Question Number 103986 by ajfour last updated on 18/Jul/20

Commented byajfour last updated on 18/Jul/20

If both coloured regions have equal  area, find x_A , x_B  .

Answered by mr W last updated on 19/Jul/20

B(−q,q^2 )  A(p,p^2 )  −(1/(tan θ))=−2q ⇒tan θ=(1/(2q))  eqn. of BA:  y=(1/(2q))(x+q)+q^2   p^2 =(1/(2q))(p+q)+q^2   2qp^2 −p−q(2q^2 +1)=0  ⇒p=q+(1/(2q))  −(1/(tan ϕ))=2p=2q+(1/q)=((2q^2 +1)/q)  ⇒tan ϕ=−(q/(2q^2 +1))  eqn. of AC:  y=−(q/(2q^2 +1))(x−p)+p^2   y=−(q/(2q^2 +1))(x−q−(1/(2q)))+(q+(1/(2q)))^2   area under AC=A_1   area under BA=A_2   x^2 +(q/(2q^2 +1))x−((q/(2q^2 +1))+q+(1/(2q)))(q+(1/(2q)))=0  Δ=((q/(2q^2 +1)))^2 +4((q/(2q^2 +1))+q+(1/(2q)))(q+(1/(2q)))  =((q/(2q^2 +1))+2q+(1/q))^2   6A_1 =(Δ)^(3/2) =((q/(2q^2 +1))+2q+(1/q))^3     x^2 −(1/(2q))x−((1/2)+q^2 )=0  Δ=(1/(4q^2 ))+2+4q^2 =((1/(2q))+2q)^2   6A_2 =(Δ)^(3/2) =((1/(2q))+2q)^3   A_1 =2A_2   ⇒(q/(2q^2 +1))+2q+(1/q)=(2)^(1/3) ((1/(2q))+2q)  ⇒8((2)^(1/3) −1)q^4 −2(5−3(2)^(1/3) )q^2 −(2−(2)^(1/3) )=0  ⇒q=(√((2−(2)^(1/3) )/(2((2)^(1/3) −1))))≈1.193173

Commented bymr W last updated on 18/Jul/20

Commented byajfour last updated on 19/Jul/20

Fabulous Sir, let me some time  comprehending it entirely..