Question-80550

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Question Number 80550 by ajfour last updated on 04/Feb/20

Commented by ajfour last updated on 04/Feb/20

$${If}\:{both}\:{coloured}\:{areas}\:{are} \\ $$$${equal},\:{find}\:{a}. \\ $$

Answered by mr W last updated on 04/Feb/20

Commented by jagoll last updated on 04/Feb/20

$${waw}…{integral}\:{sesion} \\ $$

Commented by ajfour last updated on 04/Feb/20

$${yes}\:{sir},\:{your}\:{solution}\:{is}\:{perfect}! \\ $$

Commented by mr W last updated on 04/Feb/20

$${i}\:{found}\:{some}\:{typos},\:{now}\:{fixed}.\: \\ $$$${i}\:{think}\:{your}\:{equation}\:{arrives}\:{at}\:{the} \\ $$$${same}\:{result}. \\ $$

Commented by john santu last updated on 04/Feb/20

let P(c,c^2 ) area blue = (1/2)(a−c).c^2 +(1/3)c^3 = (1/2)ac^2 −(1/6)c^3 . let Q(−b,b^2 ). line PQ : y=−((b^2 /(a+b)))(x−a) area green = ∫_(−b) ^( c) −((b^2 /(a+b)))(x−a)−x^2 dx

$${let}\:{P}\left({c},{c}^{\mathrm{2}} \right) \\ $$$${area}\:{blue}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{c}\right).{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{c}^{\mathrm{3}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}{ac}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{6}}{c}^{\mathrm{3}} \:. \\ $$$${let}\:{Q}\left(−{b},{b}^{\mathrm{2}} \right).\:{line}\:{PQ}\::\:{y}=−\left(\frac{{b}^{\mathrm{2}} }{{a}+{b}}\right)\left({x}−{a}\right) \\ $$$${area}\:{green}\:=\:\int_{−{b}} ^{\:{c}} −\left(\frac{{b}^{\mathrm{2}} }{{a}+{b}}\right)\left({x}−{a}\right)−{x}^{\mathrm{2}} \:{dx} \\ $$

Commented by ajfour last updated on 04/Feb/20

$${looks}\:{promising}\:{sir}! \\ $$

Commented by mr W last updated on 04/Feb/20

P(p,p^2 ), Q(q,q^2 ), R(a,0) y′=2x=2p eqn. of normal to parabola at P: y=p^2 −(1/(2p))(x−p) point Q: q^2 =p^2 −(1/(2p))(q−p) 2pq^2 +q−p(2p^2 +1)=0 ⇒q=((−1±(4p^2 +1))/(4p))= { ((p ⇒no solution)),((−((1+2p^2 )/(2p))=−(1/(2p))−p)) :} point R: 0=p^2 −(1/(2p))(a−p) ⇒a=p(2p^2 +1) A_(blue) =(p^3 /3)+(((a−p)p^2 )/2)=((3ap^2 −p^3 )/6) A_(green) =(((p−q)(p^2 +q^2 ))/2)−(p^3 /3)+(q^3 /3) A_(green) =((p^3 −3p^2 q+3pq^2 −q^3 )/6) A_(blue) =A_(green) 3ap^2 −p^3 =p^3 −3p^2 q+3pq^2 −q^3 2p^3 −3(a+q)p^2 +(3p−q)q^2 =0 48p^8 −48p^6 −48p^4 −12p^2 −1=0 with t=p^2 >0 ⇒48t^4 −48t^3 −48t^2 −12t−1=0 t=(1/4)(1+(√3)+(√((42+8(√3))/3)))≈1.761750 ⇒p=(√t)≈1.327309 ⇒a=p(2p^2 +1)≈6.004086

$${P}\left({p},{p}^{\mathrm{2}} \right),\:{Q}\left({q},{q}^{\mathrm{2}} \right),\:{R}\left({a},\mathrm{0}\right) \\ $$$${y}'=\mathrm{2}{x}=\mathrm{2}{p} \\ $$$${eqn}.\:{of}\:{normal}\:{to}\:{parabola}\:{at}\:{P}: \\ $$$${y}={p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}{p}}\left({x}−{p}\right) \\ $$$${point}\:{Q}: \\ $$$${q}^{\mathrm{2}} ={p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}{p}}\left({q}−{p}\right) \\ $$$$\mathrm{2}{pq}^{\mathrm{2}} +{q}−{p}\left(\mathrm{2}{p}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{q}=\frac{−\mathrm{1}\pm\left(\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{4}{p}}=\begin{cases}{{p}\:\Rightarrow{no}\:{solution}}\\{−\frac{\mathrm{1}+\mathrm{2}{p}^{\mathrm{2}} }{\mathrm{2}{p}}=−\frac{\mathrm{1}}{\mathrm{2}{p}}−{p}}\end{cases} \\ $$$${point}\:{R}: \\ $$$$\mathrm{0}={p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}{p}}\left({a}−{p}\right) \\ $$$$\Rightarrow{a}={p}\left(\mathrm{2}{p}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$ \\ $$$${A}_{{blue}} =\frac{{p}^{\mathrm{3}} }{\mathrm{3}}+\frac{\left({a}−{p}\right){p}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}{ap}^{\mathrm{2}} −{p}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${A}_{{green}} =\frac{\left({p}−{q}\right)\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)}{\mathrm{2}}−\frac{{p}^{\mathrm{3}} }{\mathrm{3}}+\frac{{q}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${A}_{{green}} =\frac{{p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} {q}+\mathrm{3}{pq}^{\mathrm{2}} −{q}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${A}_{{blue}} ={A}_{{green}} \\ $$$$\mathrm{3}{ap}^{\mathrm{2}} −{p}^{\mathrm{3}} ={p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} {q}+\mathrm{3}{pq}^{\mathrm{2}} −{q}^{\mathrm{3}} \\ $$$$\mathrm{2}{p}^{\mathrm{3}} −\mathrm{3}\left({a}+{q}\right){p}^{\mathrm{2}} +\left(\mathrm{3}{p}−{q}\right){q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{48}{p}^{\mathrm{8}} −\mathrm{48}{p}^{\mathrm{6}} −\mathrm{48}{p}^{\mathrm{4}} −\mathrm{12}{p}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${with}\:{t}={p}^{\mathrm{2}} \:>\mathrm{0} \\ $$$$\Rightarrow\mathrm{48}{t}^{\mathrm{4}} −\mathrm{48}{t}^{\mathrm{3}} −\mathrm{48}{t}^{\mathrm{2}} −\mathrm{12}{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\sqrt{\mathrm{3}}+\sqrt{\frac{\mathrm{42}+\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}}\right)\approx\mathrm{1}.\mathrm{761750} \\ $$$$\Rightarrow{p}=\sqrt{{t}}\approx\mathrm{1}.\mathrm{327309} \\ $$$$\Rightarrow{a}={p}\left(\mathrm{2}{p}^{\mathrm{2}} +\mathrm{1}\right)\approx\mathrm{6}.\mathrm{004086} \\ $$

Commented by mr W last updated on 04/Feb/20