A cone has its guiding curve to the circle $x^2+y^2+2ax+2by=0$ and passes through a fixed point $(0,0,c)$. If the section of the cone by plane $y=0$ is a rectangular hyperbola. Prove that the vertex lies on fixed circle $x^2+y^2+z^2+2ax+2by=0$ and $2ax+2by+cz=0$.
Attempt:
Using equation of circle and fixed point, equation of cone can be found out, it comes:
$$cx^2+cy^2-2axz-2byz+2acx+2bcy=0$$
Its section by $y=0$ plane comes out to be
$$cx^2-2axz+2ax=0$$
I am unable to proceed further. Please help
Best Answer
Let $V=(x_0,y_0,z_0)$ be the vertex of the cone. The cone is composed of all lines passing through $V$ and a point of circle $\gamma$ of equation $x^2+y^2+2ax+2by=0$ in the $xy$ plane; the equation of the cone is then: $$ (z_0x-x_0z)^2+(z_0y-y_0z)^2+2a(z_0x-x_0z)(z_0-z)+2b(z_0y-y_0z)(z_0-z)=0. $$ Intersecting this with plane $y=0$ gives the equation of a conic in the $xz$ plane: $$ (z_0x-x_0z)^2+y_0^2z^2+2a(z_0x-x_0z)(z_0-z)-2by_0z(z_0-z)=0. $$ This represents a rectangular hyperbola if the coefficients of $x^2$ and $z^2$ in that equation are opposite, which leads to the equation: $$ \tag{1} x_0^2+y_0^2+z_0^2+2ax_0+2by_0=0. $$ This is the equation of the sphere having $\gamma$ as a great circle.
We know, on the other hand, that point $P=(0,0,c)$ lies on the cone, implying that $V$ belongs to the cone having $P$ as vertex and $\gamma$ as guiding curve. Hence the coordinates of $V$ must satisfy the equation: $$ \tag{2} cx_0^2+cy_0^2+2ax_0(c-z_0)+2by_0(c-z_0)=0. $$ Combining equations $(1)$ and $(2)$ we obtain the equation of a plane, to which $V$ must then belong: $$ \tag{3} 2ax_0+2by_0+cz_0=0. $$ Vertex $V$ must then lie on the intersection of sphere $(1)$ and plane $(3)$, which is exactly the circle we were required to find.