Question

The foci of the ellipse x²/16 + y²/b² = 1 and the hyperbola x²/144 - y² /81 = 1/25 coincide . then the value of b² is

Answer 1

Hi !!!

Solution :-

x² /144 - y²/81 = 1/25 a = √144/25 , b = √81/25 e = ( 1 + 81/144) = 5/4

•°• foci = ( +- 3 , 0)

•°• foci of ellipse = foci of hyperbola

•°• for ellipise e = 3 but a = 4

•°• e = 3/4 THEN , b² = a² ( 1 - e²)

=> b² = 16 ( 1- 9/16 ) = 7 Answer ✔

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Hope it helps !!

Rajukumar111 ❤