Let H₁: x²/a²-y²/b²=1 and H₂:-x²/A²+y²/B²=1 be two hyperbolas having length of latus rectums 15 √2 and 12 √5 respectively. Let their ecentricities be e₁=√(5/2) and e₂ respectively. If the product of the lengths of their transverse axes is 100 √10, then 25 e₂² is equal to _________ .

Solution

Maths sir · 4 steps

For H₂ the transverse axis is vertical, so B takes the role that a plays in H₁.

1. Solve H1. e₁² = 1 + b²/a² = 5/2 gives b² = 3/2a². Then the latus rectum 2b²/a = 3a = 15√2, so a = 5√2 and the transverse axis is 2a = 10√2.
2. Transverse axis of H2. H₂ opens along the y-axis, so its transverse axis is 2B. From 10√2 × 2B = 100√10 we get 2B = 10√5, so B = 5√5 and B² = 125.
3. Latus rectum of H2. Here the latus rectum is 2A²/B = 12√5. So A² = 6√5 × 5√5 = 150.

   e₂² = 1 + A²/B² = 1 + 150/125 = 11/5

4. So. 25e₂² = 25 × 11/5 = 55, so the answer is 55.

The idea to keep. In a conjugate-type hyperbola -x²/A²+y²/B²=1, swap the roles: transverse axis 2B, latus rectum 2A²/B, and e² = 1 + A²/B².

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