Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 105456 by 1549442205PVT last updated on 29/Jul/20

$$\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{curve}\mathrm{y}={\mathrm{x}}^4+{3\mathrm{x}}^2+2\mathrm{x}$$$$\mathrm{does}\mathrm{not}\mathrm{meet}\mathrm{the}\mathrm{straight}\mathrm{line}$$$$\mathrm{y}=2\mathrm{x}-1\mathrm{and}\mathrm{find}\mathrm{the}\mathrm{distace}\mathrm{between}$$$$\mathrm{their}\mathrm{nearest}\mathrm{points}.\left(\mathrm{Answer}\frac{1}{\sqrt{5}}\right)$$

Answered by ajfour last updated on 29/Jul/20

$$lety=2x+cbetangenttocurve$$$$\Rightarrow \frac{dy}{dx}=4x^3+6x+2=\frac{d}{dx}(2x-1)$$$$\Rightarrow x=0,y=0$$$$Now\mathrm{\perp }distanceof(0,0)from$$$$y=2x-1is$$$$p=\frac{1}{\sqrt{5}}.$$

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

$$\mathrm{Thank}\mathrm{you}\mathrm{both}\mathrm{sir}.$$

Answered by 1549442205PVT last updated on 29/Jul/20

$$\mathrm{i})\mathrm{Abscissa}\mathrm{of}\mathrm{intersection}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{curve}$$$$\mathrm{y}={\mathrm{x}}^4+{3\mathrm{x}}^2+2\mathrm{x}\left(\mathrm{C}\right)\mathrm{and}\mathrm{the}\mathrm{line}\mathrm{y}=2\mathrm{x}-1\left(\mathrm{L}\right)$$$$\mathrm{must}\mathrm{be}\mathrm{a}\mathrm{real}\mathrm{root}\mathrm{of}\mathrm{the}\mathrm{equation}:$$$${\mathrm{x}}^4+{3\mathrm{x}}^2+2\mathrm{x}=2\mathrm{x}-1\iff {\mathrm{x}}^4+{3\mathrm{x}}^2+1=0$$$$\mathrm{But}\mathrm{this}\mathrm{equation}\mathrm{has}\mathrm{no}\mathrm{real}\mathrm{roots}$$$$\mathrm{simce}{\mathrm{x}}^4+{3\mathrm{x}}^2+1⩾1\mathrm{\forall }\mathrm{x}\in \mathbb{R}.\mathrm{Hence}\mathrm{the}$$$$\mathrm{curve}\left(\mathrm{C}\right){\mathrm{don}}^{\prime }\mathrm{t}\mathrm{meet}\mathrm{the}\mathrm{line}\left(\mathrm{L}\right)(\mathrm{q}.\mathrm{e}.\mathrm{d})$$$$\mathrm{ii})\mathrm{Denote}\mathrm{by}\mathrm{A}({\mathrm{x}}_0,{\mathrm{y}}_0)\mathrm{be}\mathrm{a}\mathrm{point}\mathrm{at}$$$$\mathrm{which}\mathrm{the}\mathrm{tangent}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{parallel}$$$$\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{y}=2\mathrm{x}+1.\mathrm{Then}$$$${({\mathrm{x}}^4+{3\mathrm{x}}^2+2\mathrm{x})}^{{}^{\prime }}{\mid }_{{\mathrm{x}}_0}=2\iff {4\mathrm{x}}_0^3+{6\mathrm{x}}_0+2=2$$$$\iff {2\mathrm{x}}_0({2\mathrm{x}}_0^2+3)=0\iff {\mathrm{x}}_0=0\Rightarrow {\mathrm{y}}_0=0.\mathrm{Then}$$$$\mathrm{the}\mathrm{nearest}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{curve}$$$$\mathrm{and}\mathrm{the}\mathrm{line}\mathrm{be}\mathrm{the}\mathrm{distance}\mathrm{between}$$$$\mathrm{A}({\mathrm{x}}_0,{\mathrm{y}}_0)\mathrm{and}\mathrm{the}\mathrm{line}\mathrm{y}=2\mathrm{x}-1\mathrm{which}$$$$\mathrm{means}{\mathrm{d}}_0=\frac{\mid 2.0-1.0\mid }{\sqrt{2^2+1^2}}=\frac{1}{\sqrt{5}}$$

Commented by Ari last updated on 29/Jul/20

you are right

Terms of Service

Privacy Policy

Contact: info@tinkutara.com