Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 39591 by Rio Mike last updated on 08/Jul/18

Given the lines   l_1 :−3mx + 3y = 9   and l_(2 ) : y = mx + c  find the value of  m and c if  the point (1,2) lie on both lines.  hence the tangent of the  curve y = (mx + c)^2   when it moves across the x−axis

$${Given}\:{the}\:{lines}\: \\ $$$${l}_{\mathrm{1}} :−\mathrm{3}{mx}\:+\:\mathrm{3}{y}\:=\:\mathrm{9}\: \\ $$$${and}\:{l}_{\mathrm{2}\:} :\:{y}\:=\:{mx}\:+\:{c} \\ $$$${find}\:{the}\:{value}\:{of}\:\:{m}\:{and}\:{c}\:{if} \\ $$$${the}\:{point}\:\left(\mathrm{1},\mathrm{2}\right)\:{lie}\:{on}\:{both}\:{lines}. \\ $$$${hence}\:{the}\:{tangent}\:{of}\:{the} \\ $$$${curve}\:{y}\:=\:\left({mx}\:+\:{c}\right)^{\mathrm{2}} \\ $$$${when}\:{it}\:{moves}\:{across}\:{the}\:{x}−{axis} \\ $$

Answered by MJS last updated on 08/Jul/18

 ((1),(2) ) on l_1 : −3m+6=9 ⇒ m=−1   ((1),(2) ) on l_2 : 2=−1×1+c ⇒ c=3  y=(3−x)^2   (3−x)^2 =0 ⇒ x=3  curve touches x−axis ⇒ tangent=x−axis  in point  ((3),(0) ) ⇒ t: y=0  if you mean curve moves across y−axis:  y=(3−0)^2 =9  tangent in  ((0),(9) ) :  y=kx+d  d=9  f(x)=(3−x)^2   f′(x)=2x−6  k=f′(0)=−6  t: y=−6x+9

$$\begin{pmatrix}{\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}\:\mathrm{on}\:{l}_{\mathrm{1}} :\:−\mathrm{3}{m}+\mathrm{6}=\mathrm{9}\:\Rightarrow\:{m}=−\mathrm{1} \\ $$$$\begin{pmatrix}{\mathrm{1}}\\{\mathrm{2}}\end{pmatrix}\:\mathrm{on}\:{l}_{\mathrm{2}} :\:\mathrm{2}=−\mathrm{1}×\mathrm{1}+{c}\:\Rightarrow\:{c}=\mathrm{3} \\ $$$${y}=\left(\mathrm{3}−{x}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{3}−{x}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{x}=\mathrm{3} \\ $$$$\mathrm{curve}\:\mathrm{touches}\:{x}−\mathrm{axis}\:\Rightarrow\:\mathrm{tangent}={x}−\mathrm{axis} \\ $$$$\mathrm{in}\:\mathrm{point}\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\end{pmatrix}\:\Rightarrow\:{t}:\:{y}=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{curve}\:\mathrm{moves}\:\mathrm{across}\:{y}−\mathrm{axis}: \\ $$$${y}=\left(\mathrm{3}−\mathrm{0}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{tangent}\:\mathrm{in}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{9}}\end{pmatrix}\:: \\ $$$${y}={kx}+{d} \\ $$$${d}=\mathrm{9} \\ $$$${f}\left({x}\right)=\left(\mathrm{3}−{x}\right)^{\mathrm{2}} \\ $$$${f}'\left({x}\right)=\mathrm{2}{x}−\mathrm{6} \\ $$$${k}={f}'\left(\mathrm{0}\right)=−\mathrm{6} \\ $$$${t}:\:{y}=−\mathrm{6}{x}+\mathrm{9} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com