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

Commented by ajfour last updated on 18/Jul/20

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

$${If}\:{both}\:{coloured}\:{regions}\:{have}\:{equal} \\ $$$${area},\:{find}\:{x}_{{A}} ,\:{x}_{{B}} \:. \\ $$

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

$${B}\left(−{q},{q}^{\mathrm{2}} \right) \\ $$$${A}\left({p},{p}^{\mathrm{2}} \right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{tan}\:\theta}=−\mathrm{2}{q}\:\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{2}{q}} \\ $$$${eqn}.\:{of}\:{BA}: \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}{q}}\left({x}+{q}\right)+{q}^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}{q}}\left({p}+{q}\right)+{q}^{\mathrm{2}} \\ $$$$\mathrm{2}{qp}^{\mathrm{2}} −{p}−{q}\left(\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}={q}+\frac{\mathrm{1}}{\mathrm{2}{q}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}=\mathrm{2}{p}=\mathrm{2}{q}+\frac{\mathrm{1}}{{q}}=\frac{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}{{q}} \\ $$$$\Rightarrow\mathrm{tan}\:\varphi=−\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}} \\ $$$${eqn}.\:{of}\:{AC}: \\ $$$${y}=−\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}\left({x}−{p}\right)+{p}^{\mathrm{2}} \\ $$$${y}=−\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}\left({x}−{q}−\frac{\mathrm{1}}{\mathrm{2}{q}}\right)+\left({q}+\frac{\mathrm{1}}{\mathrm{2}{q}}\right)^{\mathrm{2}} \\ $$$${area}\:{under}\:{AC}={A}_{\mathrm{1}} \\ $$$${area}\:{under}\:{BA}={A}_{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}{x}−\left(\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}+{q}+\frac{\mathrm{1}}{\mathrm{2}{q}}\right)\left({q}+\frac{\mathrm{1}}{\mathrm{2}{q}}\right)=\mathrm{0} \\ $$$$\Delta=\left(\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}+{q}+\frac{\mathrm{1}}{\mathrm{2}{q}}\right)\left({q}+\frac{\mathrm{1}}{\mathrm{2}{q}}\right) \\ $$$$=\left(\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{q}+\frac{\mathrm{1}}{{q}}\right)^{\mathrm{2}} \\ $$$$\mathrm{6}{A}_{\mathrm{1}} =\left(\Delta\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\left(\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{q}+\frac{\mathrm{1}}{{q}}\right)^{\mathrm{3}} \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}{q}}{x}−\left(\frac{\mathrm{1}}{\mathrm{2}}+{q}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{4}{q}^{\mathrm{2}} }+\mathrm{2}+\mathrm{4}{q}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}{q}}+\mathrm{2}{q}\right)^{\mathrm{2}} \\ $$$$\mathrm{6}{A}_{\mathrm{2}} =\left(\Delta\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\left(\frac{\mathrm{1}}{\mathrm{2}{q}}+\mathrm{2}{q}\right)^{\mathrm{3}} \\ $$$${A}_{\mathrm{1}} =\mathrm{2}{A}_{\mathrm{2}} \\ $$$$\Rightarrow\frac{{q}}{\mathrm{2}{q}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}{q}+\frac{\mathrm{1}}{{q}}=\sqrt[{\mathrm{3}}]{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}{q}}+\mathrm{2}{q}\right) \\ $$$$\Rightarrow\mathrm{8}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right){q}^{\mathrm{4}} −\mathrm{2}\left(\mathrm{5}−\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}\right){q}^{\mathrm{2}} −\left(\mathrm{2}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow{q}=\sqrt{\frac{\mathrm{2}−\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\mathrm{2}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)}}\approx\mathrm{1}.\mathrm{193173} \\ $$

Commented by mr W last updated on 18/Jul/20

Commented by ajfour last updated on 19/Jul/20

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

$${Fabulous}\:{Sir},\:{let}\:{me}\:{some}\:{time} \\ $$$${comprehending}\:{it}\:{entirely}.. \\ $$

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