Question-50963

Question Number 50963 by ajfour last updated on 22/Dec/18

Commented by ajfour last updated on 22/Dec/18

$${Find}\:{equation}\:{of}\:{parabola}\:{for}\:{the} \\ $$$${two}\:{coloured}\:{areas}\:{to}\:{be}\:{equal}. \\ $$

Answered by mr W last updated on 23/Dec/18

parabola: y=cx^2 line: (x/a)+(y/b)=1 (x/a)+((cx^2 )/b)=1 ⇒acx^2 +bx−ab=0 ⇒x=h=((−b+(√(b^2 +4a^2 bc)))/(2ac)) A_(green) =∫_0 ^h (b−((bx)/a)−cx^2 )dx=((ab)/4) bh−((bh^2 )/(2a))−((ch^3 )/3)=((ab)/4) h[b−h((b/(2a))+((ch)/3))]=((ab)/4) ⇒solve for c.... example: a=4,b=3⇒c=0.6211⇒y=0.6211x^2

$${parabola}:\:{y}={cx}^{\mathrm{2}} \\ $$$${line}:\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$$\frac{{x}}{{a}}+\frac{{cx}^{\mathrm{2}} }{{b}}=\mathrm{1} \\ $$$$\Rightarrow{acx}^{\mathrm{2}} +{bx}−{ab}=\mathrm{0} \\ $$$$\Rightarrow{x}={h}=\frac{−{b}+\sqrt{{b}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} {bc}}}{\mathrm{2}{ac}} \\ $$$${A}_{{green}} =\int_{\mathrm{0}} ^{{h}} \left({b}−\frac{{bx}}{{a}}−{cx}^{\mathrm{2}} \right){dx}=\frac{{ab}}{\mathrm{4}} \\ $$$${bh}−\frac{{bh}^{\mathrm{2}} }{\mathrm{2}{a}}−\frac{{ch}^{\mathrm{3}} }{\mathrm{3}}=\frac{{ab}}{\mathrm{4}} \\ $$$${h}\left[{b}−{h}\left(\frac{{b}}{\mathrm{2}{a}}+\frac{{ch}}{\mathrm{3}}\right)\right]=\frac{{ab}}{\mathrm{4}} \\ $$$$\Rightarrow{solve}\:{for}\:{c}…. \\ $$$$ \\ $$$${example}: \\ $$$${a}=\mathrm{4},{b}=\mathrm{3}\Rightarrow{c}=\mathrm{0}.\mathrm{6211}\Rightarrow{y}=\mathrm{0}.\mathrm{6211}{x}^{\mathrm{2}} \\ $$

Commented by ajfour last updated on 23/Dec/18

$${Thank}\:{you}\:{Sir}. \\ $$

Answered by ajfour last updated on 23/Dec/18

let y=cx^2 ⇒ (h/a)+((ch^2 )/b)=1 ⇒ bh+ach^2 −ab = 0 ch^2 = b(1−(h/a)) ....(i) h = (((√(b^2 +4a^2 bc))−b)/(2ac)) A_(pink) = ((ch^3 )/3)+((ch^2 )/2)(a−h) = ((ab)/4) ⇒ (1−(h/a))((h/3)+(a/2)−(h/2))=(a/4) ⇒ let (h/a) = λ , then (1−λ)((1/2)−(λ/6))=(1/4) ⇒ (1−λ)(3−λ)= (3/2) ⇒ 2λ^2 −8λ+3 = 0 λ = 2−((√(10))/2) from (i) c = ((b(1−λ))/(a^2 λ^2 )) = ((b((√(10))−2))/(a^2 (13−4(√(10))))) c = ((b(14+5(√(10))))/(9a^2 )) . and if a=4, b=3 c = ((3((√(10))−2))/(16(13−4(√(10))))) = ((((√(10))−2)(13+4(√(10))))/(48)) = ((14+5(√(10)))/(48)) ≈ 0.6211 .

$${let}\:\:{y}={cx}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\frac{{h}}{{a}}+\frac{{ch}^{\mathrm{2}} }{{b}}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:{bh}+{ach}^{\mathrm{2}} −{ab}\:=\:\mathrm{0} \\ $$$$\:\:\:{ch}^{\mathrm{2}} =\:{b}\left(\mathrm{1}−\frac{{h}}{{a}}\right)\:\:\:\:\:….\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{h}\:=\:\frac{\sqrt{{b}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} {bc}}−{b}}{\mathrm{2}{ac}} \\ $$$${A}_{{pink}} \:=\:\frac{{ch}^{\mathrm{3}} }{\mathrm{3}}+\frac{{ch}^{\mathrm{2}} }{\mathrm{2}}\left({a}−{h}\right)\:=\:\frac{{ab}}{\mathrm{4}} \\ $$$$\:\:\:\Rightarrow\:\left(\mathrm{1}−\frac{{h}}{{a}}\right)\left(\frac{{h}}{\mathrm{3}}+\frac{{a}}{\mathrm{2}}−\frac{{h}}{\mathrm{2}}\right)=\frac{{a}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:{let}\:\:\frac{{h}}{{a}}\:=\:\lambda\:,\:{then} \\ $$$$\:\:\:\:\:\:\left(\mathrm{1}−\lambda\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\lambda}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:\:\left(\mathrm{1}−\lambda\right)\left(\mathrm{3}−\lambda\right)=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}\lambda^{\mathrm{2}} −\mathrm{8}\lambda+\mathrm{3}\:=\:\mathrm{0} \\ $$$$\lambda\:=\:\mathrm{2}−\frac{\sqrt{\mathrm{10}}}{\mathrm{2}} \\ $$$${from}\:\left({i}\right) \\ $$$$\:\:\:\:\:\:{c}\:=\:\frac{{b}\left(\mathrm{1}−\lambda\right)}{{a}^{\mathrm{2}} \lambda^{\mathrm{2}} }\:=\:\frac{{b}\left(\sqrt{\mathrm{10}}−\mathrm{2}\right)}{{a}^{\mathrm{2}} \left(\mathrm{13}−\mathrm{4}\sqrt{\mathrm{10}}\right)}\:\: \\ $$$$\:\:\:{c}\:=\:\frac{{b}\left(\mathrm{14}+\mathrm{5}\sqrt{\mathrm{10}}\right)}{\mathrm{9}{a}^{\mathrm{2}} }\:. \\ $$$${and}\:\:{if}\:\:{a}=\mathrm{4},\:{b}=\mathrm{3} \\ $$$$\:\:\:\:\:{c}\:=\:\frac{\mathrm{3}\left(\sqrt{\mathrm{10}}−\mathrm{2}\right)}{\mathrm{16}\left(\mathrm{13}−\mathrm{4}\sqrt{\mathrm{10}}\right)}\: \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\left(\sqrt{\mathrm{10}}−\mathrm{2}\right)\left(\mathrm{13}+\mathrm{4}\sqrt{\mathrm{10}}\right)}{\mathrm{48}} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\mathrm{14}+\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{48}}\:\approx\:\mathrm{0}.\mathrm{6211}\:\:. \\ $$

Commented by mr W last updated on 23/Dec/18

$${very}\:{nice}\:{solution}! \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply