Question

Consider the hyperbola $\frac{x^2}{(\cos \theta)^2} - \frac{y^2}{(\sin \theta)^2} = 1$. If $\theta$ changes, which of the following remains constant?}

• A. vertices
• B. directrix
• C. foci
• D. eccentricity

Hint:

This is a classic question. Whenever the denominator terms of a conic are $\cos^2 \theta$ and $\sin^2 \theta$, their sum is always constant (1), which directly keeps the foci fixed.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we can determine which of its geometric parameters remain invariant as its parameter $\theta$ varies.

Step 2: Detailed Explanation:
Let us identify the parameters of the given hyperbola:
\[ a^2 = \cos^2 \theta \quad \text{and} \quad b^2 = \sin^2 \theta \]
Let us compute the coordinates of the foci, which are given by:
\[ F = (\pm ae, 0) \]
We know the relation between the semi-axes and eccentricity ($e$) of a hyperbola:
\[ b^2 = a^2(e^2 - 1) \implies a^2 e^2 = a^2 + b^2 \]
Substitute the values of $a^2$ and $b^2$ into this equation:
\[ a^2 e^2 = \cos^2 \theta + \sin^2 \theta \]
By the fundamental trigonometric identity, $\cos^2 \theta + \sin^2 \theta = 1$:
\[ a^2 e^2 = 1 \implies ae = 1 \]
Substitute $ae = 1$ back into the coordinates of the foci:
\[ F = (\pm 1, 0) \]
Notice that the coordinates $(\pm 1, 0)$ are entirely numeric and independent of the angle $\theta$.
Therefore, even if $\theta$ changes, the foci of the hyperbola remain constant.

Step 3: Final Answer
The correct option is (C).