Skip to content

Ellipse

Hardmathematics

An ellipse has foci separated by a distance of 6 units and a sum of focal radii equal to 10 for every point; what is its standard equation centred at the origin?

Select the correct option:

🔒 Solution Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

About This Question

Subject
mathematics
Chapter
coordinate geometry
Topic
ellipse
Difficulty
Hard
Year
2025
Tags
advanced-calculus-drillellipsefocal radiidefinitionstandard equation

Solution

Correct Answer:

An ellipse is defined by a constant sum of distances to two foci equal to 2a, while the focal separation is 2c, and the semi-minor axis follows from b^2 = a^2 - c^2. The sum of focal radii is 10, so 2a = 10 and a = 5, giving a^2 = 25. The foci are 6 apart, so 2c = 6 and c = 3. Then b^2 = a^2 - c^2 = 25 - 9 = 16. With foci on the x-axis, the standard equation is x^2/25 + y^2/16 = 1. Option x^2/25 + y^2/9 = 1 wrongly sets b^2 = c^2. Option x^2/16 + y^2/25 = 1 places the major axis vertically, contradicting horizontal foci. Option x^2/100 + y^2/64 = 1 doubles the axes by misreading 2a as a. This is the standard JEE Advanced definition-to-equation construction. Plausibility check: eccentricity e = c/a = 3/5 lies strictly between 0 and 1, and b^2 = 16 is positive, confirming a genuine, properly oriented ellipse.

This hard difficulty mathematics question is from the chapter coordinate geometry, covering the topic of ellipse. It appeared in the 2025 exam.

Looking for more practice? Explore all mathematics questions or browse coordinate geometry questions on RankGuru.