Question: Show that the generators through any one of the ends of an equi-conjugate diameter of the principal elliptic section of the hyperboloid $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2} = 1$ are inclined to each other at an angle of $60 ^\circ$ if $a^2+b^2=6c^2$.
Solution: I took the point on the diameter as $(a\cos\theta,b\sin\theta,0)$ . So the two generators through this point are $$\begin{align} \frac{x-a\cos\theta}{a\sin\theta}=\frac{y-b\sin\theta}{-b\cos\theta}=\frac{z}{\pm c} \end{align}$$
So the direction ratios of two generators are $(a\sin\theta,-b\cos\theta,c)$ and $(a\sin\theta, -b\cos\theta,-c)$ respectively.
$$\begin{align} \cos\theta &=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{{l_1}^2+{m_1}^2+{n_1}^2}\sqrt{{l_2}^2+{m_2}^2+{n_3}^2}} \\ \implies \cos\theta &=\frac{a^2\sin^2\theta+b^2\cos^2\theta-c^2}{a^2\sin^2\theta+b^2\cos^2\theta+c^2} \end{align}$$
Putting $\theta = 60^\circ$ and solving I got $\,3a^2+b^2 = 12c^2$
I am wondering whether the question is wrong or I missed anything or my approach itself is wrong. Can some one please help?