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Step-by-Step Solution
Step 1: Identify the standard form of the hyperbola
The given hyperbola is
$ \frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1 $.
If we compare this with the standard form of the hyperbola
$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $,
we see that
$ a^2 = \cos^2 \alpha \quad \text{and} \quad b^2 = \sin^2 \alpha. $
Step 2: Relate $b^2$ and $a^2$ through eccentricity $e$
For a hyperbola of the form
$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, $
the relationship involving eccentricity is
$ b^2 = a^2 \left(e^2 - 1\right). $
Substituting
$ a^2 = \cos^2 \alpha $
and
$ b^2 = \sin^2 \alpha, $
we get
$ \sin^2 \alpha = \cos^2 \alpha (e^2 - 1). $
Step 3: Solve for the eccentricity $e$
Rearrange the above equation to find $e^2$:
$ \sin^2 \alpha + \cos^2 \alpha = \cos^2 \alpha \, e^2
\quad \Rightarrow \quad
\sin^2 \alpha + \cos^2 \alpha = 1 = \cos^2 \alpha \, e^2.
$
Thus,
$ e^2 = \frac{1}{\cos^2 \alpha} = \sec^2 \alpha
\quad \Rightarrow \quad
e = \sec \alpha. $
Step 4: Find the coordinates of the foci
The coordinates of the foci for the hyperbola
$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $
are
$ (\pm a e, \, 0). $
From above,
$ a = \cos \alpha $
and
$ e = \sec \alpha, $
so
$ a e = \cos \alpha \times \sec \alpha = 1. $
Hence, the foci have coordinates
$ (\pm 1, 0). $
Step 5: Conclude the abscissae remain constant
Since the coordinates of the foci are
$ (\pm 1,0) $
and do not depend on $ \alpha, $
the abscissae of the foci remain constant irrespective of the value of $ \alpha. $
Therefore, the answer is that the abscissae of the foci remain constant.
