Hyperbola
Latus Rectum
Grade 11
Question:
<p>Let \(0 < \theta < \pi/2\). If the eccentricity of the hyperbola \(\dfrac{x^2}{\cos^2\theta} - \dfrac{y^2}{\sin^2\theta} = 1\) is greater than 2, then the length of its latus rectum lies in the interval:</p>
<p>\((3, \infty)\)</p>
<p>\(\left(\dfrac{3}{2}, 2\right]\)</p>
<p>\((2, 3]\)</p>
<p>\(\left(1, \dfrac{3}{2}\right]\)</p>
Step-by-Step Solution
Key Concept: For a hyperbola with equation x²/a² - y²/b² = 1, the eccentricity e = √(1 + b²/a²). When b = a (rectangular hyperbola), e = √2. The constraint 0 < b/a < 1 restricts eccentricity to the range 1 < e < √2.
<p><strong>Step 1:</strong> For hyperbola x²/a² - y²/b² = 1, eccentricity e = c/a where c² = a² + b²</p><p><strong>Step 2:</strong> Therefore e² = (a² + b²)/a² = 1 + (b/a)²</p><p><strong>Step 3:</strong> Given 0 < b/a < 1, we have 0 < (b/a)² < 1</p><p><strong>Step 4:</strong> Adding 1: 1 < 1 + (b/a)² < 2, which gives 1 < e² < 2</p><p><strong>Step 5:</strong> Taking square roots: 1 < e < √2</p><p>∴ Answer: A</p>
Correct Answer: A