The locus of P(x,y) such that (√(x^2 +y^2 +8y+16))−(√(x^2 +y^2 −6x+9))=5 is?


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Question Number 48143 by rahul 19 last updated on 20/Nov/18

$T h e l o c u s o f P (x, y) s u c h t h a t$ $\sqrt{x^{2} + y^{2} + 8 y + 16} - \sqrt{x^{2} + y^{2} - 6 x + 9} = 5 i s ?$
Commented by rahul 19 last updated on 20/Nov/18

$I^{'} m g e t t i n g p a r a b o l a b u t a n s . i s i n f i n i t e$ $r a y .$
	Answered by MJS last updated on 20/Nov/18

	$\sqrt{A} - \sqrt{B} = 5$ $A + B - 2 \sqrt{A} \sqrt{B} = 25$ $2 \sqrt{A} \sqrt{B} = A + B - 25$ $4 A B = A^{2} + 2 A B + B^{2} - 50 A - 50 B + 625$ $A^{2} - 2 A B + B^{2} - 50 A - 50 B + 625 = 0$ $in our case$ $16 x^{2} - 24 x y + 9 y^{2} - 96 x + 72 y + 144 = 0$ ${(4 x - 3 y - 12)}^{2} = 0$ $\Rightarrow y = \frac{4}{3} x - 4$
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