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1)

The foci of ellipse  x216+y2b2=1 and the hyperbola  x2144y281=125 coincide then value of b2 is 


A) 1

B) 5

C) 7

D) 9

Answer:

Option C

Explanation:

Given Ellipse   x216+y2b2=1 

Now b2=a2(1e2)

 b2=16(1e2),b216=1e2

e2=1b216=16b216

e=16b24

  Foci =  (±ae,0)=(±16b2,0)

 Given hyperbola : x214+y281=125 

 x2(125)2y2(95)2=1

Now,   b2=a2(e21)

(95)2=(125)2(e21)

 (95)2=(e21)

 e2=1+81144=144+81144

 e=1512=54

  Foci =  (±ae,0)= (±3,0)

 Since foci of the given ellipse and hyperbola coincide , therefore

 16b2=316b2=9

   b2 =7