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Ellipse: JEE Main Previous Year Questions (2021-2026)
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Page 1 JEE Main Previous Year Questions (2021-2026): Ellipse (January 2026) Q1: An ellipse has its center at (1, -2), one focus at (3, -2) and one vertex at (5, -2). Then the length of its latus rectum is: A: 6 B: 6v3 C: 16/v3 D: 4v3 Answer: A Explanation: Given Center of ellipse : (1, -2) Focus : (3, -2) Vertex : (5, -2) The center, focus, and vertex all have the same y-coordinate (-2), which means the major axis of the ellipse is horizontal (parallel to the x-axis). Find a (semi-major axis) Distance from center to vertex : a = |5 - 1| = 4 Page 2 JEE Main Previous Year Questions (2021-2026): Ellipse (January 2026) Q1: An ellipse has its center at (1, -2), one focus at (3, -2) and one vertex at (5, -2). Then the length of its latus rectum is: A: 6 B: 6v3 C: 16/v3 D: 4v3 Answer: A Explanation: Given Center of ellipse : (1, -2) Focus : (3, -2) Vertex : (5, -2) The center, focus, and vertex all have the same y-coordinate (-2), which means the major axis of the ellipse is horizontal (parallel to the x-axis). Find a (semi-major axis) Distance from center to vertex : a = |5 - 1| = 4 Find c (distance from center to focus) c = |3 - 1| = 2 Find b using ellipse relation c² = a² - b² ? 2² = 4² - b² ? 4 = 16 - b² ? b² = 12 ? b = 2v3 Length of latus rectum For an ellipse : Length of latus rectum = 2b²/a = 2(12)/4 = 6 Q2: Let the length of the latus rectum of an ellipse x²/a² + y²/b² = 1, (a > b), be 30. If its eccentricity is the maximum value of the function is equal to A: 276 B: 516 C: 256 D: 496 Answer: D Explanation: Page 3 JEE Main Previous Year Questions (2021-2026): Ellipse (January 2026) Q1: An ellipse has its center at (1, -2), one focus at (3, -2) and one vertex at (5, -2). Then the length of its latus rectum is: A: 6 B: 6v3 C: 16/v3 D: 4v3 Answer: A Explanation: Given Center of ellipse : (1, -2) Focus : (3, -2) Vertex : (5, -2) The center, focus, and vertex all have the same y-coordinate (-2), which means the major axis of the ellipse is horizontal (parallel to the x-axis). Find a (semi-major axis) Distance from center to vertex : a = |5 - 1| = 4 Find c (distance from center to focus) c = |3 - 1| = 2 Find b using ellipse relation c² = a² - b² ? 2² = 4² - b² ? 4 = 16 - b² ? b² = 12 ? b = 2v3 Length of latus rectum For an ellipse : Length of latus rectum = 2b²/a = 2(12)/4 = 6 Q2: Let the length of the latus rectum of an ellipse x²/a² + y²/b² = 1, (a > b), be 30. If its eccentricity is the maximum value of the function is equal to A: 276 B: 516 C: 256 D: 496 Answer: D Explanation: = (16)² + 15 × 16 = 256 + 240 = 496 Q3: Let each of the two ellipses E 1 : x²/a² + y²/b² = 1, (a > b) and E 2 : x²/A² + y²/B² = 1, (A < B) have eccentricity 4/5. Let the lengths of the latus recta of E 1 and E 2 be l 1 and l 2, respectively, such that 2l 1² = 9l 2. If the distance between the foci of E 1 is 8, then the distance between the foci of E 2 is A: 96/5 B: 8/5 C: 16/5 D: 32/5 Page 4 JEE Main Previous Year Questions (2021-2026): Ellipse (January 2026) Q1: An ellipse has its center at (1, -2), one focus at (3, -2) and one vertex at (5, -2). Then the length of its latus rectum is: A: 6 B: 6v3 C: 16/v3 D: 4v3 Answer: A Explanation: Given Center of ellipse : (1, -2) Focus : (3, -2) Vertex : (5, -2) The center, focus, and vertex all have the same y-coordinate (-2), which means the major axis of the ellipse is horizontal (parallel to the x-axis). Find a (semi-major axis) Distance from center to vertex : a = |5 - 1| = 4 Find c (distance from center to focus) c = |3 - 1| = 2 Find b using ellipse relation c² = a² - b² ? 2² = 4² - b² ? 4 = 16 - b² ? b² = 12 ? b = 2v3 Length of latus rectum For an ellipse : Length of latus rectum = 2b²/a = 2(12)/4 = 6 Q2: Let the length of the latus rectum of an ellipse x²/a² + y²/b² = 1, (a > b), be 30. If its eccentricity is the maximum value of the function is equal to A: 276 B: 516 C: 256 D: 496 Answer: D Explanation: = (16)² + 15 × 16 = 256 + 240 = 496 Q3: Let each of the two ellipses E 1 : x²/a² + y²/b² = 1, (a > b) and E 2 : x²/A² + y²/B² = 1, (A < B) have eccentricity 4/5. Let the lengths of the latus recta of E 1 and E 2 be l 1 and l 2, respectively, such that 2l 1² = 9l 2. If the distance between the foci of E 1 is 8, then the distance between the foci of E 2 is A: 96/5 B: 8/5 C: 16/5 D: 32/5 Answer: D Explanation: Given, for ellipse E 1 and E 2 eccentricity is 4/5. E 1 : x²/a² + y²/b² = 1 a > b distance between foci = 8 2ae = 8 ? ae = 4 ? a × 4/5 = 4 ? a = 5 Length of latus rectum for E 1, l 1 = 2b²/a l 1 = 2 × 3²/5 = 18/5 For E 2 : x²/A² + y²/B² = 1 (A < B) eccentricity e = 4/5 Page 5 JEE Main Previous Year Questions (2021-2026): Ellipse (January 2026) Q1: An ellipse has its center at (1, -2), one focus at (3, -2) and one vertex at (5, -2). Then the length of its latus rectum is: A: 6 B: 6v3 C: 16/v3 D: 4v3 Answer: A Explanation: Given Center of ellipse : (1, -2) Focus : (3, -2) Vertex : (5, -2) The center, focus, and vertex all have the same y-coordinate (-2), which means the major axis of the ellipse is horizontal (parallel to the x-axis). Find a (semi-major axis) Distance from center to vertex : a = |5 - 1| = 4 Find c (distance from center to focus) c = |3 - 1| = 2 Find b using ellipse relation c² = a² - b² ? 2² = 4² - b² ? 4 = 16 - b² ? b² = 12 ? b = 2v3 Length of latus rectum For an ellipse : Length of latus rectum = 2b²/a = 2(12)/4 = 6 Q2: Let the length of the latus rectum of an ellipse x²/a² + y²/b² = 1, (a > b), be 30. If its eccentricity is the maximum value of the function is equal to A: 276 B: 516 C: 256 D: 496 Answer: D Explanation: = (16)² + 15 × 16 = 256 + 240 = 496 Q3: Let each of the two ellipses E 1 : x²/a² + y²/b² = 1, (a > b) and E 2 : x²/A² + y²/B² = 1, (A < B) have eccentricity 4/5. Let the lengths of the latus recta of E 1 and E 2 be l 1 and l 2, respectively, such that 2l 1² = 9l 2. If the distance between the foci of E 1 is 8, then the distance between the foci of E 2 is A: 96/5 B: 8/5 C: 16/5 D: 32/5 Answer: D Explanation: Given, for ellipse E 1 and E 2 eccentricity is 4/5. E 1 : x²/a² + y²/b² = 1 a > b distance between foci = 8 2ae = 8 ? ae = 4 ? a × 4/5 = 4 ? a = 5 Length of latus rectum for E 1, l 1 = 2b²/a l 1 = 2 × 3²/5 = 18/5 For E 2 : x²/A² + y²/B² = 1 (A < B) eccentricity e = 4/5 Q4: If the points of intersection of the ellipses x² + 2y² - 6x - 12y + 23 = 0 and 4x² + 2y² - 20x - 12y + 35 = 0 lie on a circle of radius r and centre (a, b), then the value of ab + 18r² is: A: 53 B: 52 C: 55 D: 51 Answer: C Explanation: Let the given ellipses be : S 1 = x² + 2y² - 6x - 12y + 23 = 0 S 2 = 4x² + 2y² - 20x - 12y + 35 = 0 The equation of a curve passing through the intersection of these two ellipses is given by the family of curves S 1 + ?S 2 = 0 : (x² + 2y² - 6x - 12y + 23) + ?(4x² + 2y² - 20x - 12y + 35) = 0Read More
FAQs on Ellipse: JEE Main Previous Year Questions (2021-2026)
| 1. What are the key formulas and properties of ellipse that appear repeatedly in JEE Main exams? |  |
Ans. Standard ellipse equation is x²/a² + y²/b² = 1, where a and b are semi-major and semi-minor axes. Critical formulas include eccentricity (e = √(1 - b²/a²)), focal distance (2ae), and latus rectum (2b²/a). Understanding parametric forms and tangent-normal equations proves essential for solving JEE Main previous year questions on ellipse effectively.
| 2. How do I identify whether a point lies inside, on, or outside an ellipse in JEE problems? | |
Ans. Substitute the point's coordinates into the ellipse equation x²/a² + y²/b² = 1. If the result equals 1, the point is on the ellipse; if less than 1, it's inside; if greater than 1, it's outside. This fundamental concept appears frequently across JEE Main question papers and helps solve intersection and position-based problems quickly.
| 3. What's the difference between finding the equation of tangent to an ellipse at a given point versus from an external point? | |
Ans. For tangents at a given point on the ellipse, use xx₁/a² + yy₁/b² = 1. For tangents from external points, apply the condition that the line touches the ellipse, creating a quadratic with discriminant zero. JEE Main exams test both approaches; mastering each method is crucial for maximising marks in ellipse problems.
| 4. Why do eccentricity values and focal chord calculations cause confusion in ellipse exam questions? | |
Ans. Eccentricity determines the ellipse's shape; students often confuse it with the relationship between semi-major and semi-minor axes. Focal chord length varies based on position, making calculations error-prone. Practising JEE Main previous year questions clarifies these relationships and builds confidence with standard results for focal chords and auxiliary circles.
| 5. How can I quickly solve problems involving auxiliary circles and director circles in ellipse questions? | |
Ans. The auxiliary circle has equation x² + y² = a², while the director circle is x² + y² = a² + b². These geometric properties simplify problems involving normals, chords, and pole-polar relationships. Familiarity with these standard results from JEE Main solved examples reduces calculation time and improves accuracy in competitive exams significantly.
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