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Prove-that-the-curve-y-x-4-3x-2-2x-does-not-meet-the-straight-line-y-2x-1-and-find-the-distace-between-their-nearest-points-Answer-1-5-

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Question Number 105456 by 1549442205PVT last updated on 29/Jul/20
Prove that the curve y=x^4 +3x^2 +2x does not meet the straight line y=2x−1 and find the distace between their nearest points.(Answer (1/( (√5))))
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{4}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{2x} \\ $$$$\mathrm{does}\:\mathrm{not}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{y}=\mathrm{2x}−\mathrm{1}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distace}\:\mathrm{between} \\ $$$$\mathrm{their}\:\mathrm{nearest}\:\mathrm{points}.\left(\mathrm{Answer}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right) \\ $$
Answered by ajfour last updated on 29/Jul/20
let y=2x+c be tangent to curve ⇒ (dy/dx)= 4x^3 +6x+2 = (d/dx)(2x−1) ⇒ x=0 , y= 0 Now ⊥ distance of (0,0) from y=2x−1 is p=(1/( (√5))) .
$${let}\:\:{y}=\mathrm{2}{x}+{c}\:\:\:\:{be}\:{tangent}\:{to}\:{curve} \\ $$$$\Rightarrow\:\:\:\frac{{dy}}{{dx}}=\:\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}+\mathrm{2}\:=\:\frac{{d}}{{dx}}\left(\mathrm{2}{x}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\:\:{x}=\mathrm{0}\:\:,\:\:{y}=\:\mathrm{0} \\ $$$${Now}\:\:\bot\:{distance}\:{of}\:\left(\mathrm{0},\mathrm{0}\right)\:{from} \\ $$$$\:\:{y}=\mathrm{2}{x}−\mathrm{1}\:\:\:{is} \\ $$$$\:\:\:\:\:\:{p}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\:. \\ $$
Commented by Ari last updated on 29/Jul/20
Sir, if you can, clarify through actions your reasoning that there is ambiguity
Commented by 1549442205PVT last updated on 29/Jul/20
Thank you both sir.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{both}\:\mathrm{sir}. \\ $$
Answered by 1549442205PVT last updated on 29/Jul/20
i)Abscissa of intersection point of the curve y=x^(4 ) +3x^(2 ) +2x (C)and the line y=2x−1(L) must be a real root of the equation: x^(4 ) +3x^(2 ) +2x=2x−1⇔x^4 +3x^2 +1=0 But this equation has no real roots simce x^4 +3x^2 +1≥1∀x∈R.Hence the curve (C)don′t meet the line(L)(q.e.d) ii)Denote by A(x_0 ,y_0 )be a point at which the tangent of the curve parallel to the line y=2x+1 .Then (x^(4 ) +3x^(2 ) +2x)^′ ∣_x_0 =2⇔4x_0 ^3 +6x_0 +2=2 ⇔2x_0 (2x_0 ^2 +3)=0⇔x_0 =0⇒y_0 =0.Then the nearest distance between the curve and the line be the distance between A(x_0 ,y_0 ) and the line y=2x−1which means d_0 =((∣2.0−1.0∣)/( (√(2^2 +1^2 ))))=(1/( (√5)))
$$\left.\mathrm{i}\right)\mathrm{Abscissa}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{y}=\mathrm{x}^{\mathrm{4}\:} +\mathrm{3x}^{\mathrm{2}\:\:} +\mathrm{2x}\:\left(\mathrm{C}\right)\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=\mathrm{2x}−\mathrm{1}\left(\mathrm{L}\right) \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{x}^{\mathrm{4}\:} +\mathrm{3x}^{\mathrm{2}\:\:} +\mathrm{2x}=\mathrm{2x}−\mathrm{1}\Leftrightarrow\mathrm{x}^{\mathrm{4}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{But}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{no}\:\mathrm{real}\:\mathrm{roots} \\ $$$$\mathrm{simce}\:\mathrm{x}^{\mathrm{4}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{1}\forall\mathrm{x}\in\mathbb{R}.\mathrm{Hence}\:\mathrm{the} \\ $$$$\mathrm{curve}\:\left(\mathrm{C}\right)\mathrm{don}'\mathrm{t}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{line}\left(\mathrm{L}\right)\left(\mathrm{q}.\mathrm{e}.\mathrm{d}\right) \\ $$$$\left.\mathrm{ii}\right)\mathrm{Denote}\:\mathrm{by}\:\mathrm{A}\left(\mathrm{x}_{\mathrm{0}} ,\mathrm{y}_{\mathrm{0}} \right)\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{at} \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{parallel} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=\mathrm{2x}+\mathrm{1}\:.\mathrm{Then}\: \\ $$$$\left(\mathrm{x}^{\mathrm{4}\:} +\mathrm{3x}^{\mathrm{2}\:\:} +\mathrm{2x}\right)^{'} \mid_{\mathrm{x}_{\mathrm{0}} } =\mathrm{2}\Leftrightarrow\mathrm{4x}_{\mathrm{0}} ^{\mathrm{3}} +\mathrm{6x}_{\mathrm{0}} +\mathrm{2}=\mathrm{2} \\ $$$$\Leftrightarrow\mathrm{2x}_{\mathrm{0}} \left(\mathrm{2x}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{3}\right)=\mathrm{0}\Leftrightarrow\mathrm{x}_{\mathrm{0}} =\mathrm{0}\Rightarrow\mathrm{y}_{\mathrm{0}} =\mathrm{0}.\mathrm{Then}\: \\ $$$$\mathrm{the}\:\mathrm{nearest}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:\mathrm{be}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{A}\left(\mathrm{x}_{\mathrm{0}} ,\mathrm{y}_{\mathrm{0}} \right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=\mathrm{2x}−\mathrm{1which} \\ $$$$\mathrm{means}\:\mathrm{d}_{\mathrm{0}} =\frac{\mid\mathrm{2}.\mathrm{0}−\mathrm{1}.\mathrm{0}\mid}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$
Commented by Ari last updated on 29/Jul/20
you are right

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