NDA Mathematics Question (2026) — Solution

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 ( 4e_1, 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 e_2 of the hyperbola is e_2 = 1 + 81/25 144/25 = 1 + 81 144 = 225 144 = 15 12 = 5 4 = 1.25. The foci of the hyperbola are ( A e_2, 0) = ( 12 5 5 4 , 0 ) = ( 3, 0). Since the foci of the ellipse and the hyperbola coincide, we have 16 - p = 3 16 - p = 9 p = 7. The eccentricity of the ellipse is e_1 = 1 - 7 16 = 9 16 = 3 4 = 0.75. The difference between the eccentricities is e_2 - e_1 = 1.25 - 0.75 = 0.5. Answer: 0 5