prove that the curve (√((x−1)^2 +y^2 ))+(√((x+1)^2 +y^2 ))=4 is an ellipse and find its semi major axis and semi minor axis.


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Question Number 196723 by mr W last updated on 30/Aug/23

$p r o v e t h a t t h e c u r v e$ $\sqrt{{(x - 1)}^{2} + y^{2}} + \sqrt{{(x + 1)}^{2} + y^{2}} = 4$ $i s a n e l l i p s e a n d f i n d i t s s e m i$ $m a j o r a x i s a n d s e m i m i n o r a x i s .$
	Answered by Frix last updated on 30/Aug/23

	$Squaring, transforming etc . leads to$ $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$ $Testing with y^{2} = \frac{12 - 3 x^{2}}{4} in the given equation$ $gives$ $∣ x + 4 ∣ + ∣ x - 4 ∣= 8; true for - 4 ⩽ x ⩽ 4$
	Commented by mr W last updated on 30/Aug/23
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	Answered by mr W last updated on 30/Aug/23

	$v = \sqrt{{(x - 1)}^{2} + y^{2}}$ $u = \sqrt{{(x + 1)}^{2} + y^{2}}$ $u + v = 4$ $u^{2} + v^{2} = 2 (x^{2} + y^{2} + 1)$ $u^{2} - v^{2} = 4 x$ $\Rightarrow u - v = x$ $\Rightarrow u = \frac{4 + x}{2}$ $\Rightarrow v = \frac{4 - x}{2}$ $u^{2} + v^{2} = \frac{2 (16 + x^{2})}{4} = 2 (x^{2} + y^{2} + 1)$ $3 x^{2} + 4 y^{2} = 12$ $\Rightarrow \frac{x^{2}}{2^{2}} + \frac{y^{2}}{{(\sqrt{3})}^{2}} = 1$ $\Rightarrow e l l i p s e w i t h a = 2, b = \sqrt{3}$
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