Let e 1 and e 2 be the eccentricities of the ellipse x 2 25 + y 2 b 2 = 1   b < 5 and the hyperbola…

Let e1 and e2 be the eccentricities of the ellipse x225+y2b2=1 b<5 and the hyperbola x216-y2b2=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:
  1. 8, 10
  2. 203, 12
  3. 8, 12
  4. 245, 10

Solution

e1=1-b225 and e2=1+b216

Given e1e2=1

e1e22=11-b2251+b216=1

b216-b225-b425×16=0

916×25b2-b425×16=0b2=9

Thus, e1=1-925=45 and e2=1+916=54

Therefore, α=25e1=8 and β=24e2=10

α, β=8, 10

Asked in: JEE Main 2020 (03 Sep Shift 2)

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