Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 11338 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Mar/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Mar/17

Answered by mrW1 last updated on 21/Mar/17

(1) let B(t,(1/t)) a=OA^(→) b=OB^(→) cos ∠AOB=((a∙b)/(∣a∣×∣b∣))=((t(√3)+(1/(t(√3))))/((√(((√3))^2 +((1/(√3)))^2 ))×(√(t^2 +(1/t^2 )))))=cos (π/3)=(1/2) ((t(√3)+(1/(t(√3))))/((√(((√3))^2 +((1/(√3)))^2 ))×(√(t^2 +(1/t^2 )))))=(1/2) ((3t+(1/t))/((√(10))×(√(t^2 +(1/t^2 )))))=(1/2) 3t+(1/t)=((√(10))/2)×(√(t^2 +(1/t^2 ))) 9t^2 +(1/t^2 )+6=((10)/4)(t^2 +(1/t^2 )) 9t^4 +1+6t^2 =(5/2)(t^4 +1) 13t^4 +12t^2 −3=0 t^2 =((−12+(√(144+4×13×3)))/(2×13))=((5(√3)−6)/(13)) t=(√((5(√3)−6)/(13)))

$$\left(\mathrm{1}\right) \\ $$$${let}\:{B}\left({t},\frac{\mathrm{1}}{{t}}\right) \\ $$$$\boldsymbol{{a}}=\overset{\rightarrow} {{OA}} \\ $$$$\boldsymbol{{b}}=\overset{\rightarrow} {{OB}} \\ $$$$\mathrm{cos}\:\angle{AOB}=\frac{\boldsymbol{{a}}\centerdot\boldsymbol{{b}}}{\mid\boldsymbol{{a}}\mid×\mid\boldsymbol{{b}}\mid}=\frac{{t}\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{{t}\sqrt{\mathrm{3}}}}{\sqrt{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }×\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}}=\mathrm{cos}\:\frac{\pi}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{t}\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{{t}\sqrt{\mathrm{3}}}}{\sqrt{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }×\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}{t}+\frac{\mathrm{1}}{{t}}}{\sqrt{\mathrm{10}}×\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{3}{t}+\frac{\mathrm{1}}{{t}}=\frac{\sqrt{\mathrm{10}}}{\mathrm{2}}×\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }} \\ $$$$\mathrm{9}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{6}=\frac{\mathrm{10}}{\mathrm{4}}\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right) \\ $$$$\mathrm{9}{t}^{\mathrm{4}} +\mathrm{1}+\mathrm{6}{t}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{2}}\left({t}^{\mathrm{4}} +\mathrm{1}\right) \\ $$$$\mathrm{13}{t}^{\mathrm{4}} +\mathrm{12}{t}^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} =\frac{−\mathrm{12}+\sqrt{\mathrm{144}+\mathrm{4}×\mathrm{13}×\mathrm{3}}}{\mathrm{2}×\mathrm{13}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{13}} \\ $$$${t}=\sqrt{\frac{\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{13}}} \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Apr/17

part#1 AB^2 =oA^2 +oB^2 −2oA.oB.cos∡AoB (s−t)^2 +((1/s)−(1/t))^2 =(s^2 +(1/s^2 ))+(t^2 +(1/t^2 ))−2((√(s^4 +1))/s).((√(t^4 +1))/t)((1/2))⇒ −2(st+(1/(st)))=−(((√(s^4 +1)).(√(t^4 +1)))/(st))⇒ 2(s^2 t^2 +1)=(√(s^4 +1)).(√(t^4 +1)) 4s^4 t^4 +8s^2 t^2 +4=(s^4 +1)(t^4 +1)=s^4 t^4 +s^4 +t^4 +1⇒ (3s^4 −1)t^4 +8s^2 t^2 +3−s^4 =0⇒ t^2 =((−8s^2 +(√(64s^4 −4(3s^4 −1)(3−s^4 ))))/(2(3s^4 −1)))= t^2 =((−8s^2 +(√(12(s^4 +1)^2 )))/(2(3s^4 −1)))=(((√3)(s^4 +1)−4s^2 )/(3s^4 −1)) t^2 =(((s^2 −(√3))((√3)s^2 −1))/(((√3)s^2 +1)((√3)s^2 −1)))=((s^2 −(√3))/((√3)s^2 +1))(in general) i).(√3)s^2 −1≠0⇒s≠(1/(3)^(1/4) ). ii).s^2 −(√3)>0⇒s>(3)^(1/4) . iii).s=(√3)⇒t^2 =((3−(√3))/(3(√)3+1)) t=(√((3−(√3))/(3(√3)+1))).B((((3−(√3))/(3(√3)+1)))^(1/�) ,(√((3(√3)+1)/(3−(√3))))). 2.m_(AB) =((y_B −y_A )/(x_B −x_A ))=(((1/t)−(1/s))/(t−s))=((−1)/(st)) y−y_B =m_(AB) (x−x_B )⇒y−(1/t)=((−1)/(st))(x−t) sty−s=−(x−t)⇒sty−s=−x+t AB: sty=−x+s+t. oN=(√(1^2 +1^2 ))=(√2) intersect of AB and circle not lie on y=x in all cases.

$${part}#\mathrm{1} \\ $$$${AB}^{\mathrm{2}} ={oA}^{\mathrm{2}} +{oB}^{\mathrm{2}} −\mathrm{2}{oA}.{oB}.{cos}\measuredangle{AoB}\:\:\:\:\: \\ $$$$\left({s}−{t}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{s}}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} =\left({s}^{\mathrm{2}} +\frac{\mathrm{1}}{{s}^{\mathrm{2}} }\right)+\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)−\mathrm{2}\frac{\sqrt{{s}^{\mathrm{4}} +\mathrm{1}}}{{s}}.\frac{\sqrt{{t}^{\mathrm{4}} +\mathrm{1}}}{{t}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Rightarrow \\ $$$$−\mathrm{2}\left({st}+\frac{\mathrm{1}}{{st}}\right)=−\frac{\sqrt{{s}^{\mathrm{4}} +\mathrm{1}}.\sqrt{{t}^{\mathrm{4}} +\mathrm{1}}}{{st}}\Rightarrow \\ $$$$\mathrm{2}\left({s}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}\right)=\sqrt{{s}^{\mathrm{4}} +\mathrm{1}}.\sqrt{{t}^{\mathrm{4}} +\mathrm{1}} \\ $$$$\mathrm{4}{s}^{\mathrm{4}} {t}^{\mathrm{4}} +\mathrm{8}{s}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{4}=\left({s}^{\mathrm{4}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{1}\right)={s}^{\mathrm{4}} {t}^{\mathrm{4}} +{s}^{\mathrm{4}} +{t}^{\mathrm{4}} +\mathrm{1}\Rightarrow \\ $$$$\left(\mathrm{3}{s}^{\mathrm{4}} −\mathrm{1}\right){t}^{\mathrm{4}} +\mathrm{8}{s}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{3}−{s}^{\mathrm{4}} =\mathrm{0}\Rightarrow \\ $$$${t}^{\mathrm{2}} =\frac{−\mathrm{8}{s}^{\mathrm{2}} +\sqrt{\mathrm{64}{s}^{\mathrm{4}} −\mathrm{4}\left(\mathrm{3}{s}^{\mathrm{4}} −\mathrm{1}\right)\left(\mathrm{3}−{s}^{\mathrm{4}} \right)}}{\mathrm{2}\left(\mathrm{3}{s}^{\mathrm{4}} −\mathrm{1}\right)}= \\ $$$${t}^{\mathrm{2}} =\frac{−\mathrm{8}{s}^{\mathrm{2}} +\sqrt{\mathrm{12}\left({s}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{2}\left(\mathrm{3}{s}^{\mathrm{4}} −\mathrm{1}\right)}=\frac{\sqrt{\mathrm{3}}\left({s}^{\mathrm{4}} +\mathrm{1}\right)−\mathrm{4}{s}^{\mathrm{2}} }{\mathrm{3}{s}^{\mathrm{4}} −\mathrm{1}} \\ $$$${t}^{\mathrm{2}} =\frac{\left({s}^{\mathrm{2}} −\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{3}}{s}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\sqrt{\mathrm{3}}{s}^{\mathrm{2}} +\mathrm{1}\right)\left(\sqrt{\mathrm{3}}{s}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{{s}^{\mathrm{2}} −\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}{s}^{\mathrm{2}} +\mathrm{1}}\left({in}\:{general}\right) \\ $$$$\left.{i}\right).\sqrt{\mathrm{3}}{s}^{\mathrm{2}} −\mathrm{1}\neq\mathrm{0}\Rightarrow{s}\neq\frac{\mathrm{1}}{\sqrt[{\mathrm{4}}]{\mathrm{3}}}. \\ $$$$\left.{ii}\right).{s}^{\mathrm{2}} −\sqrt{\mathrm{3}}>\mathrm{0}\Rightarrow{s}>\sqrt[{\mathrm{4}}]{\mathrm{3}}\:. \\ $$$$\left.{iii}\right).{s}=\sqrt{\mathrm{3}}\Rightarrow{t}^{\mathrm{2}} =\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{}\mathrm{3}+\mathrm{1}} \\ $$$$\boldsymbol{{t}}=\sqrt{\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}}}.\boldsymbol{{B}}\left(\sqrt[{}]{\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}}},\sqrt{\frac{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{3}−\sqrt{\mathrm{3}}}}\right). \\ $$$$\mathrm{2}.{m}_{{AB}} =\frac{{y}_{{B}} −{y}_{{A}} }{{x}_{{B}} −{x}_{{A}} }=\frac{\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{{s}}}{{t}−{s}}=\frac{−\mathrm{1}}{{st}} \\ $$$${y}−{y}_{{B}} ={m}_{{AB}} \left({x}−{x}_{{B}} \right)\Rightarrow{y}−\frac{\mathrm{1}}{{t}}=\frac{−\mathrm{1}}{{st}}\left({x}−{t}\right) \\ $$$${sty}−{s}=−\left({x}−{t}\right)\Rightarrow{sty}−{s}=−{x}+{t} \\ $$$${AB}:\:{sty}=−{x}+{s}+{t}. \\ $$$${oN}=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\sqrt{\mathrm{2}} \\ $$$${intersect}\:{of}\:{AB}\:{and}\:{circle}\:{not}\:{lie}\:{on}\: \\ $$$${y}={x}\:{in}\:{all}\:{cases}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Mar/17

((3−(√3))/(3(√3)+1))×((3(√3)−1)/(3(√3)−1))=((9(√3)−3−9+(√3))/(27−1))=((10(√3)−12)/(26))=((5(√3)−6)/(13)).

$$\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}}×\frac{\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{1}}=\frac{\mathrm{9}\sqrt{\mathrm{3}}−\mathrm{3}−\mathrm{9}+\sqrt{\mathrm{3}}}{\mathrm{27}−\mathrm{1}}=\frac{\mathrm{10}\sqrt{\mathrm{3}}−\mathrm{12}}{\mathrm{26}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{13}}. \\ $$

Commented by mrW1 last updated on 22/Mar/17

Good!

$$\mathscr{G}{ood}! \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Apr/17

part#2. but in a special case,that point M is located on:y=x,we have: x_M =y_M ⇒stx_M =−x_M +s+t⇒ x_M =y_M =((s+t)/(st+1)). in this case AB⊥(y=x)⇒m_(AB) ×m_(y=x) =−1 ⇒(−(1/(st)))(1)=(−1)⇒st=1⇒s=(1/t) ⇒x_M =y_M =((s+t)/(st+1))=(1/2)(s+t)⇒AM=MB A(s,(1/s)),B(t,(1/t))=B((1/s),s) OM=(√(x_M ^2 +y_M ^2 ))=(√2)x_M =((√2)/2)(s+(1/s)) OA=OB=(√(s^2 +(1/s^2 )))=((√(s^4 +1))/s) AB=(√((s−(1/s))^2 +((1/s)−s)^2 ))=((√2)/s)(s^2 −1). (1/2)OA.OB.sin((π/3))=(1/2)AB.OM⇒ (1/2)((√(s^4 +1))/s).((√(s^4 +1))/s).((√3)/2)=(1/2).((√2)/s)(s^2 −1).((√2)/2).((s^2 +1)/s) (√3)(s^4 +1)=2(s^4 −1)⇒s^4 =((2+(√3))/(2−(√3))) s^4 =(2+(√3))^2 ⇒s=(√(2+(√3))) . x_M =y_M =(1/2)(s+t)=((s^2 +1)/(2s))=((3+(√3))/(2(√(2+(√3))))) = (((√3)((√3)+1))/(2(√((((√3)+1)^2 )/2))))=(((√3)((√3)+1))/(2((((√3)+1))/(√2))))=((√6)/2) . A((√(2+(√3))),(√(2−(√3)))),B((√(2−(√3))),(√(2+(√3)))) ON+2r=OM⇒(√2)+2r=(√2).x_M ⇒ r=((√2)/2)(x_M −1)=((√2)/2).(((√6)/2)−1) =(((√3)−(√2))/2) ■

$${part}#\mathrm{2}.\: \\ $$$$\:{but}\:{in}\:{a}\:{special}\:{case},{that}\:{point}\:{M}\:{is} \\ $$$${located}\:{on}:{y}={x},{we}\:{have}: \\ $$$${x}_{{M}} ={y}_{{M}} \Rightarrow{stx}_{{M}} =−{x}_{{M}} +{s}+{t}\Rightarrow \\ $$$${x}_{{M}} ={y}_{{M}} =\frac{{s}+{t}}{{st}+\mathrm{1}}. \\ $$$${in}\:{this}\:{case}\:{AB}\bot\left({y}={x}\right)\Rightarrow{m}_{{AB}} ×{m}_{{y}={x}} =−\mathrm{1} \\ $$$$\Rightarrow\left(−\frac{\mathrm{1}}{{st}}\right)\left(\mathrm{1}\right)=\left(−\mathrm{1}\right)\Rightarrow{st}=\mathrm{1}\Rightarrow{s}=\frac{\mathrm{1}}{{t}} \\ $$$$\Rightarrow{x}_{{M}} ={y}_{{M}} =\frac{{s}+{t}}{{st}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\left({s}+{t}\right)\Rightarrow{AM}={MB} \\ $$$${A}\left({s},\frac{\mathrm{1}}{{s}}\right),{B}\left({t},\frac{\mathrm{1}}{{t}}\right)={B}\left(\frac{\mathrm{1}}{{s}},{s}\right) \\ $$$${OM}=\sqrt{{x}_{{M}} ^{\mathrm{2}} +{y}_{{M}} ^{\mathrm{2}} }=\sqrt{\mathrm{2}}{x}_{{M}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({s}+\frac{\mathrm{1}}{{s}}\right) \\ $$$${OA}={OB}=\sqrt{{s}^{\mathrm{2}} +\frac{\mathrm{1}}{{s}^{\mathrm{2}} }}=\frac{\sqrt{{s}^{\mathrm{4}} +\mathrm{1}}}{{s}} \\ $$$${AB}=\sqrt{\left({s}−\frac{\mathrm{1}}{{s}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{s}}−{s}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{2}}}{{s}}\left({s}^{\mathrm{2}} −\mathrm{1}\right). \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{OA}.{OB}.{sin}\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{AB}.{OM}\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\frac{\sqrt{{s}^{\mathrm{4}} +\mathrm{1}}}{{s}}.\frac{\sqrt{{s}^{\mathrm{4}} +\mathrm{1}}}{{s}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{{s}}\left({s}^{\mathrm{2}} −\mathrm{1}\right).\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\frac{{s}^{\mathrm{2}} +\mathrm{1}}{{s}} \\ $$$$\sqrt{\mathrm{3}}\left({s}^{\mathrm{4}} +\mathrm{1}\right)=\mathrm{2}\left({s}^{\mathrm{4}} −\mathrm{1}\right)\Rightarrow{s}^{\mathrm{4}} =\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${s}^{\mathrm{4}} =\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \Rightarrow{s}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:. \\ $$$${x}_{{M}} ={y}_{{M}} =\frac{\mathrm{1}}{\mathrm{2}}\left({s}+{t}\right)=\frac{{s}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{s}}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}\:= \\ $$$$\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}\sqrt{\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}}}=\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\sqrt{\mathrm{2}}}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\:\:. \\ $$$${A}\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}},\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right),{B}\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}},\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right) \\ $$$${ON}+\mathrm{2}{r}={OM}\Rightarrow\sqrt{\mathrm{2}}+\mathrm{2}{r}=\sqrt{\mathrm{2}}.{x}_{{M}} \Rightarrow \\ $$$${r}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({x}_{{M}} −\mathrm{1}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\left(\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}−\mathrm{1}\right)\:=\frac{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}{\mathrm{2}}\:\blacksquare \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Apr/17

part#3. B(s,(1/s)),A(t,(1/t)) A,can move on : xy=1 until the tagent line to circle i.e: AB ⊥OB.then we have: OB^2 +AB^2 =OA^2 (s^2 +(1/s^2 ))+((s−t)^2 +((1/s)−(1/t))^2 )=(t^2 +(1/t^2 )) 2s^2 +(2/s^2 )−2st−(2/(st))=0⇒s^2 +(1/s^2 )=st+(1/(st))⇒ t(s^4 +1)=s(s^2 t^2 +1)⇒(s^3 )t^2 −(s^4 +1)t+s=0 ⇒t=(((s^4 +1)±(√((s^4 +1)^2 −4s.s^3 )))/(2s^3 ))⇒ t=((s^4 +1±(s^4 −1))/(2s^3 ))⇒t=s ,t=(1/s^3 ) ⇒A(s,(1/s)),B(s,(1/s))⇒AB=0 (impossible) A((1/s^3 ),s^3 ),B(s,(1/s))⇒AB^2 =((1/s^3 )−s)^2 +(s^3 −(1/s))^2 ⇒AB^2 =(((1−s^4 )^2 )/s^6 )+(((s^4 −1)^2 )/s^2 )=(s^4 −1)^2 (1+s^4 )/s^6 OB^2 =s^2 +(1/s^2 ) A,can move on xy=1 until the tagent line AB ⊥OB.at this time: ∡OAB=30^° or otherwise B can move on xy=1 until tangent line to circle i.e: AB⊥OA. in this two case ,we can write: tg(30^° )=((OA)/(AB)) or tg(30^° )=((OB)/(AB)) tg(30^° )=((OB)/(AB))⇒((√3)/3)=((OB)/(AB))⇒AB^2 =3OB^2 ⇒ (((s^4 −1)^2 (s^4 +1))/s^6 )=3(s^2 +(1/s^2 ))=3(((s^4 +1)/s^2 ))⇒ (s^4 −1)^2 =3s^4 ⇒s^4 ±(√3)s^2 −1=0 s^2 =((±(√3)+(√(3−4(−1))))/2)=((±(√3)+(√7))/2) s=(√(((√7)±(√3))/2))=1.48 or 0.68. from part#1,we know that: s≥(3)^(1/4) =1.31 and now we have: s≤(√(((√7)+(√3))/2))=1.48 ⇒(3)^(1/4) ≤s≤(√((((√7)+(√3))/2),))(3)^(1/4) ≤t≤(√((((√7)+(√3))/2) ))■ ⇒Δx_A =1.48−1.31=0.17=Δy_B Δy_A =(1/(1.48))−(1/(1.31))=−0.087=Δx_B

$${part}#\mathrm{3}. \\ $$$${B}\left({s},\frac{\mathrm{1}}{{s}}\right),{A}\left({t},\frac{\mathrm{1}}{{t}}\right) \\ $$$${A},{can}\:{move}\:{on}\::\:{xy}=\mathrm{1}\:{until}\:{the}\:{tagent} \\ $$$${line}\:{to}\:{circle}\:{i}.{e}:\:{AB}\:\bot{OB}.{then}\:{we}\:{have}: \\ $$$${OB}^{\mathrm{2}} +{AB}^{\mathrm{2}} ={OA}^{\mathrm{2}} \\ $$$$\left({s}^{\mathrm{2}} +\frac{\mathrm{1}}{{s}^{\mathrm{2}} }\right)+\left(\left({s}−{t}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{s}}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} \right)=\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right) \\ $$$$\mathrm{2}{s}^{\mathrm{2}} +\frac{\mathrm{2}}{{s}^{\mathrm{2}} }−\mathrm{2}{st}−\frac{\mathrm{2}}{{st}}=\mathrm{0}\Rightarrow{s}^{\mathrm{2}} +\frac{\mathrm{1}}{{s}^{\mathrm{2}} }={st}+\frac{\mathrm{1}}{{st}}\Rightarrow \\ $$$${t}\left({s}^{\mathrm{4}} +\mathrm{1}\right)={s}\left({s}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}\right)\Rightarrow\left({s}^{\mathrm{3}} \right){t}^{\mathrm{2}} −\left({s}^{\mathrm{4}} +\mathrm{1}\right){t}+{s}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\left({s}^{\mathrm{4}} +\mathrm{1}\right)\pm\sqrt{\left({s}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{s}.{s}^{\mathrm{3}} }}{\mathrm{2}{s}^{\mathrm{3}} }\Rightarrow \\ $$$${t}=\frac{{s}^{\mathrm{4}} +\mathrm{1}\pm\left({s}^{\mathrm{4}} −\mathrm{1}\right)}{\mathrm{2}{s}^{\mathrm{3}} }\Rightarrow{t}={s}\:,{t}=\frac{\mathrm{1}}{{s}^{\mathrm{3}} } \\ $$$$\Rightarrow{A}\left({s},\frac{\mathrm{1}}{{s}}\right),{B}\left({s},\frac{\mathrm{1}}{{s}}\right)\Rightarrow{AB}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\left({impossible}\right) \\ $$$${A}\left(\frac{\mathrm{1}}{{s}^{\mathrm{3}} },{s}^{\mathrm{3}} \right),{B}\left({s},\frac{\mathrm{1}}{{s}}\right)\Rightarrow{AB}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{{s}^{\mathrm{3}} }−{s}\right)^{\mathrm{2}} +\left({s}^{\mathrm{3}} −\frac{\mathrm{1}}{{s}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{AB}^{\mathrm{2}} =\frac{\left(\mathrm{1}−{s}^{\mathrm{4}} \right)^{\mathrm{2}} }{{s}^{\mathrm{6}} }+\frac{\left({s}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} }{{s}^{\mathrm{2}} }=\left({s}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+{s}^{\mathrm{4}} \right)/{s}^{\mathrm{6}} \\ $$$${OB}^{\mathrm{2}} ={s}^{\mathrm{2}} +\frac{\mathrm{1}}{{s}^{\mathrm{2}} } \\ $$$${A},{can}\:{move}\:{on}\:{xy}=\mathrm{1}\:{until}\:{the}\:{tagent} \\ $$$${line}\:{AB}\:\bot{OB}.{at}\:{this}\:{time}:\:\measuredangle{OAB}=\mathrm{30}^{°} \\ $$$${or}\:{otherwise}\:{B}\:{can}\:{move}\:\:{on}\:{xy}=\mathrm{1} \\ $$$${until}\:{tangent}\:{line}\:{to}\:{circle}\:{i}.{e}:\:{AB}\bot{OA}. \\ $$$${in}\:{this}\:{two}\:{case}\:,{we}\:{can}\:{write}: \\ $$$${tg}\left(\mathrm{30}^{°} \right)=\frac{{OA}}{{AB}}\:{or}\:{tg}\left(\mathrm{30}^{°} \right)=\frac{{OB}}{{AB}} \\ $$$${tg}\left(\mathrm{30}^{°} \right)=\frac{{OB}}{{AB}}\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}=\frac{{OB}}{{AB}}\Rightarrow{AB}^{\mathrm{2}} =\mathrm{3}{OB}^{\mathrm{2}} \Rightarrow \\ $$$$\frac{\left({s}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} \left({s}^{\mathrm{4}} +\mathrm{1}\right)}{{s}^{\mathrm{6}} }=\mathrm{3}\left({s}^{\mathrm{2}} +\frac{\mathrm{1}}{{s}^{\mathrm{2}} }\right)=\mathrm{3}\left(\frac{{s}^{\mathrm{4}} +\mathrm{1}}{{s}^{\mathrm{2}} }\right)\Rightarrow \\ $$$$\left({s}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3}{s}^{\mathrm{4}} \Rightarrow{s}^{\mathrm{4}} \pm\sqrt{\mathrm{3}}{s}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${s}^{\mathrm{2}} =\frac{\pm\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}−\mathrm{4}\left(−\mathrm{1}\right)}}{\mathrm{2}}=\frac{\pm\sqrt{\mathrm{3}}+\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$${s}=\sqrt{\frac{\sqrt{\mathrm{7}}\pm\sqrt{\mathrm{3}}}{\mathrm{2}}}=\mathrm{1}.\mathrm{48}\:{or}\:\mathrm{0}.\mathrm{68}. \\ $$$${from}\:{part}#\mathrm{1},{we}\:{know}\:{that}:\:{s}\geqslant\sqrt[{\mathrm{4}}]{\mathrm{3}}=\mathrm{1}.\mathrm{31} \\ $$$${and}\:{now}\:{we}\:{have}:\:{s}\leqslant\sqrt{\frac{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}{\mathrm{2}}}=\mathrm{1}.\mathrm{48} \\ $$$$\Rightarrow\sqrt[{\mathrm{4}}]{\mathrm{3}}\leqslant{s}\leqslant\sqrt{\frac{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}{\mathrm{2}},}\sqrt[{\mathrm{4}}]{\mathrm{3}}\leqslant{t}\leqslant\sqrt{\frac{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}{\mathrm{2}}\:}\blacksquare \\ $$$$\Rightarrow\Delta{x}_{{A}} =\mathrm{1}.\mathrm{48}−\mathrm{1}.\mathrm{31}=\mathrm{0}.\mathrm{17}=\Delta{y}_{{B}} \\ $$$$\:\:\:\:\:\Delta{y}_{{A}} =\frac{\mathrm{1}}{\mathrm{1}.\mathrm{48}}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{31}}=−\mathrm{0}.\mathrm{087}=\Delta{x}_{{B}} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Apr/17

OA=(√((1/s^6 )+s^6 ))=((√(s^(12) +1))/s^3 ) OB=(√(s^2 +(1/s^2 )))=((√(s^4 +1))/s) S_(OAB) =(1/2)OA.OB.sin(π/3)= =(1/2).((√(s^(12) +1))/s^3 ).((√(s^4 +1))/s).((√3)/2)=((√3)/(4s^4 ))(s^4 +1)(√(s^8 −s^4 +1)) s^(12) +1=(s^4 +1)(s^8 −s^4 +1) s^8 −s^4 +1=(s^8 −2s^4 +1+s^4 )= (s^4 −1)^2 +s^4 =3s^4 +s^4 =4s^4 S_(OAB) =((√3)/(4s^4 ))(s^4 +1).2s^2 =((√3)/2).((s^4 +1)/s^2 )= ((√3)/2)(7−(√(21)))=((7(√3)−3(√7))/2) . s^2 =(((√7)+(√3))/2) s^4 =((10+2(√(21)))/4) s^4 +1=((14+2(√(21)))/4)=((7+(√(21)))/2) ((.5(7+(√(21))))/(.5((√7)+(√3))))=((7(√7)−7(√3)+7(√3)−3(√7))/((√7)+(√3)))= =((4(√7))/((√7)+(√3)))=(√7)((√7)−(√3))=7−(√(21))

$${OA}=\sqrt{\frac{\mathrm{1}}{{s}^{\mathrm{6}} }+{s}^{\mathrm{6}} }=\frac{\sqrt{{s}^{\mathrm{12}} +\mathrm{1}}}{{s}^{\mathrm{3}} } \\ $$$${OB}=\sqrt{{s}^{\mathrm{2}} +\frac{\mathrm{1}}{{s}^{\mathrm{2}} }}=\frac{\sqrt{{s}^{\mathrm{4}} +\mathrm{1}}}{{s}} \\ $$$${S}_{{OAB}} =\frac{\mathrm{1}}{\mathrm{2}}{OA}.{OB}.{sin}\frac{\pi}{\mathrm{3}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\sqrt{{s}^{\mathrm{12}} +\mathrm{1}}}{{s}^{\mathrm{3}} }.\frac{\sqrt{{s}^{\mathrm{4}} +\mathrm{1}}}{{s}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}{s}^{\mathrm{4}} }\left({s}^{\mathrm{4}} +\mathrm{1}\right)\sqrt{{s}^{\mathrm{8}} −{s}^{\mathrm{4}} +\mathrm{1}} \\ $$$${s}^{\mathrm{12}} +\mathrm{1}=\left({s}^{\mathrm{4}} +\mathrm{1}\right)\left({s}^{\mathrm{8}} −{s}^{\mathrm{4}} +\mathrm{1}\right) \\ $$$${s}^{\mathrm{8}} −{s}^{\mathrm{4}} +\mathrm{1}=\left({s}^{\mathrm{8}} −\mathrm{2}{s}^{\mathrm{4}} +\mathrm{1}+{s}^{\mathrm{4}} \right)= \\ $$$$\left({s}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{2}} +{s}^{\mathrm{4}} =\mathrm{3}{s}^{\mathrm{4}} +{s}^{\mathrm{4}} =\mathrm{4}{s}^{\mathrm{4}} \\ $$$${S}_{{OAB}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}{s}^{\mathrm{4}} }\left({s}^{\mathrm{4}} +\mathrm{1}\right).\mathrm{2}{s}^{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\frac{{s}^{\mathrm{4}} +\mathrm{1}}{{s}^{\mathrm{2}} }= \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{7}−\sqrt{\mathrm{21}}\right)=\frac{\mathrm{7}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\:\:. \\ $$$${s}^{\mathrm{2}} =\frac{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{s}^{\mathrm{4}} =\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{21}}}{\mathrm{4}} \\ $$$${s}^{\mathrm{4}} +\mathrm{1}=\frac{\mathrm{14}+\mathrm{2}\sqrt{\mathrm{21}}}{\mathrm{4}}=\frac{\mathrm{7}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\frac{.\mathrm{5}\left(\mathrm{7}+\sqrt{\mathrm{21}}\right)}{.\mathrm{5}\left(\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}\right)}=\frac{\mathrm{7}\sqrt{\mathrm{7}}−\mathrm{7}\sqrt{\mathrm{3}}+\mathrm{7}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{7}}}{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}= \\ $$$$=\frac{\mathrm{4}\sqrt{\mathrm{7}}}{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}=\sqrt{\mathrm{7}}\left(\sqrt{\mathrm{7}}−\sqrt{\mathrm{3}}\right)=\mathrm{7}−\sqrt{\mathrm{21}} \\ $$

Commented by ridwan balatif last updated on 09/Apr/17

Ngomong opo sih kalian ini?

$$\mathrm{Ngomong}\:\mathrm{opo}\:\mathrm{sih}\:\mathrm{kalian}\:\mathrm{ini}? \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Apr/17

2(s^2 t^2 +1)=(√3)(√(s^4 +1))(√(t^4 +1)) 4(s^4 t^4 +2s^2 t^2 +1)=3(s^4 +1)(t^4 +1) 4s^4 t^4 +8s^2 t^2 +4=3s^4 t^4 +3s^4 +3t^4 +3 (s^4 −3)t_ ^4 +8s^2 t^2 +(1−3s^4 )=0 t^2 =((−8s^2 +(√(64s^4 −4(s^4 −3)(1−3s^4 ))))/(2(s^4 −3))) 64s^4 −4(s^4 −3s^8 −3+9s^4 )=24s^4 +12s^8 +12 =12(s^4 +1)^2 t^2 =((−8s^2 +2(√3)(s^4 +1))/(2(s^4 −3)))=(((√3)(s^4 +1)−4s^2 )/(s^4 −3)) t^2 =(((s^2 −(√3))((√3)s^2 −1))/((s^2 −(√3))(s^2 +(√3))))=(((√3)s^2 −1)/(s^2 +(√3))) i)s^2 −(√3)≠0⇒s≠(3)^(1/4) . ii)(√3)s^2 −1≥0⇒s^2 ≥(1/(√3))⇒s≥(1/(3)^(1/4) ) . iii)s=(√3)⇒t^2 =(((√3)×3−1)/(3+(√3)))=((3(√3)−1)/(3+(√3))) . t^2 =((9(√3)−9−3+(√3))/6)=(1/3)(5(√3)−6) t=(√(((5(√3)−6)/3) )) .(if ∡AoB=(π/6) only part#1)

$$\mathrm{2}\left({s}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}\right)=\sqrt{\mathrm{3}}\sqrt{{s}^{\mathrm{4}} +\mathrm{1}}\sqrt{{t}^{\mathrm{4}} +\mathrm{1}} \\ $$$$\mathrm{4}\left({s}^{\mathrm{4}} {t}^{\mathrm{4}} +\mathrm{2}{s}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{3}\left({s}^{\mathrm{4}} +\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{1}\right) \\ $$$$\mathrm{4}{s}^{\mathrm{4}} {t}^{\mathrm{4}} +\mathrm{8}{s}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{4}=\mathrm{3}{s}^{\mathrm{4}} {t}^{\mathrm{4}} +\mathrm{3}{s}^{\mathrm{4}} +\mathrm{3}{t}^{\mathrm{4}} +\mathrm{3} \\ $$$$\left({s}^{\mathrm{4}} −\mathrm{3}\right){t}_{} ^{\mathrm{4}} +\mathrm{8}{s}^{\mathrm{2}} {t}^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{3}{s}^{\mathrm{4}} \right)=\mathrm{0} \\ $$$${t}^{\mathrm{2}} =\frac{−\mathrm{8}{s}^{\mathrm{2}} +\sqrt{\mathrm{64}{s}^{\mathrm{4}} −\mathrm{4}\left({s}^{\mathrm{4}} −\mathrm{3}\right)\left(\mathrm{1}−\mathrm{3}{s}^{\mathrm{4}} \right)}}{\mathrm{2}\left({s}^{\mathrm{4}} −\mathrm{3}\right)} \\ $$$$\mathrm{64}{s}^{\mathrm{4}} −\mathrm{4}\left({s}^{\mathrm{4}} −\mathrm{3}{s}^{\mathrm{8}} −\mathrm{3}+\mathrm{9}{s}^{\mathrm{4}} \right)=\mathrm{24}{s}^{\mathrm{4}} +\mathrm{12}{s}^{\mathrm{8}} +\mathrm{12} \\ $$$$=\mathrm{12}\left({s}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} =\frac{−\mathrm{8}{s}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{3}}\left({s}^{\mathrm{4}} +\mathrm{1}\right)}{\mathrm{2}\left({s}^{\mathrm{4}} −\mathrm{3}\right)}=\frac{\sqrt{\mathrm{3}}\left({s}^{\mathrm{4}} +\mathrm{1}\right)−\mathrm{4}{s}^{\mathrm{2}} }{{s}^{\mathrm{4}} −\mathrm{3}} \\ $$$${t}^{\mathrm{2}} =\frac{\left({s}^{\mathrm{2}} −\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{3}}{s}^{\mathrm{2}} −\mathrm{1}\right)}{\left({s}^{\mathrm{2}} −\sqrt{\mathrm{3}}\right)\left({s}^{\mathrm{2}} +\sqrt{\mathrm{3}}\right)}=\frac{\sqrt{\mathrm{3}}{s}^{\mathrm{2}} −\mathrm{1}}{{s}^{\mathrm{2}} +\sqrt{\mathrm{3}}} \\ $$$$\left.{i}\right){s}^{\mathrm{2}} −\sqrt{\mathrm{3}}\neq\mathrm{0}\Rightarrow{s}\neq\sqrt[{\mathrm{4}}]{\mathrm{3}}. \\ $$$$\left.{ii}\right)\sqrt{\mathrm{3}}{s}^{\mathrm{2}} −\mathrm{1}\geqslant\mathrm{0}\Rightarrow{s}^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\Rightarrow{s}\geqslant\frac{\mathrm{1}}{\sqrt[{\mathrm{4}}]{\mathrm{3}}}\:\:. \\ $$$$\left.{iii}\right){s}=\sqrt{\mathrm{3}}\Rightarrow{t}^{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}×\mathrm{3}−\mathrm{1}}{\mathrm{3}+\sqrt{\mathrm{3}}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{3}+\sqrt{\mathrm{3}}}\:\:. \\ $$$${t}^{\mathrm{2}} =\frac{\mathrm{9}\sqrt{\mathrm{3}}−\mathrm{9}−\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{6}\right) \\ $$$${t}=\sqrt{\frac{\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{3}}\:\:}\:\:.\left({if}\:\measuredangle{AoB}=\frac{\pi}{\mathrm{6}}\:{only}\:{part}#\mathrm{1}\right) \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com