Given the points A(0, 4) and B(0, -4) the equation of the locus of the point P(x, y) such that | AP - BP | = 6 is

9 x^{2} - 7 y^{2} + 63 = 0

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9 x^{2} - 7 y^{2} - 63 = 0

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7 x^{2} - 9 y^{2} + 63 = 0

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7 x^{2} - 9 y^{2} - 63 = 0

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Solution

The correct option is C $9 x^{2} - 7 y^{2} + 63 = 0$
$G i v e n - P (x, y) i s a p o i n t s u c h t h a t t h e d i f f e r e n c e o f i t s d i s t a n c e s f r o m A (0, 4) & B (0, - 4) i s A P - B P = 6. T o o b t a i n - t h e e q u a t i o n o f t h e l o c u s o f P (x, y) = ? S o l u t i o n - W e s h a l l u s e d i s t a n c e f o r m u l a d = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} t o o b t a i n A P, B P . A P = \sqrt{{(x - 0)}^{2} + {(y - 4)}^{2}} = a n d B P = \sqrt{{(x - 0)}^{2} + {(y + 4)}^{2}} . N o w A P - B P = 6 \sqrt{{(x - 0)}^{2} + {(y - 4)}^{2}} - \sqrt{{(x - 0)}^{2} + {(y + 4)}^{2}} = 6 ⟹ \sqrt{{(x - 0)}^{2} + {(y - 4)}^{2}} = 6 + \sqrt{{(x - 0)}^{2} + {(y + 4)}^{2}} ⟹ x^{2} + {(y - 4)}^{2} = 36 + x^{2} + {(y + 4)}^{2} + 12 \sqrt{{(x - 0)}^{2} + {(y + 4)}^{2}} ⟹ - 16 y - 36 = 12 \sqrt{{(x - 0)}^{2} + {(y + 4)}^{2}} ⟹ - 4 y - 9 = 3 \sqrt{{(x - 0)}^{2} + {(y + 4)}^{2}} ⟹ 16 y^{2} + 72 y + 81 = 9 x^{2} + 9 y^{2} + 72 y + 144 ⟹ 9 x^{2} - 7 y^{2} + 63 = 0. T h i s i s t h e l o c u s o f P . A n s - O p t i o n A .$

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