JEE Main 2020 · 7 January · morning · Single correct · +4 / −1

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is :

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

Answer (b) 3√2

Solution

Maths sir · 4 steps

Multiply ae by a/e and the e cancels, so a² comes straight out of the two given distances.

  1. Read the data. The foci are at (± ae, 0), so 2ae = 6 and ae = 3. The directrices are x = ± a/e, so 2a/e = 12 and a/e = 6.
  2. Find a and e. Multiplying gives ae × a/e = a² = 18, so a = 3√2. Dividing gives e² = 3/6 = 1/2.
  3. Find b squared. b² = a²(1 - e²) = 18 × 1/2 = 9.

    Latus rectum = 2b²/a = (2 × 9)/3√2 = 6/√2 = 3√2

  4. So. The length of the latus rectum is 3√2, option (b).

The idea to keep. Focal distance gives ae, directrix distance gives a/e. Their product is a² and their ratio is e². Then b² = a²(1-e²) and the latus rectum is 2b²/a.

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

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