Condition for the inclination of generators through the principal elliptic section of a hyperboloid 
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?

Continuing your solution, 
$$\begin{align} 
\cos\phi &=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{{l_1}^2+{m_1}^2+{n_1}^2}\sqrt{{l_2}^2+{m_2}^2+{n_3}^2}} \\
\implies \cos\phi &=\frac{a^2\sin^2\theta+b^2\cos^2\theta-c^2}{a^2\sin^2\theta+b^2\cos^2\theta+c^2}
\end{align}$$
Putting $\phi = 60^\circ$ and solving, we get 
$$a^2sin^2\theta+b^2cos^2\theta= 3c^2$$
Equiconjugate diameters of ellipse implies $\theta=\pi/4, 3\pi/4$. Put this to get the answer.
