Exams › JEE Advanced › Maths

AB is a double ordinate of the hyperbola x²/a² - y²/b² = 1 such that triangle AOB (O is the origin) is equilateral. Which condition does the eccentricity e satisfy?

1. e > sqrt(3)
2. 1 < e < 2/sqrt(3)
3. e = 2/sqrt(3)
4. e > 2/sqrt(3)

Correct answer: e > 2/sqrt(3)

Solution

The equilateral condition gives b²/a² < 1/3, leading via e² = 1 + b²/a² to e > 2/sqrt(3).

Related JEE Advanced Maths questions

• For the circle represented by (x + c)² + y² = a² and the ellipse given as (x - h)² / b² + y² / a² = 1 (where a, b, c, and h are all positive), if they share a tangent that is horizontal, which condition must be satisfied?
• The expression √(x² + (y - 1)²) - √(x² + (y + 1)²) = K describes a hyperbola when:
• When the circle defined by x² + y² = 1 intersects the rectangular hyperbola xy = 1 at four points (x_i, y_i) for i = 1, 2, 3, 4, which of the following is true?
• The line 3x + 4y = 24 meets the x-axis and y-axis at points A and B, respectively, while the line 4x + 3y = 24 intersects at points C and D. The four points A, B, C, and D are located on which type of curve?
• The curve y = f(x), representing a parabola, is tangent to the line y = x at the point where x = 1. Which of the following is true?
• Given (x, y) ∈ R, where x² + y² = 16, we know y = ±√(16 − x²). When x = 0, y equals ±4. For x = ±4, y equals 0. It is observed that no other integer pairs (x, y) satisfy x² + y² = 16. The set R is defined as {(0, 4), (0, −4), (4, 0), (−4, 0)}. What is one value in the domain of R?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →