Ellipse

For an ellipse x^2/a^2 + y^2/b^2 = 1 with a > b, the eccentricity satisfies b^2 = a^2(1 - e^2) and the foci lie at (±ae, 0); identifying which denominator is larger fixes the orientation. Here a^2 = 25 and b^2 = 16, so a = 5 > b = 4 and the major axis is horizontal. Then e^2 = 1 - b^2/a^2 = 1 - 16/25 = 9/25, giving e = 3/5. The foci are at (±ae, 0) = (±5 · 3/5, 0) = (±3, 0). Option e = 4/5 misuses b/a as the eccentricity. Option foci (0, ±3) wrongly orients the major axis vertically. Option e = 1/5 comes from subtracting under the wrong fraction. This is the standard JEE Advanced ellipse parameter extraction. Plausibility check: since 0 < e = 3/5 < 1 the curve is a proper ellipse, and c = ae = 3 satisfies c^2 = a^2 - b^2 = 25 - 16 = 9, fully consistent with the focal distance.