Let the length of a latus rectum of an ellipse x²/a² + y²/b² = 1 be 10. If its eccentricity is the minimum value of the function f(t) = t² + t + 11/12, t ∈ ℝ, then a² + b² is equal to :

Solution

Maths sir · 4 steps

Get e from the vertex of the quadratic, then the latus rectum turns b² into 5a.

1. Find the eccentricity. The minimum of t² + t + 11/12 is at the vertex t = -1/2. There f = 1/4 - 1/2 + 11/12 = (3 - 6 + 11)/12 = 2/3, so e = 2/3.
2. Use the latus rectum. For this ellipse with a > b, the latus rectum has length 2b²/a = 10. So b² = 5a.

   e² = 1 - b²/a² ⇒ 4/9 = 1 - 5a/a² ⇒ 5/a = 5/9

3. Find a and b. This gives a = 9, so a² = 81 and b² = 5 × 9 = 45.
4. So. a² + b² = 81 + 45 = 126, which is option (d).

The idea to keep. For an ellipse, the latus rectum 2b²/a and the relation b² = a²(1 - e²) are your two equations. Write b² in terms of a first, then substitute.

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