# For the hyperbola (x^2)/(cos^2 a) - (y^2)/(sin^2 a) = 1, which one of the following remain constant with change of a? A. Abscissa of vertices B. Abscissa of foci C. Eccentricity D. Directrix

## Question

For the hyperbola $\frac{x^{2}}{\cos^{2} a} - \frac{y^{2}}{\sin^{2} a} = 1$, which one of the following remain constant with change of $a$?

A. Abscissa of vertices

B. Abscissa of foci

C. Eccentricity

D. Directrix

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## Answer

The correct response is **Option B: Abscissa of foci**.

Given the hyperbola equation: $$ \frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1 $$

This can be compared to the standard form of a hyperbola, where we derive: $$ a^2 = \cos^2 \alpha, \quad b^2 = \sin^2 \alpha $$

Key to this question is identifying which properties of the hyperbola remain consistent regardless of changes in $\alpha$. By exploring the equation's dependence on $\alpha$, we observe:

The terms

**$a^2 = \cos^2 \alpha$**and**$b^2 = \sin^2 \alpha$**clearly change with $\alpha$.

Focusing on the eccentricty ($e$): $$ e^2 - 1 = \frac{b^2}{a^2} = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \tan^2 \alpha $$ So, $$ e = |\sec \alpha| $$

Considering the **abscissa of the foci** ($\pm a e$):
$$
ae = \cos \alpha \cdot |\sec \alpha| = \pm 1
$$

This calculation shows that the **abscissa of the foci remains constant** as $1$ and is independent of $\alpha$.

Therefore, the abscissa of the foci does not change with the angle $\alpha$ and is the correct answer to the question.

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