1. If (√((x+2)^(2 ) +y^2 ))+(√((x−2)^2 +y^2 ))=6 show that when the equation is simplified, it can be express as (x^2 /9)+(y^2 /5)=1 2. find the value of n such that the linear factors of the form x+ay+b and x+cy+d with integer coefficients have the product x^2 +5xy+x+ny−n sir please help


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Question Number 53500 by Otchere Abdullai last updated on 22/Jan/19

$1 . I f \sqrt{{(x + 2)}^{2} + y^{2}} + \sqrt{{(x - 2)}^{2} + y^{2}} = 6$ $s h o w t h a t w h e n t h e e q u a t i o n i s$ $s i m p l i f i e d, i t c a n b e e x p r e s s a s$ $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ $2 . f i n d t h e v a l u e o f n s u c h t h a t t h e$ $l i n e a r f a c t o r s o f t h e f o r m x + a y + b$ $a n d x + c y + d w i t h i n t e g e r c o e f f i c i e n t s$ $h a v e t h e p r o d u c t x^{2} + 5 x y + x + n y - n$ $s i r p l e a s e h e l p$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

	$\sqrt{{(x + 2)}^{2} + y^{2}} = 6 - \sqrt{{(x - 2)}^{2} + y^{2}}$ $x^{2} + 4 x + 4 + y^{2} = 36 - 12 \sqrt{{(x - 2)}^{2} + y^{2}} + x^{2} - 4 x + 4 + y^{2}$ $c a n c e l l i n g x^{2}, y^{2}, 4 f r o m b o t h s i d e$ $4 x = 36 - 12 \sqrt{x^{2} - 4 x + 4 + y^{2}} - 4 x$ $8 x - 36 = - 12 \sqrt{x^{2} - 4 x + 4 + y^{2}}$ $2 x - 9 = - 3 \sqrt{x^{2} - 4 x + 4 + y^{2}}$ $4 x^{2} - 36 x + 81 = 9 x^{2} - 36 x + 36 + 9 y^{2}$ $5 x^{2} + 9 y^{2} = 45$ $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$
	Commented by Otchere Abdullai last updated on 22/Jan/19

	$w o w! t h a n k y o u s i r G o d b l e s s y o u$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

	$2) (x + a y + b) (x + c y + d) = x^{2} + 5 x y + x + n y - n$ $x^{2} + c x y + x d + a x y + {a c y}^{2} + a y d + b x + b c y + b d = x^{2} + 5 x y + x + n y - n$ $x^{2} + x y (a + c) + x (b + d) + y (a d + b c) + {a c y}^{2} + b d = x^{2} + 5 x y + x + n y - n$ $s o$ $a + c = 5$ $b + d = 1$ $a d + b c = n$ $a c = 0$ $b d = - n$ $a c = 0$ $a (5 - a) = 0 e i t h e r a = 0 o r a = 5$ $w h e n a = 0 c = 5 a n d w h e n a = 5 c = 0$ $n o w l e t a = 0 c . = 5$ $a d + b c = n$ $0 \times d + b \times 5 = n [b = \frac{n}{5}]$ $a = 0$ $b = \frac{n}{5}$ $c = 5$ $d = 1 - b = 1 - \frac{n}{5}$ $b d = - n$ $(\frac{n}{5}) \times (1 - \frac{n}{5}) = - n$ $\frac{1}{5} \times (\frac{5 - n}{5}) = - 1$ $5 - n = - 25$ $- n = - 30 . . . n = 30$
	Commented by Otchere Abdullai last updated on 22/Jan/19

	$p o w e r f u l M a t h e m a t i c i a n! W e t h a n k$ $G o d f o r h a v i n g y o u i n t h e g r o u p$ $G o d b l e s s y o u s i r!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19

	$m a t h e m a t i c s . . . a c t i v a t e s o n e s m i n d t o t h i n k . . .$
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