JEE Main 2020 · 2 September · evening · Single correct · +4 / −1

For some θ ∈ ( 0,π/2 ), if the eccentricity of the
hyperbola, x²–y²sec²θ = 10 is √5 times the
eccentricity of the ellipse, x²sec²θ + y² = 5, then the length of the latus rectum of the ellipse, is :

  1. (a) √30
  2. (b) 2√6
  3. (c) 4√5/3
  4. (d) 2√5/3
Answer and solution

Answer (c) 4√5/3

Solution

Maths sir · 4 steps

Since sec² θ > 1, this ellipse is taller than it is wide, so its major axis lies along y.

  1. Hyperbola eccentricity. Divide by 10: x²/10 - y²/(10 cos² θ) = 1. So eₕ² = 1 + (10 cos² θ)/10 = 1 + cos² θ.
  2. Ellipse eccentricity. Divide by 5: x²/(5 cos² θ) + y²/5 = 1. Since 5 cos² θ < 5, the semi-major axis is √5 along y and the semi-minor axis is √5 cos θ. So eₑ² = 1 - cos² θ = sin² θ.

    eₕ² = 5eₑ² ⇒ 1 + cos² θ = 5 - 5 cos² θ ⇒ cos² θ = 2/3

  3. Latus rectum. The length is (2 × (semi-minor)²)/(semi-major) = (2 × 5 cos² θ)/√5 = 2√5 × 2/3.
  4. So. The latus rectum of the ellipse is 4√5/3, which is option (c).

The idea to keep. Before using 2b²/a, check which denominator is bigger. The latus rectum is twice the square of the semi-minor axis divided by the semi-major axis, whichever axis it lies on.

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

Study Ellipse

Asked about 100 times in JEE Main since 2020. This question is from 2020.