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Eliminate the parameter

1) x=3-2t
    y=2+3t
 
 
2) x=sqrt(t)
     y=1-t
 
 
 
3) x=4+2cos(theta)
    y=1+2sin(theta)
 
 
4) x=sec(theta)
    y=cos(theta)
 
 
 
 
5) x=e^2t
    y=e^t
 
 
                          can someone help me with just one problem so i can do the rest?
                            thank you :)
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3 Answers

1)  x=3-2t  --->  t= (3-x)/2   
     Replace t in the second equation   y= 2+3((3-x)/2) = 2+ 9/2-3x/2 = -3x/2 + 13/2 = -1/2(3x-13)
     y=-1/2(3x-13), where x is any real number
2)  Replace t=1-y from the second equation, into the first eq.
     x= sqrt(1-y), where 1-y >=0, Y<=1 and x>=0
     x^2 = 1-y  --->  y= 1- x^2, where x>=0
3)  cos(theta) =(x-4)/2  --->  theta = arccos((x-4)/2), where 0 <= (x-4)/2 <= pi  --->  4<= x <=2pi+4
     y= 1+2sin(arccos((x-4)/2), where 4<= x <=2pi+4
4)  theta= arcsec(x), 
     Y=cos(arcsec(x))  where x<=-1 or x>=1
5) x=e^(2t)  --->  ln x= ln e^(2t) = 2t ln(e) --->  t= (ln x)/2, where x>0
    y=e^((ln x)/2) =e^ln (sqrt(x)) = sqrt (x), x>o  
or easier way
    x= e^(2t) = (e^t)^2 = y^2 , where x>0
   or  y= sqrt(x), where x>0 
  
    
Daisy,
          The idea is to solve one of the parametric equations for t in terms of x or y and then substitute that in the other equation giving a relationship between x and y.
 
1. t=(y-2)/3 and x=3-2((y-2)/3)=3-(2/3)y+4/3=(9/3)+(4/3)-(2/3)y=(13/3)-(2/3)y
 
         so (2/3)y+x=(13/3)
 
The other easy one is 5. x=y2 by inspection or ln(y)=t and ln(x) =2t so ln(x)=2ln(y) or x=y2.
 
Jim
 
 
1) x=3-2t
y=2+3t
 
Set one equation equal to t.Then plug what t equals into the other equation. You are trying to get rid of the t and only be left with x and y.
 
X=3-2t
 
x-3= -2t
divide -2 by both sides
 
-0.5x+1.5=t
 
plug t into the y equation
 
y=2+3t
 
y=2+ 3(-0.5x+1.5)
 
y=2+ (-1.5x+4.5)
 
y=2-1.5x+4.5
 
y= -1.5x+6.5