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For a point P = (x, y), let the foci be F1 = (3, 0) and F2 = (-3, 0). If the point satisfies 16x² + 25y² = 400, then the value of PF1 + PF2 is
- 8
- 6
- 10
- 12
Correct answer: 10
Solution
Dividing by 400 gives x^2/25 + y^2/16 = 1 with a=5, b=4, c=3, so foci are (+-3,0) as given. For any point on the ellipse PF1+PF2 = 2a = 10.
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