Question Number 49392 by ajfour last updated on 06/Dec/18
Answered by mr W last updated on 08/Dec/18
![let m=tan θ eqn. of line: y=m(x+a) kx^2 =m(x+a) kx^2 −mx−ma=0 x=((m±(√(m(m+4ak))))/(2k)) x_Q =((m−(√(m(m+4ak))))/(2k)) x_P =((m+(√(m(m+4ak))))/(2k)) A_(blue) =∫_x_Q ^x_P (mx+ma−kx^2 )dx =(m/2)(x_P ^2 −x_Q ^2 )+ma(x_P −x_Q )−(k/3)(x_P ^3 −x_Q ^3 ) =(x_P −x_Q ){(m/2)(x_P +x_Q )+ma−(k/3)[(x_P −x_Q )^2 +3x_P x_Q ]} =((√(m(m+4ak)))/k){(m/2)×(m/k)+ma−(k/3)[(m^2 /k^2 )+((ma)/k)]} =(([m(m+4ak)]^(3/2) )/(6k^2 )) A_(yellow) =((ma^2 )/2)−∫_x_Q ^0 (mx+ma−kx^2 )dx =((ma^2 )/2)+(m/2)x_Q ^2 +max_Q −(k/3)x_Q ^3 A_(yellow) =((m(a+x_Q )^2 )/2)+∫_(−x_Q ) ^0 kx^2 dx =(m/2)(a+x_Q )^2 −((kx_Q ^3 )/3) =(m/2)[a+((m−(√(m(m+4ak))))/(2k))]^2 −(k/3)[((m−(√(m(m+4ak))))/(2k))]^3 =(m/(8k^2 ))[m+2ak−(√(m(m+4ak)))]^2 −(1/(24k^2 ))[m−(√(m(m+4ak)))]^3 A_(blue) =A_(yellow) (([m(m+4ak)]^(3/2) )/(6k^2 ))=(m/(8k^2 ))[m+2ak−(√(m(m+4ak)))]^2 −(1/(24k^2 ))[m−(√(m(m+4ak)))]^3 (([m(m+4ak)]^(3/2) )/3)=(m/4)[m+2ak−(√(m(m+4ak)))]^2 −(1/(12))[m−(√(m(m+4ak)))]^3 let λ=(√(m(m+4ak))) ⇒16mλ^3 −(λ−m)^3 (3λ+m)=0 with θ=53° ⇒λ=10.2745 k=(1/(4a))((λ^2 /m)−m)=4.8889](Q49571.png)
$${let}\:{m}=\mathrm{tan}\:\theta \\ $$$${eqn}.\:{of}\:{line}: \\ $$$${y}={m}\left({x}+{a}\right) \\ $$$${kx}^{\mathrm{2}} ={m}\left({x}+{a}\right) \\ $$$${kx}^{\mathrm{2}} −{mx}−{ma}=\mathrm{0} \\ $$$${x}=\frac{{m}\pm\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}}{\mathrm{2}{k}} \\ $$$${x}_{{Q}} =\frac{{m}−\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}}{\mathrm{2}{k}} \\ $$$${x}_{{P}} =\frac{{m}+\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}}{\mathrm{2}{k}} \\ $$$${A}_{{blue}} =\int_{{x}_{{Q}} } ^{{x}_{{P}} } \left({mx}+{ma}−{kx}^{\mathrm{2}} \right){dx} \\ $$$$=\frac{{m}}{\mathrm{2}}\left({x}_{{P}} ^{\mathrm{2}} −{x}_{{Q}} ^{\mathrm{2}} \right)+{ma}\left({x}_{{P}} −{x}_{{Q}} \right)−\frac{{k}}{\mathrm{3}}\left({x}_{{P}} ^{\mathrm{3}} −{x}_{{Q}} ^{\mathrm{3}} \right) \\ $$$$=\left({x}_{{P}} −{x}_{{Q}} \right)\left\{\frac{{m}}{\mathrm{2}}\left({x}_{{P}} +{x}_{{Q}} \right)+{ma}−\frac{{k}}{\mathrm{3}}\left[\left({x}_{{P}} −{x}_{{Q}} \right)^{\mathrm{2}} +\mathrm{3}{x}_{{P}} {x}_{{Q}} \right]\right\} \\ $$$$=\frac{\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}}{{k}}\left\{\frac{{m}}{\mathrm{2}}×\frac{{m}}{{k}}+{ma}−\frac{{k}}{\mathrm{3}}\left[\frac{{m}^{\mathrm{2}} }{{k}^{\mathrm{2}} }+\frac{{ma}}{{k}}\right]\right\} \\ $$$$=\frac{\left[{m}\left({m}+\mathrm{4}{ak}\right)\right]^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{6}{k}^{\mathrm{2}} } \\ $$$$ \\ $$$${A}_{{yellow}} =\frac{{ma}^{\mathrm{2}} }{\mathrm{2}}−\int_{{x}_{{Q}} } ^{\mathrm{0}} \left({mx}+{ma}−{kx}^{\mathrm{2}} \right){dx} \\ $$$$=\frac{{ma}^{\mathrm{2}} }{\mathrm{2}}+\frac{{m}}{\mathrm{2}}{x}_{{Q}} ^{\mathrm{2}} +{max}_{{Q}} −\frac{{k}}{\mathrm{3}}{x}_{{Q}} ^{\mathrm{3}} \\ $$$$ \\ $$$${A}_{{yellow}} =\frac{{m}\left({a}+{x}_{{Q}} \right)^{\mathrm{2}} }{\mathrm{2}}+\int_{−{x}_{{Q}} } ^{\mathrm{0}} {kx}^{\mathrm{2}} {dx} \\ $$$$=\frac{{m}}{\mathrm{2}}\left({a}+{x}_{{Q}} \right)^{\mathrm{2}} −\frac{{kx}_{{Q}} ^{\mathrm{3}} }{\mathrm{3}} \\ $$$$=\frac{{m}}{\mathrm{2}}\left[{a}+\frac{{m}−\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}}{\mathrm{2}{k}}\right]^{\mathrm{2}} −\frac{{k}}{\mathrm{3}}\left[\frac{{m}−\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}}{\mathrm{2}{k}}\right]^{\mathrm{3}} \\ $$$$=\frac{{m}}{\mathrm{8}{k}^{\mathrm{2}} }\left[{m}+\mathrm{2}{ak}−\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}\right]^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{24}{k}^{\mathrm{2}} }\left[{m}−\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}\right]^{\mathrm{3}} \\ $$$${A}_{{blue}} ={A}_{{yellow}} \\ $$$$\frac{\left[{m}\left({m}+\mathrm{4}{ak}\right)\right]^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{6}{k}^{\mathrm{2}} }=\frac{{m}}{\mathrm{8}{k}^{\mathrm{2}} }\left[{m}+\mathrm{2}{ak}−\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}\right]^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{24}{k}^{\mathrm{2}} }\left[{m}−\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}\right]^{\mathrm{3}} \\ $$$$\frac{\left[{m}\left({m}+\mathrm{4}{ak}\right)\right]^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}=\frac{{m}}{\mathrm{4}}\left[{m}+\mathrm{2}{ak}−\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}\right]^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{12}}\left[{m}−\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)}\right]^{\mathrm{3}} \\ $$$${let}\:\lambda=\sqrt{{m}\left({m}+\mathrm{4}{ak}\right)} \\ $$$$\Rightarrow\mathrm{16}{m}\lambda^{\mathrm{3}} −\left(\lambda−{m}\right)^{\mathrm{3}} \left(\mathrm{3}\lambda+{m}\right)=\mathrm{0} \\ $$$${with}\:\theta=\mathrm{53}° \\ $$$$\Rightarrow\lambda=\mathrm{10}.\mathrm{2745} \\ $$$${k}=\frac{\mathrm{1}}{\mathrm{4}{a}}\left(\frac{\lambda^{\mathrm{2}} }{{m}}−{m}\right)=\mathrm{4}.\mathrm{8889} \\ $$
Commented by ajfour last updated on 08/Dec/18
$${Excellent}\:{solution}\:{Sir}! \\ $$$${brilliant}\:{image}\:{too}. \\ $$
Commented by mr W last updated on 07/Dec/18
