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Jee Main 2025 — Medium JEE math MCQ

Let \( \frac{x^2}{16} + \frac{y^2}{25} = 1 \), \( z \in C \), be the equation of a circle with center at \( C \). If the area of the triangle, whose vertices are at the points \( (0, 0) \), \( C \) and \( (\alpha, 0) \) is 11 square units, then \( \alpha^2 \) equals:
  1. A. 50 -
  2. B. 100 -
  3. C. \( \frac{81}{25} \) -
  4. D. \( \frac{121}{25} \)

Solution

The correct option is **B**. (B. 100 -)

MATH

medium • PYQ Reworded
Question
Read carefully, then pick the best option.
Let x216+y225=1 \frac{x^2}{16} + \frac{y^2}{25} = 1 , z∈C z \in C , be the equation of a circle with center at C C . If the area of the triangle, whose vertices are at the points (0,0) (0, 0) , C C and (α,0) (\alpha, 0) is 11 square units, then α2 \alpha^2 equals:
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Jee Main 2025 — Medium JEE Mathematics MCQ | MyGoalPrep