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Question Number 103751 by hardylanes last updated on 17/Jul/20
  given that the parametric equation of   a curvd are x=(1/(t−1))  y=(1/(t+1)) obtaun a   cartesian equation of the curve. Hemce find  an equqtion of tbe nirmal to the curve at the   point t=2
$$ \\ $$$${given}\:{that}\:{the}\:{parametric}\:{equation}\:{of}\: \\ $$$${a}\:{curvd}\:{are}\:{x}=\frac{\mathrm{1}}{{t}−\mathrm{1}}\:\:{y}=\frac{\mathrm{1}}{{t}+\mathrm{1}}\:{obtaun}\:{a}\: \\ $$$${cartesian}\:{equation}\:{of}\:{the}\:{curve}.\:{Hemce}\:{find} \\ $$$${an}\:{equqtion}\:{of}\:{tbe}\:{nirmal}\:{to}\:{the}\:{curve}\:{at}\:{the}\: \\ $$$${point}\:{t}=\mathrm{2} \\ $$
Answered by bobhans last updated on 17/Jul/20
x = (1/(t−1)) →t−1 = (1/x); t+1 = (1/x)+2  y = (1/(t+1)) = (x/(1+2x)) . ⇒y′ = (1/((2x+1)^2 ))  slope normal line ⇒m = −(1/(y′)) = −(2x+1)^2   at point (x,y)=(1,(1/3)) ⇒m = −9  eq of the normal line ⇒9x+y = 9.1+1.(1/3)  9x+y = ((28)/3) or 27x + 3y = 28 ⊕
$${x}\:=\:\frac{\mathrm{1}}{{t}−\mathrm{1}}\:\rightarrow{t}−\mathrm{1}\:=\:\frac{\mathrm{1}}{{x}};\:{t}+\mathrm{1}\:=\:\frac{\mathrm{1}}{{x}}+\mathrm{2} \\ $$$${y}\:=\:\frac{\mathrm{1}}{{t}+\mathrm{1}}\:=\:\frac{{x}}{\mathrm{1}+\mathrm{2}{x}}\:.\:\Rightarrow{y}'\:=\:\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${slope}\:{normal}\:{line}\:\Rightarrow{m}\:=\:−\frac{\mathrm{1}}{{y}'}\:=\:−\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${at}\:{point}\:\left({x},{y}\right)=\left(\mathrm{1},\frac{\mathrm{1}}{\mathrm{3}}\right)\:\Rightarrow{m}\:=\:−\mathrm{9} \\ $$$${eq}\:{of}\:{the}\:{normal}\:{line}\:\Rightarrow\mathrm{9}{x}+{y}\:=\:\mathrm{9}.\mathrm{1}+\mathrm{1}.\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{9}{x}+{y}\:=\:\frac{\mathrm{28}}{\mathrm{3}}\:{or}\:\mathrm{27}{x}\:+\:\mathrm{3}{y}\:=\:\mathrm{28}\:\oplus\: \\ $$
Commented by bobhans last updated on 17/Jul/20
what do you meant ???
$${what}\:{do}\:{you}\:{meant}\:??? \\ $$

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