To determine which of the following is independent of \(\alpha\) in the given hyperbola equation \(\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1\), we must first understand the components and properties of a hyperbola. The standard form of a hyperbola is:
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
Where:
- \(a\) is the semi-major axis.
- \(b\) is the semi-minor axis.
In this hyperbola, comparing with the standard form, we have:
- \(a^2 = \cos^2 \alpha\)
- \(b^2 = \sin^2 \alpha\)
Evaluation of Different Properties:
- Eccentricity: The eccentricity \((e)\) of a hyperbola is given by \(e = \sqrt{1 + \frac{b^2}{a^2}}\).
- Abscissa of Foci: The foci of a hyperbola are located at \((\pm ae, 0)\), where \(e\) is the eccentricity.
- Directrix: The equation of directrix in terms of \(\alpha\) is \(x = \pm \frac{a}{e}\).
- Vertex: The vertices are at \((\pm a, 0)\).
Independent Part:
Let's focus on calculating these specifications and determining their dependencies:
- Vertices: \(x = \pm a = \pm \cos \alpha\), dependent on \(\alpha\).
- Directrix: The directrix depends on both \(a\) and \(e\), hence dependent on \(\alpha\).
- Eccentricity: Computed as \(e = \sqrt{1 + \frac{\sin^2 \alpha}{\cos^2 \alpha}} = \sec \alpha\), also dependent on \(\alpha\).
- Abscissa of Foci: The expression becomes \(ae = \cos \alpha (\sec \alpha)\), simplifying to\)
Therefore, the correct answer is the Abscissa of foci, as it remains constant and independent of \(\alpha\).