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Coordinate Geometry
Ellipse, Eccentricity
jee_main_2026_april_8_shift_1
Grade None

Question:

The eccentricity of the ellipse x²/49 + y²/33 = 1 is:
A. 4/7
B. 5/7
C. 6/7
D. 7/4

Step-by-Step Solution

Key Concept: Ellipse eccentricity: e = √(1 - b²/a²).
Step 1: a² = 49, b² = 33. Step 2: e = √(1 - 33/49) = √(16/49) = 4/7.
Correct Answer: A
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