Locus of the point z satisfying the equation ∣iz − 1∣ + ∣z − i∣ = 2 is


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Question Number 19738 by Tinkutara last updated on 15/Aug/17

$Locus of the point z satisfying the$ $equation ∣ i z - 1 ∣ + ∣ z - i ∣ = 2 is$
Commented by math khazana by abdo last updated on 22/Jun/18
$let z =x+iy (e) ⇔∣ix−y−1∣ +∣x+i(y−1)∣=2 ⇒(√(x^2 +(y+1)^2 )) +(√(x^2 +(y−1)^2 ))=2 ⇒ (√(x^2 +(y+1)^2 )) =(2−(√(x^2 +(y−1)^2 ))) ⇒ x^2 +(y+1)^2 =4 −4(√(x^2 +(y−1)^2 )) +x^2 +(y−1)^2 ⇒ 4(√(x^2 +(y−1)^2 ))=(y−1)^2 −(y+1)^2 +4 ⇒ 4(√(x^2 +(y−1)^2 ))=y^2 −2y +1 −y^2 −2y −1 +4 =4 −4y ⇒ x^2 +(y−1)^2 =(1−y)^2 ⇒ x=0 ⇒ {z∈C/∣iz−∣+∣z−i∣=2}=iR .$
$l e t z = x + i y (e) \Leftrightarrow∣ i x - y - 1 ∣ + ∣ x + i (y - 1) ∣= 2$ $\Rightarrow \sqrt{x^{2} + {(y + 1)}^{2}} + \sqrt{x^{2} + {(y - 1)}^{2}} = 2 \Rightarrow$ $\sqrt{x^{2} + {(y + 1)}^{2}} = (2 - \sqrt{x^{2} + {(y - 1)}^{2}}) \Rightarrow$ $x^{2} + {(y + 1)}^{2} = 4 - 4 \sqrt{x^{2} + {(y - 1)}^{2}} + x^{2} + {(y - 1)}^{2} \Rightarrow$ $4 \sqrt{x^{2} + {(y - 1)}^{2}} = {(y - 1)}^{2} - {(y + 1)}^{2} + 4 \Rightarrow$ $4 \sqrt{x^{2} + {(y - 1)}^{2}} = y^{2} - 2 y + 1 - y^{2} - 2 y - 1 + 4$ $= 4 - 4 y \Rightarrow x^{2} + {(y - 1)}^{2} = {(1 - y)}^{2} \Rightarrow x = 0 \Rightarrow$ ${z \in C / ∣ i z - ∣ + ∣ z - i ∣= 2} = i R .$
	Answered by ajfour last updated on 15/Aug/17

	$\Rightarrow ∣ z + i ∣ + ∣ z - i ∣= 2$ $ellipse with origin as centre .$ $focii : z_{1} = - i; z_{2} = i$ $shall try yo obtain equation in$ $the form : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . .$ $if equation of ellipse be$ $∣ z - z_{1} ∣ + ∣ z - z_{2} ∣= d$ $with z_{1}, and z_{2} being focii$ $then 2 a = d$ $and b^{2} = a^{2} - \frac{∣ z_{1} - z_{2} ∣^{2}}{4}$ $so eccentricity e = \frac{∣ z_{1} - z_{2} ∣}{d} .$
	Commented by ajfour last updated on 16/Aug/17

	$H e r e m a j o r a x i s i s t h e I m a g i n a r y$ $a x i s, s o 2 b = d a n d a^{2} = b^{2} - \frac{∣ z_{1} - z_{2} ∣^{2}}{4}$ $d = 2, s o b = 1 z_{1} = i, z_{2} = - i$ $s o a^{2} = 1 - \frac{∣ 2 i ∣^{2}}{4} = 0$ $h e n c e e l l i p s e w i t h m i n o r a x i s$ $l e n g t h z e r o a n d m a j o r a x i s l e n g t h$ $e q u a l t o 2 . e x t e n d s f r o m z = - i t o$ $z = i, a d o u b l e l i n e s e g m e n t .$ $(s p e c i a l c a s e o f e l l i p s e) .$
	Commented by Tinkutara last updated on 16/Aug/17

	$Sorry, I checked the answer today and$ $it was given a straight line!$
	Commented by Tinkutara last updated on 16/Aug/17

	$So we will consider it ellipse or a line ?$
	Commented by ajfour last updated on 16/Aug/17

	$c a n t s a y!$
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