Let H:x²/a² - y²/b² = 1, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 4(2√2 + √14 ). If the eccentricity H is √11/2, then the value of a² + b² is equal to __________.

Solution

Maths sir · 4 steps

Use the eccentricity to write b in terms of a, and the axes sum then gives a directly.

1. Link b to a. For this hyperbola e² = 1 + b²/a². With e² = 11/4 we get b²/a² = 7/4, so b = √7/2a.
2. Axes sum. The transverse axis has length 2a and the conjugate axis has length 2b. So 2a + 2b = 4(2√2 + √14), which gives a + b = 4√2 + 2√14 = 2√2(2 + √7).

   a(1 + √7/2) = a(2+√7)/2 = 2√2(2+√7) ⇒ a = 4√2

3. Find both squares. a² = 32 and b² = 7/4 × 32 = 56. As a check, a² + b² = a²e² = 32 × 11/4 = 88.
4. So. a² + b² = 32 + 56 = 88.

The common slip: 352. Taking the axis lengths as a and b instead of 2a and 2b doubles a and b, which makes a² + b² four times too big: 352 instead of 88.

The idea to keep. Eccentricity fixes the ratio b/a, so one more condition on the axes pins down both. Remember the axis lengths are 2a and 2b, and a² + b² = a²e².

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