Question
Passage: The foci of the ellipse px^2+16y^2=16p and the foci of the hyperbola 25(81x^2-144y^2)=11664 coincide (assume p Question: What is the value of p ?
NDA
Mathematics
Ellipse
2026
NDA 2026 (Phase 1)
NDA Mathematics Question (2026) — Solution
Options
- A. 7
- B. 3
- C. 7
- D. 9
Answer
C. 7
Step-by-step solution
The equation of the ellipse is x^2 16 + y^2 p = 1. Since p The equation of the hyperbola is 25(81x^2 - 144y^2) = 11664, which can be rewritten as x^2 144/25 - y^2 81/25 = 1. For the hyperbola, a^2 = 144 25 and b^2 = 81 25 . The foci are given by ( a^2 + b^2 , 0) = ( 144 25 + 81 25 , 0 ) = ( 225 25 , 0 ) = ( 3, 0). Since the foci of the ellipse and the hyperbola coincide, we have 16 - p = 3. Squaring both sides, we get 16 - p = 9 p = 7. Answer: 7
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