Let x²/a² + y²/b² = 1 (a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function,
φ ( t ) = 5/12 + t - t², then a² + b² is equal to :

Solution

Maths sir · 4 steps

Find e from the quadratic's peak, then let the latus rectum fix a; b follows at once.

1. Maximise the quadratic. φ(t) = 5/12 + t - t² is a downward parabola, so its peak is at t = 1/2. The maximum is 5/12 + 1/2 - 1/4 = 8/12 = 2/3, so e = 2/3.
2. Use the latus rectum. For this ellipse the latus rectum is 2b²/a = 10, so b² = 5a.

   b² = a²(1 - e²) = a²(1 - 4/9) = 5a²/9

3. Solve for a. Equate the two expressions: 5a = 5a²/9, so a = 9. Then a² = 81 and b² = 5 × 9 = 45.
4. So. a² + b² = 81 + 45 = 126, which is option (b).

The idea to keep. In an ellipse, the latus rectum 2b²/a and b² = a²(1 - e²) give two expressions for b². Set them equal and a falls out.

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