For the hyperbola $\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{{\sin }^2\alpha }=1$, which of the following remains constant when $\alpha$ varies?

The standard form of a hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.Here $a^2={\cos }^2\alpha$ and $b^2={\sin }^2\alpha$.The eccentricity $e$ is given by $b^2=a^2(e^2-1)$:
$${\sin }^2\alpha ={\cos }^2\alpha (e^2-1)$$
$${\tan }^2\alpha =e^2-1⟹e^2=1+{\tan }^2\alpha ={\sec }^2\alpha ⟹e=\sec \alpha$$
Eccentricity depends on $\alpha$.Abscissae of vertices are $(\pm a,0)=(\pm \cos \alpha ,0)$, which depend on $\alpha$.Abscissae of foci are $(\pm ae,0)$:
$$ae=\cos \alpha \times \sec \alpha =1$$
So, the foci are $(\pm 1,0)$. These remain constant regardless of the value of $\alpha$.