Official solution
Given, equation of hyperbola is \[{{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1\]
We know that the equation of hyperbola is
\[{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1\] Here, \[{a^2} = {\cos ^2}\alpha \] and \[{b^2} = {\sin ^2}\alpha \]
We know that, \[{b^2} = {a^2}\left( {{e^2} - 1} \right)\]
\[ \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right)\]
\[ \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2}\]
\[ \Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \Rightarrow e = \sec \,\alpha \]
\[\therefore\] \[ae = \cos \alpha \,\,.\,\,{1 \over {\cos \alpha }} = 1\]
Co-ordinate of foci are \[\left( { \pm \alpha e,0} \right)\,\,\] i.e. \[\left( { \pm 1,0} \right)\]
Hence, abscissae of foci remain constant when \[\alpha \] varies.