Question

the foci of the ellipse x2/16+y2/b2=1 & the hperbola x2/144-y2/81=1/25 coincide then the value of b2 is​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

For hperbola

x²/144-y²/81=1/25

x²/144 ×25 -y²/81 ×25=1

a²=144/25, b²=81/25

e²=1+b²/a²

e²=1+144/25/81/25

e²=1+144/81

e²=225/144

e=15/12

e 5/4

foci=(±ae,0)

      =(±12/5×5/4,0)

      =(±3,0)

for ellipe

ae=3

a²e²=9

b²=a²(1-e²)

   =16(1-e²)

   =a²-a²e²

   =16-7

    =7