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

If 3x + 4y = 12√2 is a tangent to the ellipse
x²/a² + y²/9 = 1 for some a ∈ R, then the distance between the foci of the ellipse is :

  1. (a) 2√5
  2. (b) 2√7
  3. (c) 4
  4. (d) 2√2
Answer and solution

Answer (b) 2√7

Solution

Maths sir · 4 steps

Put the line in y = mx + c form, use c² = a²m² + b² to find a, then the foci follow.

  1. Slope form. Rewrite the line as y = -3/4x + 3√2. So m = -3/4 and c = 3√2. Here b² = 9.
  2. Tangency condition. The line y = mx + c touches x²/a² + y²/b² = 1 exactly when c² = a²m² + b².

    18 = a² · 9/16 + 9 ⇒ a² = 16

  3. Locate the foci. Since a² = 16 > b² = 9, the major axis lies along the x-axis. The focus distance from the centre is √(a² - b²) = √(16 - 9) = √7.
  4. So. The foci are at (±√7, 0). The distance between them is 2√7, option (b).

The idea to keep. For a tangent to an ellipse, write the line as y = mx + c and use c² = a²m² + b². Then check which of a² and b² is bigger before finding the foci.

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.