Passage: The foci of the ellipse px^2+16y^2=16p and the foci of the hyperbola 25(81x^2-144y^2)=11664 coincide (assume p Question: What is the difference between the eccentricities of the hyperbola and the ellipse ?

Step-by-step solution

The equation of the ellipse can be written as x^2 16 + y^2 p = 1 . Since p The foci of the ellipse are ( 4e₁, 0) = ( 4 1 - p 16 , 0 ) = ( 16 - p , 0) . The equation of the hyperbola is 25(81x^2 - 144y^2) = 11664 . Dividing by 11664 , we get 25x^2 144 - 25y^2 81 = 1 x^2 144/25 - y^2 81/25 = 1 . The eccentricity e₂ of the hyperbola is e₂ = 1 + 81/25 144/25 = 1 + 81 144 = 225 144 = 15 12 = 5 4 = 1.25 . The foci of the hyperbola are ( A e₂, 0) = ( 12 5 5 4 , 0 ) = ( 3, 0) . Since the foci of the ellipse and the hyperbo