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Question-44763




Question Number 44763 by ajfour last updated on 04/Oct/18
Commented by ajfour last updated on 04/Oct/18
Equation of ellipse :  (x^2 /a^2 )+(y^2 /b^2 ) = 1.  Find r in terms of a, b.
$${Equation}\:{of}\:{ellipse}\::\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}. \\ $$$${Find}\:\boldsymbol{{r}}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}},\:\boldsymbol{{b}}. \\ $$
Answered by ajfour last updated on 04/Oct/18
   let point of contact of circle  and ellipse be (x_0 , y_0 )  let     y_0 = bsin 𝛗 = r(1+sin 𝛉)  ...(i)     x_0 = acos 𝛗 = r(1+cos 𝛉) ...(ii)    slope of common tangent:     ((bcos 𝛗)/(βˆ’asin 𝛗)) = βˆ’cot 𝛉  β‡’  bcos 𝛗sin 𝛉 = asin 𝛗cos 𝛉  ..(iii)  β‡’  tan 𝛗 = ((bsin ΞΈ)/(acos ΞΈ)) = (a/b)(((1+sin 𝛉)/(1+cos 𝛉)))  we must have a double root for  this eq. in ΞΈ ;  let tan 𝛉 = m  let  tan (ΞΈ/2) = t , then    2t = m(1βˆ’t^2 )   and also get     (b/a)Γ—((2t)/(1βˆ’t^2 )) = (a/b)Γ—((1+t^2 +2t)/2)  β‡’  4b^2 t = a^2 (1βˆ’t^2 )(1+t)^2   β‡’ 4b^2 t = a^2 (1+2t+t^2 βˆ’t^2 βˆ’2t^3 βˆ’t^4 )  let  ((4b^2 )/a^2 )βˆ’2= 𝛒   (ρ > 0 if  b(√2) > a )  β‡’ t^4 +2t^3 +ρtβˆ’1=0  this eq. must have two real   roots corresponding to tan ΞΈ = m  and two complex roots;   say    (mt^2 +2tβˆ’m)(t^2 +pt+q)=0   t^4 +2t^3 +ρtβˆ’1 ⇔ (t^2 +((2t)/m)βˆ’1)(t^2 +pt+q)  β‡’ for t=0   βˆ’q = βˆ’1     ....(1)   for t=1    2+𝛒 = (2/m)(1+p+q)  ...(2)    for t=βˆ’1    βˆ’2βˆ’π›’ =βˆ’(2/m)(1βˆ’p+q)   ...(3)  from  eq.(2)  &  eq.(3) we infer     p = 0 and therefore from (2)           (2/m)(1+0+1) = 2+𝛒  β‡’  m= tan ΞΈ = (4/(2+ρ)) = (a^2 /b^2 )  And as   mt^2 +2tβˆ’m = 0  with appropriate t=tan (ΞΈ/2)  being +ve  and < 1 , we conclude          t = βˆ’(1/m)+(√(1+(1/m^2 )))    and  with  m= (a^2 /b^2 )       t = βˆ’(b^2 /a^2 )+((√(a^4 +b^4 ))/a^2 )  β‡’    from  (i) & (ii)  (in the beginning),     r^2 [(((1+sin ΞΈ)^2 )/b^2 )+(((1+cos ΞΈ)^2 )/a^2 )]=1   β‡’ r^2  = ((a^2 b^2 )/(a^2 (1+sin 𝛉)^2 +b^2 (1+cos 𝛉)^2 ))    = ((a^2 b^2 )/(a^2 (1+((2t)/(1+t^2 )))^2 +b^2 (1+((1βˆ’t^2 )/(1+t^2 )))^2 ))     { ((r^2 = ((a^2 b^2 (1+t^2 )^2 )/(a^2 (1+t)^4 +4b^2 )))),((with    t=βˆ’(b^2 /a^2 )+((√(a^4 +b^4 ))/a^2 ))) :}   As a check , if  a=b=R  we have   t= (√2)βˆ’1     and    r = ((ab(1+t^2 ))/( (√(a^2 (1+t)^4 +4b^2 ))))    β‡’  r = (((4βˆ’2(√2))R)/(2(√2)))  = (((2βˆ’(√2))R)/( (√2)))              = ((2R)/( (√2)(2+(√2)))) = (R/(1+(√2)))       (as it should be !) .
$$\:\:\:{let}\:{point}\:{of}\:{contact}\:{of}\:{circle} \\ $$$${and}\:{ellipse}\:{be}\:\left({x}_{\mathrm{0}} ,\:{y}_{\mathrm{0}} \right) \\ $$$$\boldsymbol{{let}} \\ $$$$\:\:\:\boldsymbol{{y}}_{\mathrm{0}} =\:\boldsymbol{{b}}\mathrm{sin}\:\boldsymbol{\phi}\:=\:\boldsymbol{{r}}\left(\mathrm{1}+\mathrm{sin}\:\boldsymbol{\theta}\right)\:\:…\left({i}\right) \\ $$$$\:\:\:\boldsymbol{{x}}_{\mathrm{0}} =\:\boldsymbol{{a}}\mathrm{cos}\:\boldsymbol{\phi}\:=\:\boldsymbol{{r}}\left(\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}\right)\:…\left({ii}\right) \\ $$$$\:\:{slope}\:{of}\:{common}\:{tangent}: \\ $$$$\:\:\:\frac{\boldsymbol{{b}}\mathrm{cos}\:\boldsymbol{\phi}}{βˆ’\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\phi}}\:=\:βˆ’\mathrm{cot}\:\boldsymbol{\theta} \\ $$$$\Rightarrow\:\:\boldsymbol{{b}}\mathrm{cos}\:\boldsymbol{\phi}\mathrm{sin}\:\boldsymbol{\theta}\:=\:\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\phi}\mathrm{cos}\:\boldsymbol{\theta}\:\:..\left({iii}\right) \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\boldsymbol{\phi}\:=\:\frac{{b}\mathrm{sin}\:\theta}{{a}\mathrm{cos}\:\theta}\:=\:\frac{{a}}{{b}}\left(\frac{\mathrm{1}+\mathrm{sin}\:\boldsymbol{\theta}}{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}}\right) \\ $$$${we}\:{must}\:{have}\:{a}\:{double}\:{root}\:{for} \\ $$$${this}\:{eq}.\:{in}\:\theta\:;\:\:{let}\:\mathrm{tan}\:\boldsymbol{\theta}\:=\:\boldsymbol{{m}} \\ $$$${let}\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:=\:{t}\:,\:{then} \\ $$$$\:\:\mathrm{2}\boldsymbol{{t}}\:=\:\boldsymbol{{m}}\left(\mathrm{1}βˆ’\boldsymbol{{t}}^{\mathrm{2}} \right)\:\:\:{and}\:{also}\:{get} \\ $$$$\:\:\:\frac{{b}}{{a}}Γ—\frac{\mathrm{2}{t}}{\mathrm{1}βˆ’{t}^{\mathrm{2}} }\:=\:\frac{{a}}{{b}}Γ—\frac{\mathrm{1}+{t}^{\mathrm{2}} +\mathrm{2}{t}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{4}{b}^{\mathrm{2}} {t}\:=\:{a}^{\mathrm{2}} \left(\mathrm{1}βˆ’{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{4}{b}^{\mathrm{2}} {t}\:=\:{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}{t}+{t}^{\mathrm{2}} βˆ’{t}^{\mathrm{2}} βˆ’\mathrm{2}{t}^{\mathrm{3}} βˆ’{t}^{\mathrm{4}} \right) \\ $$$${let}\:\:\frac{\mathrm{4}\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} }βˆ’\mathrm{2}=\:\boldsymbol{\rho}\:\:\:\left(\rho\:>\:\mathrm{0}\:{if}\:\:{b}\sqrt{\mathrm{2}}\:>\:{a}\:\right) \\ $$$$\Rightarrow\:{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +\rho{t}βˆ’\mathrm{1}=\mathrm{0} \\ $$$${this}\:{eq}.\:{must}\:{have}\:{two}\:{real} \\ $$$$\:{roots}\:{corresponding}\:{to}\:\mathrm{tan}\:\theta\:=\:{m} \\ $$$${and}\:{two}\:{complex}\:{roots}; \\ $$$$\:{say}\:\:\:\:\left({mt}^{\mathrm{2}} +\mathrm{2}{t}βˆ’{m}\right)\left({t}^{\mathrm{2}} +{pt}+{q}\right)=\mathrm{0} \\ $$$$\:{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +\rho{t}βˆ’\mathrm{1}\:\Leftrightarrow\:\left({t}^{\mathrm{2}} +\frac{\mathrm{2}{t}}{{m}}βˆ’\mathrm{1}\right)\left({t}^{\mathrm{2}} +{pt}+{q}\right) \\ $$$$\Rightarrow\:{for}\:{t}=\mathrm{0} \\ $$$$\:βˆ’\boldsymbol{{q}}\:=\:βˆ’\mathrm{1}\:\:\:\:\:….\left(\mathrm{1}\right) \\ $$$$\:{for}\:{t}=\mathrm{1} \\ $$$$\:\:\mathrm{2}+\boldsymbol{\rho}\:=\:\frac{\mathrm{2}}{\boldsymbol{{m}}}\left(\mathrm{1}+\boldsymbol{{p}}+\boldsymbol{{q}}\right)\:\:…\left(\mathrm{2}\right) \\ $$$$\:\:{for}\:{t}=βˆ’\mathrm{1} \\ $$$$\:\:βˆ’\mathrm{2}βˆ’\boldsymbol{\rho}\:=βˆ’\frac{\mathrm{2}}{\boldsymbol{{m}}}\left(\mathrm{1}βˆ’\boldsymbol{{p}}+\boldsymbol{{q}}\right)\:\:\:…\left(\mathrm{3}\right) \\ $$$${from}\:\:{eq}.\left(\mathrm{2}\right)\:\:\&\:\:{eq}.\left(\mathrm{3}\right)\:{we}\:{infer} \\ $$$$\:\:\:\boldsymbol{{p}}\:=\:\mathrm{0}\:{and}\:{therefore}\:{from}\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\boldsymbol{{m}}}\left(\mathrm{1}+\mathrm{0}+\mathrm{1}\right)\:=\:\mathrm{2}+\boldsymbol{\rho} \\ $$$$\Rightarrow\:\:{m}=\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{4}}{\mathrm{2}+\rho}\:=\:\frac{\boldsymbol{{a}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} } \\ $$$${And}\:{as}\:\:\:\boldsymbol{{mt}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{t}}βˆ’\boldsymbol{{m}}\:=\:\mathrm{0} \\ $$$${with}\:{appropriate}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$${being}\:+{ve}\:\:{and}\:<\:\mathrm{1}\:,\:{we}\:{conclude} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{t}}\:=\:βˆ’\frac{\mathrm{1}}{\boldsymbol{{m}}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{m}}^{\mathrm{2}} }} \\ $$$$\:\:{and}\:\:{with}\:\:\boldsymbol{{m}}=\:\frac{\boldsymbol{{a}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:{t}\:=\:βˆ’\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\sqrt{{a}^{\mathrm{4}} +{b}^{\mathrm{4}} }}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\: \\ $$$${from}\:\:\left({i}\right)\:\&\:\left({ii}\right)\:\:\left({in}\:{the}\:{beginning}\right), \\ $$$$\:\:\:\boldsymbol{{r}}^{\mathrm{2}} \left[\frac{\left(\mathrm{1}+\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]=\mathrm{1}\: \\ $$$$\Rightarrow\:\boldsymbol{{r}}^{\mathrm{2}} \:=\:\frac{\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}\:\boldsymbol{\theta}\right)^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}\right)^{\mathrm{2}} } \\ $$$$\:\:=\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}βˆ’{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\begin{cases}{\boldsymbol{{r}}^{\mathrm{2}} =\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{2}} }}\\{{with}\:\:\:\:{t}=βˆ’\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\sqrt{{a}^{\mathrm{4}} +{b}^{\mathrm{4}} }}{{a}^{\mathrm{2}} }}\end{cases} \\ $$$$\:{As}\:{a}\:{check}\:,\:{if}\:\:{a}={b}=\boldsymbol{{R}} \\ $$$${we}\:{have}\:\:\:\boldsymbol{{t}}=\:\sqrt{\mathrm{2}}βˆ’\mathrm{1} \\ $$$$\:\:\:{and}\:\:\:\:\boldsymbol{{r}}\:=\:\frac{{ab}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\:\sqrt{{a}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{2}} }} \\ $$$$\:\:\Rightarrow\:\:{r}\:=\:\frac{\left(\mathrm{4}βˆ’\mathrm{2}\sqrt{\mathrm{2}}\right){R}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:=\:\frac{\left(\mathrm{2}βˆ’\sqrt{\mathrm{2}}\right){R}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}{R}}{\:\sqrt{\mathrm{2}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}\:=\:\frac{{R}}{\mathrm{1}+\sqrt{\mathrm{2}}}\: \\ $$$$\:\:\:\:\left({as}\:{it}\:{should}\:{be}\:!\right)\:. \\ $$
Commented by MrW3 last updated on 04/Oct/18
fantastic solution sir!
$${fantastic}\:{solution}\:{sir}! \\ $$

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