If the focii of the ellipse x^2 25 + y^2 16 =1 and the hyperbola x^2 4 - y^2 b^2 =1 coincide, then b^2 is equal to

Step-by-step solution

Given, equation of ellipse is x^2 25 + y^2 16 =1 and equation of hyperbola is x^2 4 - y^2 b^2 =1 eccentricity of ellipse aligned & b^2=a^2 (1-e^2 ) & 16=25 (1-e^2 ) & e^2=1- 16 25 = 9 25 & e= 3 5 & aligned Focii of the ellipse =( a e, 0)=( 3,0) which coincide with focii of the hyperbola. Let e₁ be the eccentricity of the hyperbola. array ll & a e₁= 3 & e₁= 3 2 array Now, b^2=a^2 (e₁^2-1 ) array cc & b^2=4 ( 9 4 -1 )=4 5 4 & b^2=5 array