For hyperbola `x^2sec^2alpha-ycos e c^2alpha=1,` which of the following remains constant with change in `'alpha'` abscissa of vertices (b) abscissa of foci eccentricity (d) directrix

Options

A. abscissae of vertices
B. abscissae of foci
C. eccentricity
D. directrix

Answer

To solve the problem, we start with the given hyperbola equation:

\[ x^2 \sec^2 \alpha - y^2 \csc^2 \alpha = 1 \]

### Step 1: Rewrite the equation
We can rewrite \(\sec^2 \alpha\) and \(\csc^2 \alpha\) in terms of sine and cosine:
\[ \sec^2 \alpha = \frac{1}{\cos^2 \alpha}, \quad \csc^2 \alpha = \frac{1}{\sin^2 \alpha} \]
Thus, the equation becomes:
\[ \frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1 \]

### Step 2: 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:
\[ a^2 = \cos^2 \alpha \quad \text{and} \quad b^2 = \sin^2 \alpha \]
Thus, we have:
\[ a = \cos \alpha \quad \text{and} \quad b = \sin \alpha \]

### Step 3: Find the vertices
The vertices of the hyperbola are given by the points:
\[ (\pm a, 0) = (\pm \cos \alpha, 0) \]

### Step 4: Find the foci
The foci of the hyperbola are located at:
\[ (\pm ae, 0) \]
where \(e\) (eccentricity) is given by:
\[ e = \sqrt{a^2 + b^2} / a \]
Calculating \(e\):
\[ e = \frac{\sqrt{\cos^2 \alpha + \sin^2 \alpha}}{\cos \alpha} = \frac{1}{\cos \alpha} \]
Thus, the coordinates of the foci are:
\[ (\pm a e, 0) = \left(\pm \cos \alpha \cdot \frac{1}{\cos \alpha}, 0\right) = (\pm 1, 0) \]

### Step 5: Find the directrix
The equation of the directrix is given by:
\[ x = \pm \frac{a}{e} = \pm \frac{\cos \alpha}{\frac{1}{\cos \alpha}} = \pm \cos^2 \alpha \]

### Summary of Results
- **Abscissa of vertices**: \(\pm \cos \alpha\) (depends on \(\alpha\))
- **Abscissa of foci**: \(\pm 1\) (constant)
- **Eccentricity**: \( \frac{1}{\cos \alpha} \) (depends on \(\alpha\))
- **Directrix**: \(\pm \cos^2 \alpha\) (depends on \(\alpha\))

### Conclusion
The only quantity that remains constant with changes in \(\alpha\) is the abscissa of the foci, which is \(\pm 1\).

### Answer
(b) Abscissa of foci