Prove-that-the-locus-of-a-point-which-moves-its-distance-from-the-point-b-0-is-p-times-its-distance-from-the-point-b-0-is-p-2-1-x-2-y-2-b-2-2b-p-2-1-x

Question Number 6716 by Tawakalitu. last updated on 15/Jul/16

P r o v e t h a t t h e l o c u s o f a p o i n t w h i c h m o v e s i t s d i s t a n c e f r o m

t h e p o i n t (- b, 0) i s p t i m e s i t s d i s t a n c e f r o m t h e p o i n t (b, 0) i s

(p^{2} - 1) (x^{2} + y^{2} + b^{2}) - 2 b (p^{2} + 1) x = 0

S h o w t h a t t h i s l o c u s i s a c i r c l e a n d f i n d i t s r a d i u s .

Answered by Yozzii last updated on 16/Jul/16

L e t p o i n t s P, A a n d B h a v e c o o r d i n a t e s (x, y), (- b, 0)

a n d (b, 0) r e s p e c t i v e l y .

T h e i n f o r m a t i o n g i v e s ∣ P A ∣= p ∣ P B ∣ w h e r e p > 0 .

∣ P A ∣= \sqrt{{(x + b)}^{2} + y^{2}} a n d ∣ P B ∣= \sqrt{{(x - b)}^{2} + y^{2}} .

∴ \sqrt{{(x + b)}^{2} + y^{2}} = p \sqrt{{(x - b)}^{2} + y^{2}}

\Rightarrow {(\sqrt{{(x + b)}^{2} + y^{2}})}^{2} = {(p \sqrt{{(x - b)}^{2} + y^{2}})}^{2}

{(x + b)}^{2} + y^{2} = p^{2} {(x - b)}^{2} + p^{2} y^{2}

x^{2} + 2 b x + b^{2} + y^{2} = p^{2} x^{2} - 2 p^{2} b x + p^{2} b^{2} + p^{2} y^{2}

(p^{2} - 1) x^{2} + (p^{2} - 1) y^{2} + (p^{2} - 1) b^{2} - 2 b x (1 + p^{2}) = 0

(p^{2} - 1) (x^{2} + y^{2} + b^{2}) - 2 b (p^{2} + 1) x = 0 \dots . (1)

E q u a t i o n (1) r e p r e s e n t s t h e l o c u s o f P

i n r e c t a n g u l a r c o o r d i n a t e s .

- - - - - - - - - - - - - - - - - - - - - - - -

I f p \neq \pm 1, w e c a n r e w r i t e (1) a s

x^{2} + y^{2} + b^{2} - 2 (\frac{b (p^{2} + 1)}{p^{2} - 1}) x = 0

x^{2} - 2 (\frac{b (p^{2} + 1)}{p^{2} - 1}) x + {(\frac{b (p^{2} + 1)}{p^{2} - 1})}^{2} - {(\frac{b (p^{2} + 1)}{p^{2} - 1})}^{2} + {(y - 0)}^{2} = 0

{(x - \frac{b (p^{2} + 1)}{p^{2} - 1})}^{2} + {(y - 0)}^{2} = {(\frac{b (p^{2} + 1)}{p^{2} - 1})}^{2}

\Rightarrow \sqrt{{(x - \frac{b (p^{2} + 1)}{p^{2} - 1})}^{2} + {(y - 0)}^{2}} =∣ \frac{b}{p^{2} - 1} ∣ (p^{2} + 1) \dots . (2)

T h e f o r m o f (2) i m p l i e s t h a t t h e d i s t a n c e

b e t w e e n P (x, y) a n d t h e f i x e d p o i n t (\frac{b (p^{2} + 1)}{p^{2} - 1}, 0)

i s a l w a y s c o n s t a n t a s P m o v e s a n d b, p a r e

c o n s t a n t s . H e n c e, t h e l o c u s o f P i s a

c i r c l e a n d i t s r a d i u s i s ∣ \frac{b}{p^{2} - 1} ∣ (p^{2} + 1) .

Commented by Tawakalitu. last updated on 16/Jul/16

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