Find the locus of the points of intersection of perpendicular generators of the hyperboloid $$\frac{x^2}{a^2}+ \frac{y^2}{b^2}- \frac{z^2}{c^2} =1$$.
I know the equations of the generators in paramaterized form of $\lambda$ and $\mu$, but not able to use it to get the locus. Any help is appreciated. Thanks
Knowing that the answer is director sphere $x^2+y^2+z^2=a^2+b^2-c^2$, I am able to derive the solution.
Generators are given by $$\frac{x/a+z/c}{1+y/b}=\lambda= \frac{1-y/b}{x/a-z/c}$$ $$\frac{x/a+z/c}{1-y/b}=\mu= \frac{1+y/b}{x/a-z/c}$$
the direction ratios of the lines for $\lambda$ system and $\mu$ systems can be easily found by taking cross product of the normals of the planes of these system. Skipping the calculations, they turn out to be, respectively, $$(a(\lambda^2-1), 2\lambda b, c(1+\lambda^2); (a(1-\mu^2),2\mu b, -c(1+\mu^2))$$
They will be perpendicular if dot product is zero, or, $$a^2(\lambda^2-1)(1-\mu^2)+4b^2\lambda\mu-c^2(\lambda^2+1)(\mu^2+1)=0$$
Also as $x,y,z$ can be expressed in parametric form as $$x=a\frac{1+\lambda\mu}{\lambda+\mu}; y=b\frac{\lambda-\mu}{\lambda+\mu};x=c\frac{1-\lambda\mu}{\lambda+\mu}$$
Squaring, adding and using relation of dot product being zero, we get the locus as $$x^2+y^2+z^2=a^2+b^2-c^2$$
This solution is more of a retrofitting, knowing the solution beforehand. Its not an elegant method. Appreciate any better solution.