e1=1−25b2 and e2=1+16b2
Given e1e2=1
⇒(e1e2)2=1⇒(1−25b2)(1+16b2)=1
⇒16b2−25b2−25×16b4=0
⇒16×259b2−25×16b4=0⇒b2=9
Thus, e1=1−259=54 and e2=1+169=45
Therefore, α=2(5)(e1)=8 and β=2(4)(e2)=10
∴(α,β)=(8,10)
Let e1 and e2 be the eccentricities of the ellipse 25x2+b2y2=1(b<5) and the hyperbola 16x2−b2y2=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:
Held on 3 Sept 2020 · Verified 6 Jul 2026.
(8,10)
(320,12)
(8,12)
(524,10)
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