Question Number 44691 by ajfour last updated on 03/Oct/18
Commented by ajfour last updated on 03/Oct/18
$${Find}\:{equation}\:{of}\:{the}\:{biquadratic} \\ $$$${curve}. \\ $$
Answered by ajfour last updated on 03/Oct/18
![let y=A(x−a)(x−b)^2 (x−9)+4 and also let (dy/dx) = 4A(x−2)(x−b)(x−c) at x=c , y=0 ; hence _____________________ A(c−9)(c−b)^2 (c−a)=−4 ..(i) _____________________ at x=0 , y=−2, so _____________________ 3Aab^2 = −2 ....(ii) _____________________ (dy/dx) = (d/dx)(y) implies 4A(x−b)(x−2)(x−c)= = A[(x−b)^2 (x−9)+2(x−b)(x−a)(x−9) +(x−a)(x−b)^2 ] ⇒ 4(x−2)(x−c)=4x^2 +(−b−9−2a−18 −a−b)x +(9b+18a+ab) comparing coefficients of x^1 on both sides: ⇒ 8+4c = 3a+2b+27 _____________________ ⇒ 3a+2b+19 = 4c ....(iii) _____________________ comparing constant terms of (dy/dx) & (d/dx)(y) _____________________ ab+18a+9b = 8c ....(iv) _____________________ ... ultimately eliminating A,b, c among the four equations: 24×256a(19−6a)^2 (5+a)^2 = (3a^2 −26a−9)(3a^2 +58a+19)^2 ×(171−a^2 −10a) ⇒ a= −0.0092 b = 7.63613 c = 8.5612 A = 1.242723 so eq. is y=1.242723(x+0.0092) ×(x−7.63613)^2 (x−9)+4 . Please check..](Q44694.png)
$${let}\:\:\:\boldsymbol{{y}}=\boldsymbol{{A}}\left(\boldsymbol{{x}}−{a}\right)\left(\boldsymbol{{x}}−{b}\right)^{\mathrm{2}} \left(\boldsymbol{{x}}−\mathrm{9}\right)+\mathrm{4} \\ $$$${and}\:{also}\:{let} \\ $$$$\:\:\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}\:=\:\mathrm{4}\boldsymbol{{A}}\left(\boldsymbol{{x}}−\mathrm{2}\right)\left(\boldsymbol{{x}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{x}}−\boldsymbol{{c}}\right) \\ $$$${at}\:{x}={c}\:,\:{y}=\mathrm{0}\:;\:{hence} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\boldsymbol{{A}}\left(\boldsymbol{{c}}−\mathrm{9}\right)\left(\boldsymbol{{c}}−\boldsymbol{{b}}\right)^{\mathrm{2}} \left(\boldsymbol{{c}}−\boldsymbol{{a}}\right)=−\mathrm{4}\:\:\:\:..\left({i}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${at}\:{x}=\mathrm{0}\:,\:{y}=−\mathrm{2},\:\:{so} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\mathrm{3}\boldsymbol{{Aab}}^{\mathrm{2}} \:=\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({ii}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{d}}{{dx}}\left({y}\right)\:\:\:\:\:{implies} \\ $$$$\mathrm{4}{A}\left({x}−{b}\right)\left({x}−\mathrm{2}\right)\left({x}−{c}\right)= \\ $$$$\:\:\:\:\:=\:{A}\left[\left({x}−{b}\right)^{\mathrm{2}} \left({x}−\mathrm{9}\right)+\mathrm{2}\left({x}−{b}\right)\left({x}−{a}\right)\left({x}−\mathrm{9}\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({x}−{a}\right)\left({x}−{b}\right)^{\mathrm{2}} \right] \\ $$$$\Rightarrow \\ $$$$\mathrm{4}\left({x}−\mathrm{2}\right)\left({x}−{c}\right)=\mathrm{4}{x}^{\mathrm{2}} +\left(−{b}−\mathrm{9}−\mathrm{2}{a}−\mathrm{18}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{a}−{b}\right){x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{9}{b}+\mathrm{18}{a}+{ab}\right) \\ $$$${comparing}\:{coefficients}\:{of}\:{x}^{\mathrm{1}} \:{on} \\ $$$${both}\:{sides}: \\ $$$$\Rightarrow\:\:\mathrm{8}+\mathrm{4}{c}\:=\:\mathrm{3}{a}+\mathrm{2}{b}+\mathrm{27} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\:\mathrm{3}\boldsymbol{{a}}+\mathrm{2}\boldsymbol{{b}}+\mathrm{19}\:=\:\mathrm{4}\boldsymbol{{c}}\:\:\:\:\:....\left({iii}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:{comparing}\:{constant}\:{terms}\:{of} \\ $$$$\:\frac{{dy}}{{dx}}\:\&\:\frac{{d}}{{dx}}\left({y}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\boldsymbol{{ab}}+\mathrm{18}\boldsymbol{{a}}+\mathrm{9}\boldsymbol{{b}}\:=\:\mathrm{8}\boldsymbol{{c}}\:\:\:\:\:\:\:\:\:....\left({iv}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$... \\ $$$${ultimately}\:{eliminating}\:\boldsymbol{{A}},\boldsymbol{{b}},\:\boldsymbol{{c}} \\ $$$${among}\:{the}\:{four}\:{equations}: \\ $$$$\mathrm{24}×\mathrm{256}\boldsymbol{{a}}\left(\mathrm{19}−\mathrm{6}\boldsymbol{{a}}\right)^{\mathrm{2}} \left(\mathrm{5}+\boldsymbol{{a}}\right)^{\mathrm{2}} \\ $$$$\:\:\:=\:\left(\mathrm{3}\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{26}\boldsymbol{{a}}−\mathrm{9}\right)\left(\mathrm{3}\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{58}\boldsymbol{{a}}+\mathrm{19}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left(\mathrm{171}−\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{{a}}\right) \\ $$$$\Rightarrow\:\:\boldsymbol{{a}}=\:−\mathrm{0}.\mathrm{0092} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{b}}\:=\:\mathrm{7}.\mathrm{63613} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{c}}\:=\:\mathrm{8}.\mathrm{5612} \\ $$$$\:\:\:\:\:\:\boldsymbol{{A}}\:=\:\mathrm{1}.\mathrm{242723} \\ $$$${so}\:{eq}.\:{is} \\ $$$$\:\:{y}=\mathrm{1}.\mathrm{242723}\left({x}+\mathrm{0}.\mathrm{0092}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:×\left({x}−\mathrm{7}.\mathrm{63613}\right)^{\mathrm{2}} \left({x}−\mathrm{9}\right)+\mathrm{4}\:. \\ $$$$\:\:\:{Please}\:{check}.. \\ $$