Let x²/a²+y²/b²=1, a>b be an ellipse, whose eccentricity is 1/√2 and the length of the latusrectum is √14. Then the square of the eccentricity of x²/a²-y²/b²=1 is :

Solution

Maths sir · 3 steps

Both eccentricities depend only on b²/a². Find that ratio from the ellipse and reuse it.

1. Ellipse ratio. For the ellipse with a > b, e² = 1 - b²/a². With e² = 1/2, this gives b²/a² = 1/2.
2. Latus rectum check. The latus rectum is 2b²/a = 2/a × a²/2 = a, so a = √14 and b² = 7. This fixes the size of the ellipse, but the ratio b²/a² is all you need.

   e_H² = 1 + b²/a² = 1 + 1/2 = 3/2

3. So. The hyperbola has the same a and b, so the square of its eccentricity is 3/2, which is option (c).

The idea to keep. Ellipse: e² = 1 - b²/a². Hyperbola: e² = 1 + b²/a². When the same a and b appear in both, carry the ratio b²/a² across and skip finding a and b.

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