Question Number 102838 by ajfour last updated on 11/Jul/20

Commented byajfour last updated on 11/Jul/20

If regions 1 and 2 have equal areas, find OB=y in terms of OA=a .

Commented byajfour last updated on 11/Jul/20

(x/a)+(y/b)=1 ⇒ bx+ay=ab

Answered by mr W last updated on 11/Jul/20

OB=b intersection at P(p,p^2 ) (p/a)+(p^2 /b)=1 ⇒b=((ap^2 )/(a−p)) (p^3 /3)+((p^2 (a−p))/2)=(1/2)((a/(a−p)))^2 ((p^2 (a−p))/2) ⇒2p^2 −8ap+3a^2 =0 ⇒p=(((4−(√(10)))a)/2) ⇒b=((5(√(10))−14)/6)a^2 ≈0.302a^2

Commented bymr W last updated on 11/Jul/20

Commented byajfour last updated on 11/Jul/20

Sir, (p^3 /3)+((p^2 (a−p))/2)=((ab)/4)=((a(ap^2 ))/(4(a−p))) yes its the same thing, (p/3)+((a−p)/2)=(a^2 /(4(a−p))) 2(3a−p)(a−p)=3a^2 2p^2 −8ap+3a^2 =0 p=(((8−(√(64−24)))/4))a = (((4−(√(10)))/2))a (b/a^2 )=(((26−8(√(10)))/4))(2/(((√(10))−2))) ⇒ b=(((13−4(√(10)))((√(10))+2))/6)a^2 b=(((5(√(10))−14)/6))a^2 . (Great; thanks Sir, no error! )