Let O(0,0) and A(0,1) be two fixed points. Then the locus of a point P such that the perimeter of ΔAOP is

Correct option is B.
9x² + 8y² - 8y = 16
AP + OP + AO = 4
√(h² + (k - 1)²) + √(h² + k²) + 1 = 4
√(h² + (k - 1)²) + √(h² + k²) = 3
√(h² + (k - 1)²) = 3 - √(h² + k²)
Squaring both sides:
h² + (k - 1)² = 9 + h² + k² - 6√(h² + k²)
h² + k² - 2k + 1 = 9 + h² + k² - 6√(h² + k²)
-2k - 8 = -6√(h² + k²)
2k + 8 = 6√(h² + k²)
k + 4 = 3√(h² + k²)
(k + 4)² = 9(h² + k²)
k² + 8k + 16 = 9h² + 9k²
9h² + 8k² - 8k - 16 = 0
Locus of P is 9x² + 8y² - 8y - 16 = 0