Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

$$\mathrm{Locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:{z}\:\mathrm{satisfying}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mid{iz}\:−\:\mathrm{1}\mid\:+\:\mid{z}\:−\:{i}\mid\:=\:\mathrm{2}\:\mathrm{is} \\ $$

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 .

$${let}\:{z}\:={x}+{iy}\:\:\left({e}\right)\:\Leftrightarrow\mid{ix}−{y}−\mathrm{1}\mid\:+\mid{x}+{i}\left({y}−\mathrm{1}\right)\mid=\mathrm{2} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}} \:+\left({y}+\mathrm{1}\right)^{\mathrm{2}} }\:\:+\sqrt{{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{2}\:\Rightarrow \\ $$$$\sqrt{{x}^{\mathrm{2}} \:+\left({y}+\mathrm{1}\right)^{\mathrm{2}} }\:=\left(\mathrm{2}−\sqrt{{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+\left({y}+\mathrm{1}\right)^{\mathrm{2}} \:=\mathrm{4}\:−\mathrm{4}\sqrt{{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\:\:+{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} \Rightarrow \\ $$$$\mathrm{4}\sqrt{{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} }=\left({y}−\mathrm{1}\right)^{\mathrm{2}} −\left({y}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}\:\Rightarrow \\ $$$$\mathrm{4}\sqrt{{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} }={y}^{\mathrm{2}} \:−\mathrm{2}{y}\:+\mathrm{1}\:−{y}^{\mathrm{2}} −\mathrm{2}{y}\:−\mathrm{1}\:+\mathrm{4} \\ $$$$=\mathrm{4}\:−\mathrm{4}{y}\:\Rightarrow\:{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\left(\mathrm{1}−{y}\right)^{\mathrm{2}} \:\Rightarrow\:{x}=\mathrm{0}\:\Rightarrow \\ $$$$\left\{{z}\in{C}/\mid{iz}−\mid+\mid{z}−{i}\mid=\mathrm{2}\right\}={iR}\:. \\ $$

Answered by ajfour last updated on 15/Aug/17

⇒ ∣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: (x^2 /a^2 )+(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 2a=d and b^2 =a^2 −((∣z_1 −z_2 ∣^2 )/4) so eccentricity e=((∣z_1 −z_2 ∣)/d) .

$$\Rightarrow\:\:\:\:\mid\mathrm{z}+\mathrm{i}\mid+\mid\mathrm{z}−\mathrm{i}\mid=\mathrm{2} \\ $$$$\mathrm{ellipse}\:\mathrm{with}\:\mathrm{origin}\:\mathrm{as}\:\mathrm{centre}. \\ $$$$\mathrm{focii}\::\:\:\:\:\mathrm{z}_{\mathrm{1}} =−\mathrm{i}\:\:;\:\:\mathrm{z}_{\mathrm{2}} =\mathrm{i} \\ $$$$\:\:\:\:\mathrm{shall}\:\mathrm{try}\:\mathrm{yo}\:\mathrm{obtain}\:\mathrm{equation}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{form}:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1}\:... \\ $$$$\mathrm{if}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{ellipse}\:\mathrm{be} \\ $$$$\:\:\:\mid\mathrm{z}−\mathrm{z}_{\mathrm{1}} \mid+\mid\mathrm{z}−\mathrm{z}_{\mathrm{2}} \mid=\mathrm{d} \\ $$$$\mathrm{with}\:\mathrm{z}_{\mathrm{1}} ,\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} \:\mathrm{being}\:\mathrm{focii} \\ $$$$\:\:\:\:\mathrm{then}\:\:\:\mathrm{2a}=\mathrm{d} \\ $$$$\mathrm{and}\:\:\:\:\mathrm{b}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} −\frac{\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\mathrm{so}\:\:\:\:\mathrm{eccentricity}\:\mathrm{e}=\frac{\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \mid}{\mathrm{d}}\:. \\ $$

Commented by ajfour last updated on 16/Aug/17

Here major axis is the Imaginary axis, so 2b=d and a^2 =b^2 −((∣z_1 −z_2 ∣^2 )/4) d=2 , so b=1 z_1 =i , z_2 =−i so a^2 =1−((∣2i∣^2 )/4) =0 hence ellipse with minor axis length zero and major axis length equal to 2. extends from z=−i to z=i , a double line segment. (special case of ellipse).

$${Here}\:{major}\:{axis}\:{is}\:{the}\:{Imaginary} \\ $$$${axis},\:{so}\:\mathrm{2}{b}={d}\:{and}\:{a}^{\mathrm{2}} ={b}^{\mathrm{2}} −\frac{\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid^{\mathrm{2}} }{\mathrm{4}} \\ $$$${d}=\mathrm{2}\:\:,\:{so}\:{b}=\mathrm{1}\:\:\:{z}_{\mathrm{1}} ={i}\:,\:\:{z}_{\mathrm{2}} =−{i} \\ $$$$\:\:{so}\:\:\:{a}^{\mathrm{2}} =\mathrm{1}−\frac{\mid\mathrm{2}{i}\mid^{\mathrm{2}} }{\mathrm{4}}\:=\mathrm{0} \\ $$$${hence}\:{ellipse}\:{with}\:{minor}\:{axis} \\ $$$${length}\:{zero}\:{and}\:{major}\:{axis}\:{length} \\ $$$${equal}\:{to}\:\mathrm{2}.\:{extends}\:{from}\:{z}=−{i}\:{to} \\ $$$${z}={i}\:,\:{a}\:{double}\:{line}\:{segment}. \\ $$$$\left({special}\:{case}\:{of}\:{ellipse}\right). \\ $$

Commented by Tinkutara last updated on 16/Aug/17

Sorry, I checked the answer today and it was given a straight line!

$$\mathrm{Sorry},\:\mathrm{I}\:\mathrm{checked}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{today}\:\mathrm{and} \\ $$$$\mathrm{it}\:\mathrm{was}\:\mathrm{given}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}! \\ $$

Commented by Tinkutara last updated on 16/Aug/17

So we will consider it ellipse or a line?

$$\mathrm{So}\:\mathrm{we}\:\mathrm{will}\:\mathrm{consider}\:\mathrm{it}\:\mathrm{ellipse}\:\mathrm{or}\:\mathrm{a}\:\mathrm{line}? \\ $$

Commented by ajfour last updated on 16/Aug/17

cant say !

$${cant}\:{say}\:! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com