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Question Number 109483 by ajfour last updated on 24/Aug/20

Commented by ajfour last updated on 24/Aug/20

If the cubic curve has equation, y=x^3 −21x−20 Find coordinates of P.

Ifthecubiccurvehasequation,y=x3−21x−20FindcoordinatesofP.

Commented by ajfour last updated on 24/Aug/20

eq. of tangent passing through (5,0) is simply y=−(9/4)(x−5) and that passing through (−4,0) is y=−9(x+4) Thus P (−7, 27)

eq.oftangentpassingthrough(5,0)issimplyy=−94(x−5)andthatpassingthrough(−4,0)isy=−9(x+4)ThusP(−7,27)

Answered by 1549442205PVT last updated on 25/Aug/20

Let C be the curve denote the graph of the function y=x^3 −21x−20.Then the general equation of the tangent to the curve C at point (x_0 ,y_0 ) is (d):y=y′(x_0 )(x−x_0 )+y(x_0 )(∗) Suppose (d_1 ) pass through the point (5,0) and (d_2 )−(−4,0) then we have { ((0=(3x_0 ^2 −21)(5−x_0 )+x_0 ^3 −21x_0 −20(1))),((0=(3x_1 ^2 −21)(−4−x_1 )+x_1 ^3 −21x_1 −20(2))) :} (1)⇔−3x_0 ^3 +15x_0 ^2 +21x_0 −105+x_0 ^3 −21x_0 −20=0 ⇔−2x_0 ^3 +15x_0 ^2 −125=0 ⇔2x_0 ^3 −15x_0 ^2 +125=0⇔(x_0 −5)^2 (2x_0 +5) ⇔x_0 =((−5)/2)⇒y′(x_0 )=3.(((−5)/2))^2 −21=((−9)/4) (2)⇔−3x_1 ^3 −12x_1 ^2 +21x_1 +84+x_1 ^3 −21x_1 −20 ⇔2x_1 ^3 +12x_1 ^2 −64=0⇔x_1 ^3 +6x_1 ^2 −32=0 ⇔(x_1 +4)^2 (x_1 −2)=0 ⇔x_1 =2⇒y′(2)=−9 Therefore,we have: (d_1 ):y=((−9)/4)x+((135)/8) (d_2 ):y=−9x−54 P=(d_1 )∩(d_2 )⇒the coordinates (x,y) of P are the roots of the system: { ((y=((−9x)/4)+((135)/8))),((y=−9x−54)) :}⇔ { ((x=((−21)/2))),(((81)/2)) :} Thus,we find out P(((−21)/2);((81)/2))

LetCbethecurvedenotethegraphofthefunctiony=x3−21x−20.ThenthegeneralequationofthetangenttothecurveCatpoint(x0,y0)is(d):y=y′(x0)(x−x0)+y(x0)(∗)Suppose(d1)passthroughthepoint(5,0)and(d2)−(−4,0)thenwehave{0=(3x02−21)(5−x0)+x03−21x0−20(1)0=(3x12−21)(−4−x1)+x13−21x1−20(2)(1)⇔−3x03+15x02+21x0−105+x03−21x0−20=0⇔−2x03+15x02−125=0⇔2x03−15x02+125=0⇔(x0−5)2(2x0+5)⇔x0=−52⇒y′(x0)=3.(−52)2−21=−94(2)⇔−3x13−12x12+21x1+84+x13−21x1−20⇔2x13+12x12−64=0⇔x13+6x12−32=0⇔(x1+4)2(x1−2)=0⇔x1=2⇒y′(2)=−9Therefore,wehave:(d1):y=−94x+1358(d2):y=−9x−54P=(d1)∩(d2)⇒thecoordinates(x,y)ofParetherootsofthesystem:{y=−9x4+1358y=−9x−54⇔{x=−212812Thus,wefindoutP(−212;812)

Commented by ajfour last updated on 24/Aug/20

Thanks for your method Sir, i think your finally obtained eqs. of lines are not correct, please check..

ThanksforyourmethodSir,ithinkyourfinallyobtainedeqs.oflinesarenotcorrect,pleasecheck..

Commented by 1549442205PVT last updated on 25/Aug/20

Thank Sir .I mistaked.After find out the slops x_0 =−(9/4) need must replace that value into (∗)to find free efficient of the tangent y=−(9/4)x+b b=−y′(x_0 ).+y(x_0 )=(9/4).(−(5/2))+((135)/8) =−((45)/8)+((135)/8)=((90)/8)=((45)/4)⇒y=−(9/4)x+((45)/4) For x_1 =2.we have y=−9x+c c=−y′(x_1 ).x_1 +y(x_1 )=9.2−54=36 we get y=−9x+36 Sir′s results are correct

ThankSir.Imistaked.Afterfindouttheslopsx0=−94needmustreplacethatvalueinto(∗)tofindfreeefficientofthetangenty=−94x+bb=−y′(x0).+y(x0)=94.(−52)+1358=−458+1358=908=454⇒y=−94x+454Forx1=2.wehavey=−9x+cc=−y′(x1).x1+y(x1)=9.2−54=36wegety=−9x+36Sir′sresultsarecorrect

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