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Question Number 49392 by ajfour last updated on 06/Dec/18

	Answered by mr W last updated on 08/Dec/18
	$let m=tan θ eqn. of line: y=m(x+a) kx^2 =m(x+a) kx^2 −mx−ma=0 x=((m±(√(m(m+4ak))))/(2k)) x_Q =((m−(√(m(m+4ak))))/(2k)) x_P =((m+(√(m(m+4ak))))/(2k)) A_(blue) =∫_x_Q ^x_P (mx+ma−kx^2 )dx =(m/2)(x_P ^2 −x_Q ^2 )+ma(x_P −x_Q )−(k/3)(x_P ^3 −x_Q ^3 ) =(x_P −x_Q ){(m/2)(x_P +x_Q )+ma−(k/3)[(x_P −x_Q )^2 +3x_P x_Q ]} =((√(m(m+4ak)))/k){(m/2)×(m/k)+ma−(k/3)[(m^2 /k^2 )+((ma)/k)]} =(([m(m+4ak)]^(3/2) )/(6k^2 )) A_(yellow) =((ma^2 )/2)−∫_x_Q ^0 (mx+ma−kx^2 )dx =((ma^2 )/2)+(m/2)x_Q ^2 +max_Q −(k/3)x_Q ^3 A_(yellow) =((m(a+x_Q )^2 )/2)+∫_(−x_Q ) ^0 kx^2 dx =(m/2)(a+x_Q )^2 −((kx_Q ^3 )/3) =(m/2)[a+((m−(√(m(m+4ak))))/(2k))]^2 −(k/3)[((m−(√(m(m+4ak))))/(2k))]^3 =(m/(8k^2 ))[m+2ak−(√(m(m+4ak)))]^2 −(1/(24k^2 ))[m−(√(m(m+4ak)))]^3 A_(blue) =A_(yellow) (([m(m+4ak)]^(3/2) )/(6k^2 ))=(m/(8k^2 ))[m+2ak−(√(m(m+4ak)))]^2 −(1/(24k^2 ))[m−(√(m(m+4ak)))]^3 (([m(m+4ak)]^(3/2) )/3)=(m/4)[m+2ak−(√(m(m+4ak)))]^2 −(1/(12))[m−(√(m(m+4ak)))]^3 let λ=(√(m(m+4ak))) ⇒16mλ^3 −(λ−m)^3 (3λ+m)=0 with θ=53° ⇒λ=10.2745 k=(1/(4a))((λ^2 /m)−m)=4.8889$
	$l e t m = \tan θ$ $e q n . o f l i n e :$ $y = m (x + a)$ ${k x}^{2} = m (x + a)$ ${k x}^{2} - m x - m a = 0$ $x = \frac{m \pm \sqrt{m (m + 4 a k)}}{2 k}$ $x_{Q} = \frac{m - \sqrt{m (m + 4 a k)}}{2 k}$ $x_{P} = \frac{m + \sqrt{m (m + 4 a k)}}{2 k}$ $A_{b l u e} = \int_{x_{Q}}^{x_{P}} (m x + m a - {k x}^{2}) d x$ $= \frac{m}{2} (x_{P}^{2} - x_{Q}^{2}) + m a (x_{P} - x_{Q}) - \frac{k}{3} (x_{P}^{3} - x_{Q}^{3})$ $= (x_{P} - x_{Q}) {\frac{m}{2} (x_{P} + x_{Q}) + m a - \frac{k}{3} [{(x_{P} - x_{Q})}^{2} + 3 x_{P} x_{Q}]}$ $= \frac{\sqrt{m (m + 4 a k)}}{k} {\frac{m}{2} \times \frac{m}{k} + m a - \frac{k}{3} [\frac{m^{2}}{k^{2}} + \frac{m a}{k}]}$ $= \frac{{[m (m + 4 a k)]}^{\frac{3}{2}}}{6 k^{2}}$ $A_{y e l l o w} = \frac{{m a}^{2}}{2} - \int_{x_{Q}}^{0} (m x + m a - {k x}^{2}) d x$ $= \frac{{m a}^{2}}{2} + \frac{m}{2} x_{Q}^{2} + {m a x}_{Q} - \frac{k}{3} x_{Q}^{3}$ $A_{y e l l o w} = \frac{m {(a + x_{Q})}^{2}}{2} + \int_{- x_{Q}}^{0} {k x}^{2} d x$ $= \frac{m}{2} {(a + x_{Q})}^{2} - \frac{{k x}_{Q}^{3}}{3}$ $= \frac{m}{2} {[a + \frac{m - \sqrt{m (m + 4 a k)}}{2 k}]}^{2} - \frac{k}{3} {[\frac{m - \sqrt{m (m + 4 a k)}}{2 k}]}^{3}$ $= \frac{m}{8 k^{2}} {[m + 2 a k - \sqrt{m (m + 4 a k)}]}^{2} - \frac{1}{24 k^{2}} {[m - \sqrt{m (m + 4 a k)}]}^{3}$ $A_{b l u e} = A_{y e l l o w}$ $\frac{{[m (m + 4 a k)]}^{\frac{3}{2}}}{6 k^{2}} = \frac{m}{8 k^{2}} {[m + 2 a k - \sqrt{m (m + 4 a k)}]}^{2} - \frac{1}{24 k^{2}} {[m - \sqrt{m (m + 4 a k)}]}^{3}$ $\frac{{[m (m + 4 a k)]}^{\frac{3}{2}}}{3} = \frac{m}{4} {[m + 2 a k - \sqrt{m (m + 4 a k)}]}^{2} - \frac{1}{12} {[m - \sqrt{m (m + 4 a k)}]}^{3}$ $l e t λ = \sqrt{m (m + 4 a k)}$ $\Rightarrow 16 m λ^{3} - {(λ - m)}^{3} (3 λ + m) = 0$ $w i t h θ = 53 °$ $\Rightarrow λ = 10.2745$ $k = \frac{1}{4 a} (\frac{λ^{2}}{m} - m) = 4.8889$
	Commented by ajfour last updated on 08/Dec/18

	$E x c e l l e n t s o l u t i o n S i r!$ $b r i l l i a n t i m a g e t o o .$
	Commented by mr W last updated on 07/Dec/18

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