_1=a^2. \quad (1)$$ The equations of the generator through $P$ are $$\frac{x-\alpha}{x_1-\alpha} = \frac{y-\beta}{-\beta} = \frac{z-\gamma}{z_1-\gamma}.$$ From these, $$x_1=\alpha-\beta\frac{x-\alpha}{y-\beta} =\frac{\alpha y-\beta x}{y-\beta}, \quad z_1=\gamma-\beta\frac{z-\gamma}{y-\beta} =\frac{\gamma y-\beta z}{y-\beta}.$$ Using equation (1), the equation of the first cone is obtained as $$\begin{aligned} \frac{\gamma y-\beta z}{y-\beta}\cdot \frac{\alpha y-\beta x}{y-\beta} &=a^2 \\ \Rightarrow\; (\gamma y-\beta z)(\alpha y-\beta x) &=a^2(y-\beta)^2. \quad (2) \end{aligned}$$ Similarly, the equation of the second cone is $$(\beta x-\alpha y)(\gamma x-\alpha z)=b^2(x-\alpha)^2. \quad (3)$$ The cone (2) cuts the plane $z=0$ in the conic $$\gamma y(\alpha y-\beta x)=a^2(y-\beta)^2, \quad z=0, \quad (4)$$ and the cone (3) cuts the same plane in $$\gamma x(\beta x-\alpha y)=b^2(x-\alpha)^2, \quad z=0. \quad (5)$$ These two conics intersect in four points. Any conic passing through these four points can be written as $$k_1\{\gamma y(\alpha y-\beta x)-a^2(y-\beta)^2\} + k_2\{\gamma x(\beta x-\alpha y)-b^2(x-\alpha)^2\} =0, \quad z=0.$$ This conic is given to be a circle. Hence, the coefficients of $x^2$ and $y^2$ must be equal and the coefficient of $xy$ must vanish. Therefore, $$k_2(\beta\gamma-b^2)=k_1(\alpha\gamma-a^2),$$ and $$-k_1\beta\gamma-k_2\alpha\gamma=0.$$ Eliminating $\dfrac{k_2}{k_1}$, one obtains $$\frac{\alpha\gamma-a^2}{\beta\gamma-b^2} = -\frac{\beta}{\alpha},$$ which gives $$\alpha^2\gamma-a^2\alpha+\beta^2\gamma-b^2\beta=0,$$ or $$\gamma(\alpha^2+\beta^2)=a^2\alpha+b^2\beta.$$ Hence, the vertex $(\alpha,\beta,\gamma)$ lies on the surface $$z(x^2+y^2)=a^2x+b^2y.$$ \quad $\blacksquare$