Which one of the following is independent of `(0 lt alpha lt pi//2)` for the hyperbola `(x^(2))/(cos^(2) alpha)-(y^(2))/(sin^(2)alpha)=1`

Options

A. eccentricity
B. equation of a directrix
C. absicssa of foci
D. abscissa of vertices

Answer

To solve the problem, we need to analyze the hyperbola given by the equation:

\[ \frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1 \]

We will find the eccentricity, the equation of the directrix, the coordinates of the foci, and the coordinates of the vertices to determine which of these is independent of \(\alpha\).

### Step 1: Identify \(a\) and \(b\)

From the standard form of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:

\[ a = \cos \alpha \quad \text{and} \quad b = \sin \alpha \]

### Step 2: Calculate the eccentricity \(e\)

The eccentricity \(e\) of a hyperbola is given by the formula:

\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Substituting \(a\) and \(b\):

\[ e = \sqrt{1 + \frac{\sin^2 \alpha}{\cos^2 \alpha}} = \sqrt{1 + \tan^2 \alpha} = \sec \alpha \]

This indicates that the eccentricity \(e\) is dependent on \(\alpha\).

### Step 3: Equation of the directrix

The equation of the directrix for a hyperbola is given by:

\[ x = \pm \frac{a}{e} \]

Substituting \(a\) and \(e\):

\[ x = \pm \frac{\cos \alpha}{\sec \alpha} = \pm \cos^2 \alpha \]

This shows that the directrix is also dependent on \(\alpha\).

### Step 4: Coordinates of the foci

The coordinates of the foci are given by:

\[ (\pm ae, 0) \]

Substituting \(a\) and \(e\):

\[ (\pm \cos \alpha \cdot \sec \alpha, 0) = (\pm 1, 0) \]

This indicates that the coordinates of the foci are independent of \(\alpha\).

### Step 5: Coordinates of the vertices

The coordinates of the vertices are given by:

\[ (a, 0) \text{ and } (-a, 0) \]

Substituting \(a\):

\[ (\cos \alpha, 0) \text{ and } (-\cos \alpha, 0) \]

This shows that the vertices are dependent on \(\alpha\).

### Conclusion

Among the options provided, the only quantity that is independent of \(\alpha\) is the coordinates of the foci, which are \((1, 0)\) and \((-1, 0)\).

Thus, the answer is the coordinates of the foci.

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