<p>For the hyperbola \(\dfrac{x^2}{\cos^2\alpha} - \dfrac{y^2}{\sin^2\alpha} = 1\), which of the following remains constant when \(\alpha\) varies?</p>
Step-by-Step Solution
Key Concept: For a hyperbola of the form x²/a² - y²/b² = 1, the eccentricity e = √(1 + b²/a²) depends on the ratio b²/a², not on the individual values. Here, b²/a² = sin²α/cos²α = tan²α, so e² = 1 + tan²α = sec²α, making e = |sec α|. However, the product of semi-axes ab = cos α · sin α, and more importantly, the relationship a² + b² = cos²α + sin²α = 1 remains constant, which constrains the focal distance relationship.
<p><strong>Step 1:</strong> Identify the hyperbola parameters. Here a² = cos²α and b² = sin²α.</p><p><strong>Step 2:</strong> For a hyperbola, c² = a² + b² = cos²α + sin²α = 1 (using the Pythagorean identity).</p><p><strong>Step 3:</strong> Therefore c = 1 always, regardless of the value of α.</p><p><strong>Step 4:</strong> This means the distance between foci = 2c = 2 remains constant.</p><p><strong>Step 5:</strong> Verify: Eccentricity e = c/a = 1/cos α varies with α ✗; focal distance 2c = 2 stays constant ✓</p><p>∴ <strong>Answer: The distance between foci (or focal distance 2c = 2) remains constant.</strong></p>
Correct Answer: D