Question Number 102260 by ajfour last updated on 07/Jul/20

Commented byajfour last updated on 07/Jul/20

Find r in terms of a and b (ellipse parameters).

Commented bymr W last updated on 07/Jul/20

won′t be easy sir...

Answered by mr W last updated on 07/Jul/20

μ=(b/a) touching point P(a cos θ, b sin θ) tan ϕ=y′=(dy/dθ)×(1/(dx/dθ))=−((b cos θ)/(a sin θ))=−(μ/(tan θ)) tan φ=−(1/(tan ϕ))=((tan θ)/μ) center of circle 1 (r, h) r=a cos θ_1 +r cos φ_1 ⇒r=a(1/(√(1+tan^2 θ_1 )))+r(μ/(√(μ^2 +tan^2 θ_1 ))) ⇒(1/(√(1+tan^2 θ_1 )))=(r/a)(1−(μ/(√(μ^2 +tan^2 θ_1 )))) ⇒(1/(√(1+tan^2 θ_1 )))=ξ(1−(μ/(√(μ^2 +tan^2 θ_1 )))) ⇒tan θ_1 =f_1 (ξ) ... h=b sin θ_1 +r sin φ_1 ⇒(h/a)=μ sin θ_1 +ξ((tan θ_1 )/(√(μ^2 +tan^2 θ_1 ))) ⇒(h/a)=((μ tan θ_1 )/(√(1+tan^2 θ_1 )))+((ξ tan θ_1 )/(√(μ^2 +tan^2 θ_1 ))) center of circle 1 (k, r) ⇒r=b sin θ_2 +r((tan θ_2 )/(√(μ^2 +tan^2 θ_2 ))) ⇒(μ/(√(1+tan^2 θ_2 )))=ξ((1/(tan θ_2 ))−(1/(√(μ^2 +tan^2 θ_2 )))) ⇒tan θ_2 =f_2 (ξ) ... ⇒k=a cos θ_2 +r(μ/(√(μ^2 +tan^2 θ_2 ))) ⇒(k/a)=(1/(√(1+tan^2 θ_2 )))+((μξ)/(√(μ^2 +tan^2 θ_2 ))) (k−r)^2 +(r−h)^2 =(2r)^2 ((k/a)−ξ)^2 +((h/a)−ξ)^2 =4ξ^2 ((1/(√(1+tan^2 θ_2 )))+((μξ)/(√(μ^2 +tan^2 θ_2 )))−ξ)^2 +(((μ tan θ_1 )/(√(1+tan^2 θ_1 )))+((ξ tan θ_1 )/(√(μ^2 +tan^2 θ_1 )))−ξ)^2 =4ξ^2 ⇒ξ=(r/a)=...

Commented byajfour last updated on 10/Jul/20

Thanks for the path Sir.. but its really a challenge yet..!

Answered by ajfour last updated on 08/Jul/20

x=acos θ , y=bsin θ −(dx/dy)=((a/b))((sin θ)/(cos θ)) = (a/b)tan θ=((am)/b) = s m =p, q for circle A and circle B. let center of right circle A be (h,r) and that of upper circle B be (r,k). h=acos θ_1 +(r/(√(s^2 +1))) = (a/(√(p^2 +1)))+(r/((((a^2 p^2 )/b^2 )+1))^(1/) ) r=((ap)/(√(p^2 +1)))+((r(((ap)/b)))/((((a^2 p^2 )/b^2 )+1))^(1/) ) similarly r=(a/(√(q^2 +1)))+(r/((((a^2 q^2 )/b^2 )+1))^(1/) ) k=((aq)/(√(q^2 +1)))+((r(((aq)/b)))/((((a^2 q^2 )/b^2 )+1))^(1/) ) Now (h−r)^2 +(k−r)^2 =4r^2 ..........