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Question Number 39003 by behi83417@gmail.com last updated on 01/Jul/18

Answered by MJS last updated on 01/Jul/18

A= ((a),((1/a)) ); B= ((b),((1/b)) ); b>a  ∣AB∣=(((b−a)(√(a^2 b^2 +1)))/(ab))=c  M_(AB) =((A+B)/2)= ((((a+b)/2)),(((a+b)/(2ab))) )    A= ((α),(β) )= ((α),((1/α)) )  B= ((β),(α) )= (((1/α)),(α) )  M_(AB) = ((((α^2 +1)/(2α))),(((α^2 +1)/(2α))) )  if ∣AB∣=(((b−a)(√(a^2 b^2 +1)))/(ab))=c:  ((α^2 −1)/α)(√2)=c; α>1 ⇒ α=((√2)/4)((√(c^2 +8))+c)  ((1−α^2 )/α)(√2)=c; α<1 ⇒ α=((√2)/4)((√(c^2 +8))−c)  in both cases M_(AB) = (((((√(2(c^2 +8))))/4)),((((√(2(c^2 +8))))/4)) )

$${A}=\begin{pmatrix}{{a}}\\{\frac{\mathrm{1}}{{a}}}\end{pmatrix};\:{B}=\begin{pmatrix}{{b}}\\{\frac{\mathrm{1}}{{b}}}\end{pmatrix};\:{b}>{a} \\ $$$$\mid{AB}\mid=\frac{\left({b}−{a}\right)\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{1}}}{{ab}}={c} \\ $$$${M}_{{AB}} =\frac{{A}+{B}}{\mathrm{2}}=\begin{pmatrix}{\frac{{a}+{b}}{\mathrm{2}}}\\{\frac{{a}+{b}}{\mathrm{2}{ab}}}\end{pmatrix} \\ $$$$ \\ $$$${A}=\begin{pmatrix}{\alpha}\\{\beta}\end{pmatrix}=\begin{pmatrix}{\alpha}\\{\frac{\mathrm{1}}{\alpha}}\end{pmatrix} \\ $$$${B}=\begin{pmatrix}{\beta}\\{\alpha}\end{pmatrix}=\begin{pmatrix}{\frac{\mathrm{1}}{\alpha}}\\{\alpha}\end{pmatrix} \\ $$$${M}_{{AB}} =\begin{pmatrix}{\frac{\alpha^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}\alpha}}\\{\frac{\alpha^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}\alpha}}\end{pmatrix} \\ $$$$\mathrm{if}\:\mid{AB}\mid=\frac{\left({b}−{a}\right)\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{1}}}{{ab}}={c}: \\ $$$$\frac{\alpha^{\mathrm{2}} −\mathrm{1}}{\alpha}\sqrt{\mathrm{2}}={c};\:\alpha>\mathrm{1}\:\Rightarrow\:\alpha=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\sqrt{{c}^{\mathrm{2}} +\mathrm{8}}+{c}\right) \\ $$$$\frac{\mathrm{1}−\alpha^{\mathrm{2}} }{\alpha}\sqrt{\mathrm{2}}={c};\:\alpha<\mathrm{1}\:\Rightarrow\:\alpha=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\sqrt{{c}^{\mathrm{2}} +\mathrm{8}}−{c}\right) \\ $$$$\mathrm{in}\:\mathrm{both}\:\mathrm{cases}\:{M}_{{AB}} =\begin{pmatrix}{\frac{\left.\sqrt{\mathrm{2}\left({c}^{\mathrm{2}} +\mathrm{8}\right.}\right)}{\mathrm{4}}}\\{\frac{\left.\sqrt{\mathrm{2}\left({c}^{\mathrm{2}} +\mathrm{8}\right.}\right)}{\mathrm{4}}}\end{pmatrix} \\ $$

Commented by behi83417@gmail.com last updated on 01/Jul/18

dear MJS! thanks.  but in #2,length of AB,not constant.  what about equation of locus in#1&2?

$${dear}\:{MJS}!\:{thanks}. \\ $$$${but}\:{in}\:#\mathrm{2},{length}\:{of}\:{AB},{not}\:{constant}. \\ $$$${what}\:{about}\:{equation}\:{of}\:{locus}\:{in}#\mathrm{1\&2}? \\ $$

Answered by MrW3 last updated on 02/Jul/18

(1)  A(a,(1/a)), B(b,(1/b))  AB^2 =(a−b)^2 +((1/a)−(1/b))^2 =l^2 =const.  (a−b)^2 +(((a−b)^2 )/(a^2 b^2 ))=l^2   (a−b)^2 [1+(1/(a^2 b^2 ))]=l^2   ⇒ [(a+b)^2 −4ab][1+(1/(a^2 b^2 ))]=l^2    ...(i)  M(x,y) =midpoint of AB  x=((a+b)/2)  ⇒a+b=2x    ...(ii)  y=(((1/a)+(1/b))/2)=((a+b)/(2ab))  ⇒ab=(x/y)   ...(iii)    (ii) & (iii) into (i):  [(2x)^2 −4(x/y)][1+(1/(((x/y))^2 ))]=l^2   [4x^2 −4(x/y)][1+(y^2 /x^2 )]=l^2   ⇒4(xy−1)(x^2 +y^2 )−xyl^2 =0    see graph with l=6:

$$\left(\mathrm{1}\right) \\ $$$${A}\left({a},\frac{\mathrm{1}}{{a}}\right),\:{B}\left({b},\frac{\mathrm{1}}{{b}}\right) \\ $$$${AB}^{\mathrm{2}} =\left({a}−{b}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} ={l}^{\mathrm{2}} ={const}. \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} +\frac{\left({a}−{b}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }={l}^{\mathrm{2}} \\ $$$$\left({a}−{b}\right)^{\mathrm{2}} \left[\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\right]={l}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left[\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{4}{ab}\right]\left[\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\right]={l}^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$${M}\left({x},{y}\right)\:={midpoint}\:{of}\:{AB} \\ $$$${x}=\frac{{a}+{b}}{\mathrm{2}} \\ $$$$\Rightarrow{a}+{b}=\mathrm{2}{x}\:\:\:\:...\left({ii}\right) \\ $$$${y}=\frac{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}}{\mathrm{2}}=\frac{{a}+{b}}{\mathrm{2}{ab}} \\ $$$$\Rightarrow{ab}=\frac{{x}}{{y}}\:\:\:...\left({iii}\right) \\ $$$$ \\ $$$$\left({ii}\right)\:\&\:\left({iii}\right)\:{into}\:\left({i}\right): \\ $$$$\left[\left(\mathrm{2}{x}\right)^{\mathrm{2}} −\mathrm{4}\frac{{x}}{{y}}\right]\left[\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} }\right]={l}^{\mathrm{2}} \\ $$$$\left[\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}\frac{{x}}{{y}}\right]\left[\mathrm{1}+\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right]={l}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}\left({xy}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−{xyl}^{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$$${see}\:{graph}\:{with}\:{l}=\mathrm{6}: \\ $$

Commented by MrW3 last updated on 02/Jul/18

Commented by MrW3 last updated on 02/Jul/18

(2)  A(α,(1/α)), B(β,(1/β))  M(x,y) =midpoint of AB  x=((α+β)/2)  ⇒β=2x−α  y=(((1/α)+(1/β))/2)=((α+β)/(2αβ))=(x/(αβ))=(x/(α(2x−α)))  ⇒x−α(2x−α)y=0    see graph with α=0.5:

$$\left(\mathrm{2}\right) \\ $$$${A}\left(\alpha,\frac{\mathrm{1}}{\alpha}\right),\:{B}\left(\beta,\frac{\mathrm{1}}{\beta}\right) \\ $$$${M}\left({x},{y}\right)\:={midpoint}\:{of}\:{AB} \\ $$$${x}=\frac{\alpha+\beta}{\mathrm{2}} \\ $$$$\Rightarrow\beta=\mathrm{2}{x}−\alpha \\ $$$${y}=\frac{\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}}{\mathrm{2}}=\frac{\alpha+\beta}{\mathrm{2}\alpha\beta}=\frac{{x}}{\alpha\beta}=\frac{{x}}{\alpha\left(\mathrm{2}{x}−\alpha\right)} \\ $$$$\Rightarrow{x}−\alpha\left(\mathrm{2}{x}−\alpha\right){y}=\mathrm{0} \\ $$$$ \\ $$$${see}\:{graph}\:{with}\:\alpha=\mathrm{0}.\mathrm{5}: \\ $$

Commented by MrW3 last updated on 02/Jul/18

Commented by behi83417@gmail.com last updated on 02/Jul/18

dear master!thank you very much.  alwyes the best.your answers are  the finishing bullets to the heart of  questions.being you in this forum   encouraged me.do not be far from us  any more.god bless  you.

$${dear}\:{master}!{thank}\:{you}\:{very}\:{much}. \\ $$$${alwyes}\:{the}\:{best}.{your}\:{answers}\:{are} \\ $$$${the}\:{finishing}\:{bullets}\:{to}\:{the}\:{heart}\:{of} \\ $$$${questions}.{being}\:{you}\:{in}\:{this}\:{forum}\: \\ $$$${encouraged}\:{me}.{do}\:{not}\:{be}\:{far}\:{from}\:{us} \\ $$$${any}\:{more}.{god}\:{bless}\:\:{you}. \\ $$

Commented by MrW3 last updated on 02/Jul/18

Thank you too! I′ll try not to go too far  away from this forum :)

$${Thank}\:{you}\:{too}!\:{I}'{ll}\:{try}\:{not}\:{to}\:{go}\:{too}\:{far} \\ $$$$\left.{away}\:{from}\:{this}\:{forum}\::\right) \\ $$

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