Ellipse
Given the ellipse with equation x^2/25 + y^2/16 = 1, calculate the eccentricity together with the coordinates of its two foci on the major axis.
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About This Question
- Subject
- mathematics
- Chapter
- coordinate geometry
- Topic
- ellipse
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
e=3/5,foci(±3,0)
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.
This medium difficulty mathematics question is from the chapter coordinate geometry, covering the topic of ellipse. It appeared in the 2025 exam.
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