Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 122399 by ajfour last updated on 16/Nov/20

Commented by ajfour last updated on 16/Nov/20

Find in terms of radii a and b the minimum length of common tangent EF between the axes.

$${Find}\:{in}\:{terms}\:{of}\:{radii}\:{a}\:{and}\:{b} \\ $$$${the}\:{minimum}\:{length}\:{of}\:{common} \\ $$$${tangent}\:{EF}\:{between}\:{the}\:{axes}. \\ $$

Answered by mr W last updated on 16/Nov/20

L=(a/(sin α))+(b/(tan β))+(√((b+a)^2 −(b−a)^2 )) =(a/(tan α))+(b/(tan ((π/4)−α)))+2(√(ab)) =(a/(tan α))+((b(1+tan α))/(1−tan α))+2(√(ab)) L=(a/t)+((2b)/(1−t))−b+2(√(ab)) (dL/dt)=−(a/t^2 )+((2b)/((1−t)^2 ))=0 (1−(1/t))^2 =((2b)/a) ⇒t=tan α=(1/( (√((2b)/a))+1)) L_(min) =a((√((2b)/a))+1)+2b((1/( (√((2b)/a))))+1)−b+2(√(ab)) =a+b+2(1+(√2))(√(ab))

$${L}=\frac{{a}}{\mathrm{sin}\:\alpha}+\frac{{b}}{\mathrm{tan}\:\beta}+\sqrt{\left({b}+{a}\right)^{\mathrm{2}} −\left({b}−{a}\right)^{\mathrm{2}} } \\ $$$$=\frac{{a}}{\mathrm{tan}\:\alpha}+\frac{{b}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\alpha\right)}+\mathrm{2}\sqrt{{ab}} \\ $$$$=\frac{{a}}{\mathrm{tan}\:\alpha}+\frac{{b}\left(\mathrm{1}+\mathrm{tan}\:\alpha\right)}{\mathrm{1}−\mathrm{tan}\:\alpha}+\mathrm{2}\sqrt{{ab}} \\ $$$${L}=\frac{{a}}{\mathrm{t}}+\frac{\mathrm{2}{b}}{\mathrm{1}−{t}}−{b}+\mathrm{2}\sqrt{{ab}} \\ $$$$\frac{{dL}}{{dt}}=−\frac{{a}}{{t}^{\mathrm{2}} }+\frac{\mathrm{2}{b}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} =\frac{\mathrm{2}{b}}{{a}} \\ $$$$\Rightarrow{t}=\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{2}{b}}{{a}}}+\mathrm{1}} \\ $$$${L}_{{min}} ={a}\left(\sqrt{\frac{\mathrm{2}{b}}{{a}}}+\mathrm{1}\right)+\mathrm{2}{b}\left(\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{2}{b}}{{a}}}}+\mathrm{1}\right)−{b}+\mathrm{2}\sqrt{{ab}} \\ $$$$={a}+{b}+\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\sqrt{{ab}} \\ $$

Commented by ajfour last updated on 16/Nov/20

thanks for solving sir!

$${thanks}\:{for}\:{solving}\:{sir}! \\ $$

Answered by ajfour last updated on 16/Nov/20

let L=x+y+2(√(ab)) tan^(−1) (b/x)+tan^(−1) (a/y)=(π/4) (((b/x)+(a/y))/(1−((ab)/(xy))))=1 ⇒ ax+by=xy−ab L=x+a(((x+b)/(x−b)))+2(√(ab)) = x+a+((2ab)/(x−b))+2(√(ab)) (dL/dx)=1−((2ab)/((x−b)^2 ))=0 ⇒ x−b=(√(2ab)) ⇒ x+a=a+b+(√(2ab)) L_(min) = a+b+2(√(2ab))+2(√(ab)) L_(min) = a+b+2((√2)+1)(√(ab))

$${let}\:{L}={x}+{y}+\mathrm{2}\sqrt{{ab}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{x}}+\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{y}}=\frac{\pi}{\mathrm{4}} \\ $$$$\frac{\frac{{b}}{{x}}+\frac{{a}}{{y}}}{\mathrm{1}−\frac{{ab}}{{xy}}}=\mathrm{1}\:\:\Rightarrow\:\:{ax}+{by}={xy}−{ab} \\ $$$$\:{L}={x}+{a}\left(\frac{{x}+{b}}{{x}−{b}}\right)+\mathrm{2}\sqrt{{ab}} \\ $$$$\:\:\:=\:{x}+{a}+\frac{\mathrm{2}{ab}}{{x}−{b}}+\mathrm{2}\sqrt{{ab}} \\ $$$$\frac{{dL}}{{dx}}=\mathrm{1}−\frac{\mathrm{2}{ab}}{\left({x}−{b}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:\:{x}−{b}=\sqrt{\mathrm{2}{ab}}\:\:\: \\ $$$$\Rightarrow\:\:{x}+{a}={a}+{b}+\sqrt{\mathrm{2}{ab}} \\ $$$${L}_{{min}} =\:{a}+{b}+\mathrm{2}\sqrt{\mathrm{2}{ab}}+\mathrm{2}\sqrt{{ab}} \\ $$$${L}_{{min}} =\:{a}+{b}+\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\sqrt{{ab}}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com