Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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.

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

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

Commented by jagoll last updated on 04/Feb/20

waw...integral sesion

$${waw}...{integral}\:{sesion} \\ $$

Commented by ajfour last updated on 04/Feb/20

yes sir, your solution is perfect!

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

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.

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

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!

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

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

Commented by TawaTawa last updated on 04/Feb/20

Weldone sir. God bless you more.

$$\mathrm{Weldone}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more}. \\ $$

Commented by ajfour last updated on 04/Feb/20

Thanks sir, i believe your answer, is the correct one. Let me check mine..

$${Thanks}\:{sir},\:{i}\:{believe}\:{your}\:{answer}, \\ $$$${is}\:{the}\:{correct}\:{one}.\:{Let}\:{me} \\ $$$${check}\:{mine}.. \\ $$

Commented by ajfour last updated on 04/Feb/20

go ahead santu sir..

$${go}\:{ahead}\:{santu}\:{sir}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com