About the two cones cross-section

This document prepared for "MATH for FUN - A place to post and work math problems" Yahoo Group. Initial post (anonymous author): You have two cones where the base radii is 2 and the height of each cone is 3^(1/2) and you put the bases together. The center is on the origin and the vertices are at (0,0,3^(1/2) ) and (0,0,-3^(1/2) ). The endpoints of x never change position, but the vertices of the cones can be rotated around the y axis until the vertices coincide with the y axis. The intersection of the (x,y) plane with the double cone traces the curve x^2+y^2=1 when the cusps are on the z axis. When the cusps coincide with the y axis, the curve is two joined equilateral triangles with sides=2. My question; For other tilts of the cusps between the z axis and coinciding with the y axis, does the equation of the curve that intersects the double cone through the (x,y) plane have the form ((1+x)^2+y^2) ^(n/2)+(( 1-x)^2+y^ 2)^(n/2)= 2^n where n is a function of the tilt of the cusps. n=2 (no tilt) to n=oo when the cusps coincide with the y axis?. (n=tan(f(theta) theta=0 to sqrt(3) ???)