Let . If the eccentricity of the hyperbola is greater than 2, then the length of its latus rectum lies in the interval
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
For the hyperbola x²/cos²θ - y²/sin²θ = 1, we have a² = cos²θ, b² = sin²θ.
Eccentricity e = √(1 + b²/a²) = √(1 + sin²θ/cos²θ) = √(1/cos²θ) = secθ.
Condition e > 2 means secθ > 2, so cosθ < 1/2, giving θ ∈ (π/3, π/2).
Length of latus rectum = 2b²/a = 2sin²θ/cosθ.
Let f(θ) = 2sin²θ/cosθ = 2(1-cos²θ)/cosθ. As θ → π/3⁺, cosθ → 1/2⁻, f → 2(1-1/4)/(1/2) = 2(3/4)(2) = 3. As θ → π/2⁻, cosθ → 0⁺, f → ∞.
Since f is strictly increasing on (π/3, π/2), the latus rectum lies in (3, ∞).
Eccentricity e = √(1 + b²/a²) = √(1 + sin²θ/cos²θ) = √(1/cos²θ) = secθ.
Condition e > 2 means secθ > 2, so cosθ < 1/2, giving θ ∈ (π/3, π/2).
Length of latus rectum = 2b²/a = 2sin²θ/cosθ.
Let f(θ) = 2sin²θ/cosθ = 2(1-cos²θ)/cosθ. As θ → π/3⁺, cosθ → 1/2⁻, f → 2(1-1/4)/(1/2) = 2(3/4)(2) = 3. As θ → π/2⁻, cosθ → 0⁺, f → ∞.
Since f is strictly increasing on (π/3, π/2), the latus rectum lies in (3, ∞).
[JEE Main 2019 · 9 Jan · Shift I]