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

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