JEE Main 2021 · 25 February · evening · Single correct · +4 / −1

A hyperbola passes through the foci of the ellipse x²/25 + y²/16 = 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is :

  1. (a) x²/9 - y²/4 = 1
  2. (b) x²/9 - y²/16 = 1
  3. (c) x²/9 - y²/25 = 1
  4. (d) x² - y² = 9
Answer and solution

Answer (b) x²/9 - y²/16 = 1

Solution

Maths sir · 4 steps

Find the ellipse's foci and eccentricity first; the hyperbola's a and e both come straight from them.

  1. Read the ellipse. Here a = 5 and b = 4. So e₁ = √(1 - 16/25) = 3/5, and the foci are (± a e₁, 0) = (± 3, 0).
  2. Fix the hyperbola's a. The axes match the ellipse, so the hyperbola is x²/A² - y²/B² = 1. It passes through (3, 0), which puts 9/A² = 1, so A² = 9.
  3. Use the eccentricity. From e₁ e₂ = 1 we get e₂ = 5/3. For a hyperbola B² = A²(e₂² - 1).

    B² = 9(25/9 - 1) = 9 × 16/9 = 16

  4. So. The hyperbola is x²/9 - y²/16 = 1, which is option (b).

The idea to keep. A curve passing through a point on its own transverse axis tells you that axis length directly. Then the eccentricity gives the other one: b² = a²(e² - 1) for a hyperbola, b² = a²(1 - e²) for an ellipse.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Hyperbola

Asked about 60 times in JEE Main since 2020. This question is from 2021.