Question

Let \( O (0,0) \) and \( A (0,1) \) be two fixed points. Then the locus of a point \( P \) such that the perimeter of \( \triangle A O P \) is 4, is :

(a) \( 8 x^{2}-9 y^{2}+9 y=18 \)

(b) \( 9 x^{2}-8 y^{2}+8 y=16 \)

(c) \( 9 x^{2}+8 y^{2}-8 y=16 \)

(d) \( 8 x^{2}+9 y^{2}-9 y=18 \)

Answer 1

The correct option is (C) 9x² + 8y² – 8y = 16

Squaring both the sides,

k² + 16 + 8k = 9h² + 9k²

9h² + 8k² – 8k – 16 = 0

Thus 9x² + 8y² – 8y – 16 = 0