Let e₁ and e₂ be the eccentricities of the ellipse,
x²/25 + y²/b² = 1(b < 5) and the hyperbola,
x²/16 - y²/b² = 1 respectively satisfying e₁e₂ = 1. If α
and β are the distances between the foci of the
ellipse and the foci of the hyperbola
respectively, then the ordered pair (α, β) is equal to :

Solution

Maths sir · 4 steps

Write both eccentricities in terms of b², use e₁e₂ = 1 to find b², then read off the foci.

1. Both eccentricities. For the ellipse, e₁² = 1 - b²/25. For the hyperbola, e₂² = 1 + b²/16. Both contain the same unknown b².

   (1 - b²/25)(1 + b²/16) = 1

2. Solve for b squared. Expand and cancel the 1 on each side: 9b²/400 - b⁴/400 = 0, so b²(9 - b²) = 0. Since b ≠ 0, b² = 9.
3. Find each focal distance. Ellipse: c² = 25 - 9 = 16, so c = 4 and α = 2c = 8. Hyperbola: c² = 16 + 9 = 25, so c = 5 and β = 2c = 10. Check: e₁ e₂ = 4/5 × 5/4 = 1.
4. So. (α, β) = (8, 10), which is option (a).

The idea to keep. Distance between foci is 2ae, which is 2c. For an ellipse c² = a² - b²; for a hyperbola c² = a² + b². Remember the sign flip.

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