For some �, let the eccentricity and the length of the latus rectum of the hyperbola � be � and �, respectively, and let the eccentricity and the length of the latus rectum of the ellipse � be � and �, respectively.

Hyperbola and Ellipse Latus Rectum Numerical Question | JEE Main 2026

Q. For some θ ∈ (0, π/2), let the eccentricity and the length of the latus rectum of the hyperbola x² − y²sec²θ = 8 be e₁ and l₁, respectively, and let the eccentricity and the length of the latus rectum of the ellipse x²sec²θ + y² = 6 be e₂ and l₂, respectively. If e₁² = e₂²(sec²θ + 1), then ( l₁l₂ / e₁e₂ ) tan²θ is equal to ____.

Correct Answer: 8

Explanation

Step 1: Hyperbola

Given equation: x² − y²sec²θ = 8

Divide by 8:

x²/8 − y²/(8cos²θ) = 1

So, a² = 8, b² = 8cos²θ

e₁² = 1 + b²/a² = 1 + cos²θ

Latus rectum: l₁ = 2b²/a = 2(8cos²θ)/√8 = 4√2 cos²θ

Step 2: Ellipse

x²sec²θ + y² = 6

Divide by 6:

x²/(6cos²θ) + y²/6 = 1

a² = 6, b² = 6cos²θ

e₂² = 1 − b²/a² = 1 − cos²θ = sin²θ

Latus rectum: l₂ = 2b²/a = 2(6cos²θ)/√6 = 2√6 cos²θ

Step 3: Required Expression

(l₁l₂)/(e₁e₂) tan²θ

= (4√2 cos²θ · 2√6 cos²θ)/(√(1+cos²θ) · sinθ) · tan²θ

After simplification, the value comes out to be:

Hence, the required answer is 8.

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