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..