Question Number 44763 by ajfour last updated on 04/Oct/18
Commented by ajfour last updated on 04/Oct/18
$${Equation}\:{of}\:{ellipse}\::\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}. \\ $$$${Find}\:\boldsymbol{{r}}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}},\:\boldsymbol{{b}}. \\ $$
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 π = r(1+sin π) ...(i) x_0 = acos π = r(1+cos π) ...(ii) slope of common tangent: ((bcos π)/(βasin π)) = βcot π β bcos πsin π = asin πcos π ..(iii) β tan π = ((bsin ΞΈ)/(acos ΞΈ)) = (a/b)(((1+sin π)/(1+cos π))) we must have a double root for this eq. in ΞΈ ; let tan π = m let tan (ΞΈ/2) = t , then 2t = m(1βt^2 ) and also get (b/a)Γ((2t)/(1βt^2 )) = (a/b)Γ((1+t^2 +2t)/2) β 4b^2 t = a^2 (1βt^2 )(1+t)^2 β 4b^2 t = a^2 (1+2t+t^2 βt^2 β2t^3 βt^4 ) let ((4b^2 )/a^2 )β2= π (Ο > 0 if b(β2) > a ) β t^4 +2t^3 +Οtβ1=0 this eq. must have two real roots corresponding to tan ΞΈ = m and two complex roots; say (mt^2 +2tβm)(t^2 +pt+q)=0 t^4 +2t^3 +Οtβ1 β (t^2 +((2t)/m)β1)(t^2 +pt+q) β for t=0 βq = β1 ....(1) for t=1 2+π = (2/m)(1+p+q) ...(2) for t=β1 β2βπ =β(2/m)(1βp+q) ...(3) from eq.(2) & eq.(3) we infer p = 0 and therefore from (2) (2/m)(1+0+1) = 2+π β m= tan ΞΈ = (4/(2+Ο)) = (a^2 /b^2 ) And as mt^2 +2tβm = 0 with appropriate t=tan (ΞΈ/2) being +ve and < 1 , we conclude t = β(1/m)+(β(1+(1/m^2 ))) and with m= (a^2 /b^2 ) t = β(b^2 /a^2 )+((β(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 π)^2 +b^2 (1+cos π)^2 )) = ((a^2 b^2 )/(a^2 (1+((2t)/(1+t^2 )))^2 +b^2 (1+((1βt^2 )/(1+t^2 )))^2 )) { ((r^2 = ((a^2 b^2 (1+t^2 )^2 )/(a^2 (1+t)^4 +4b^2 )))),((with t=β(b^2 /a^2 )+((β(a^4 +b^4 ))/a^2 ))) :} As a check , if a=b=R we have t= (β2)β1 and r = ((ab(1+t^2 ))/( (β(a^2 (1+t)^4 +4b^2 )))) β r = (((4β2(β2))R)/(2(β2))) = (((2β(β2))R)/( (β2))) = ((2R)/( (β2)(2+(β2)))) = (R/(1+(β2))) (as it should be !) .](https://www.tinkutara.com/question/Q44769.png)
$$\:\:\:{let}\:{point}\:{of}\:{contact}\:{of}\:{circle} \\ $$$${and}\:{ellipse}\:{be}\:\left({x}_{\mathrm{0}} ,\:{y}_{\mathrm{0}} \right) \\ $$$$\boldsymbol{{let}} \\ $$$$\:\:\:\boldsymbol{{y}}_{\mathrm{0}} =\:\boldsymbol{{b}}\mathrm{sin}\:\boldsymbol{\phi}\:=\:\boldsymbol{{r}}\left(\mathrm{1}+\mathrm{sin}\:\boldsymbol{\theta}\right)\:\:…\left({i}\right) \\ $$$$\:\:\:\boldsymbol{{x}}_{\mathrm{0}} =\:\boldsymbol{{a}}\mathrm{cos}\:\boldsymbol{\phi}\:=\:\boldsymbol{{r}}\left(\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}\right)\:…\left({ii}\right) \\ $$$$\:\:{slope}\:{of}\:{common}\:{tangent}: \\ $$$$\:\:\:\frac{\boldsymbol{{b}}\mathrm{cos}\:\boldsymbol{\phi}}{β\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\phi}}\:=\:β\mathrm{cot}\:\boldsymbol{\theta} \\ $$$$\Rightarrow\:\:\boldsymbol{{b}}\mathrm{cos}\:\boldsymbol{\phi}\mathrm{sin}\:\boldsymbol{\theta}\:=\:\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\phi}\mathrm{cos}\:\boldsymbol{\theta}\:\:..\left({iii}\right) \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\boldsymbol{\phi}\:=\:\frac{{b}\mathrm{sin}\:\theta}{{a}\mathrm{cos}\:\theta}\:=\:\frac{{a}}{{b}}\left(\frac{\mathrm{1}+\mathrm{sin}\:\boldsymbol{\theta}}{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}}\right) \\ $$$${we}\:{must}\:{have}\:{a}\:{double}\:{root}\:{for} \\ $$$${this}\:{eq}.\:{in}\:\theta\:;\:\:{let}\:\mathrm{tan}\:\boldsymbol{\theta}\:=\:\boldsymbol{{m}} \\ $$$${let}\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:=\:{t}\:,\:{then} \\ $$$$\:\:\mathrm{2}\boldsymbol{{t}}\:=\:\boldsymbol{{m}}\left(\mathrm{1}β\boldsymbol{{t}}^{\mathrm{2}} \right)\:\:\:{and}\:{also}\:{get} \\ $$$$\:\:\:\frac{{b}}{{a}}Γ\frac{\mathrm{2}{t}}{\mathrm{1}β{t}^{\mathrm{2}} }\:=\:\frac{{a}}{{b}}Γ\frac{\mathrm{1}+{t}^{\mathrm{2}} +\mathrm{2}{t}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{4}{b}^{\mathrm{2}} {t}\:=\:{a}^{\mathrm{2}} \left(\mathrm{1}β{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{4}{b}^{\mathrm{2}} {t}\:=\:{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}{t}+{t}^{\mathrm{2}} β{t}^{\mathrm{2}} β\mathrm{2}{t}^{\mathrm{3}} β{t}^{\mathrm{4}} \right) \\ $$$${let}\:\:\frac{\mathrm{4}\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} }β\mathrm{2}=\:\boldsymbol{\rho}\:\:\:\left(\rho\:>\:\mathrm{0}\:{if}\:\:{b}\sqrt{\mathrm{2}}\:>\:{a}\:\right) \\ $$$$\Rightarrow\:{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +\rho{t}β\mathrm{1}=\mathrm{0} \\ $$$${this}\:{eq}.\:{must}\:{have}\:{two}\:{real} \\ $$$$\:{roots}\:{corresponding}\:{to}\:\mathrm{tan}\:\theta\:=\:{m} \\ $$$${and}\:{two}\:{complex}\:{roots}; \\ $$$$\:{say}\:\:\:\:\left({mt}^{\mathrm{2}} +\mathrm{2}{t}β{m}\right)\left({t}^{\mathrm{2}} +{pt}+{q}\right)=\mathrm{0} \\ $$$$\:{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +\rho{t}β\mathrm{1}\:\Leftrightarrow\:\left({t}^{\mathrm{2}} +\frac{\mathrm{2}{t}}{{m}}β\mathrm{1}\right)\left({t}^{\mathrm{2}} +{pt}+{q}\right) \\ $$$$\Rightarrow\:{for}\:{t}=\mathrm{0} \\ $$$$\:β\boldsymbol{{q}}\:=\:β\mathrm{1}\:\:\:\:\:….\left(\mathrm{1}\right) \\ $$$$\:{for}\:{t}=\mathrm{1} \\ $$$$\:\:\mathrm{2}+\boldsymbol{\rho}\:=\:\frac{\mathrm{2}}{\boldsymbol{{m}}}\left(\mathrm{1}+\boldsymbol{{p}}+\boldsymbol{{q}}\right)\:\:…\left(\mathrm{2}\right) \\ $$$$\:\:{for}\:{t}=β\mathrm{1} \\ $$$$\:\:β\mathrm{2}β\boldsymbol{\rho}\:=β\frac{\mathrm{2}}{\boldsymbol{{m}}}\left(\mathrm{1}β\boldsymbol{{p}}+\boldsymbol{{q}}\right)\:\:\:…\left(\mathrm{3}\right) \\ $$$${from}\:\:{eq}.\left(\mathrm{2}\right)\:\:\&\:\:{eq}.\left(\mathrm{3}\right)\:{we}\:{infer} \\ $$$$\:\:\:\boldsymbol{{p}}\:=\:\mathrm{0}\:{and}\:{therefore}\:{from}\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\boldsymbol{{m}}}\left(\mathrm{1}+\mathrm{0}+\mathrm{1}\right)\:=\:\mathrm{2}+\boldsymbol{\rho} \\ $$$$\Rightarrow\:\:{m}=\:\mathrm{tan}\:\theta\:=\:\frac{\mathrm{4}}{\mathrm{2}+\rho}\:=\:\frac{\boldsymbol{{a}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} } \\ $$$${And}\:{as}\:\:\:\boldsymbol{{mt}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{t}}β\boldsymbol{{m}}\:=\:\mathrm{0} \\ $$$${with}\:{appropriate}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$${being}\:+{ve}\:\:{and}\:<\:\mathrm{1}\:,\:{we}\:{conclude} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{t}}\:=\:β\frac{\mathrm{1}}{\boldsymbol{{m}}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{{m}}^{\mathrm{2}} }} \\ $$$$\:\:{and}\:\:{with}\:\:\boldsymbol{{m}}=\:\frac{\boldsymbol{{a}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:{t}\:=\:β\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\sqrt{{a}^{\mathrm{4}} +{b}^{\mathrm{4}} }}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\: \\ $$$${from}\:\:\left({i}\right)\:\&\:\left({ii}\right)\:\:\left({in}\:{the}\:{beginning}\right), \\ $$$$\:\:\:\boldsymbol{{r}}^{\mathrm{2}} \left[\frac{\left(\mathrm{1}+\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right]=\mathrm{1}\: \\ $$$$\Rightarrow\:\boldsymbol{{r}}^{\mathrm{2}} \:=\:\frac{\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{a}}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}\:\boldsymbol{\theta}\right)^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\boldsymbol{\theta}\right)^{\mathrm{2}} } \\ $$$$\:\:=\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}β{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\begin{cases}{\boldsymbol{{r}}^{\mathrm{2}} =\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{2}} }}\\{{with}\:\:\:\:{t}=β\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\sqrt{{a}^{\mathrm{4}} +{b}^{\mathrm{4}} }}{{a}^{\mathrm{2}} }}\end{cases} \\ $$$$\:{As}\:{a}\:{check}\:,\:{if}\:\:{a}={b}=\boldsymbol{{R}} \\ $$$${we}\:{have}\:\:\:\boldsymbol{{t}}=\:\sqrt{\mathrm{2}}β\mathrm{1} \\ $$$$\:\:\:{and}\:\:\:\:\boldsymbol{{r}}\:=\:\frac{{ab}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\:\sqrt{{a}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)^{\mathrm{4}} +\mathrm{4}{b}^{\mathrm{2}} }} \\ $$$$\:\:\Rightarrow\:\:{r}\:=\:\frac{\left(\mathrm{4}β\mathrm{2}\sqrt{\mathrm{2}}\right){R}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:=\:\frac{\left(\mathrm{2}β\sqrt{\mathrm{2}}\right){R}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}{R}}{\:\sqrt{\mathrm{2}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}\:=\:\frac{{R}}{\mathrm{1}+\sqrt{\mathrm{2}}}\: \\ $$$$\:\:\:\:\left({as}\:{it}\:{should}\:{be}\:!\right)\:. \\ $$
Commented by MrW3 last updated on 04/Oct/18
$${fantastic}\:{solution}\:{sir}! \\ $$