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! )