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1. PARAMETRIC
EQUATIONS

2. DEFINITION: PARAMETRIC EQUATIONS
If there are functions f and g with a common domain
T, the equations x = f(t) and y = g(t), for t in T, are
parametric equations of the curve consisting of all
points ( f(t), g(t) ), for t in T. The variable t is the
parameter.
The equations x = t + 2 and y = 3t – 1
for example are parametric equations and t is the
parameter. The equations define a graph. If t is assigned
a value, corresponding values are determined for x and
y. The pair of values for x and y constitute the
coordinates of a point of the graph. The complete graph
consists of the set of all points determined in this way

3. as t varies through all its chosen values. We can
eliminate t between the equations and obtain an
equation involving x and y. Thus, solving either equation
for t and substituting in the other, we get
3x – y = 7
The graph of this equation, which also the graph of
the parametric equations, is a straight line.
Example 1: Sketch the graph of the parametric
equations x = 2 + t and y = 3 – t2
.
t -3 -2 -1 0 1 2 3
x -1 0 1 2 3 4 5
y -6 -1 2 3 2 -1 -6

4. y
x
●●
● ●
●
●●
t=-1
t=2t=-2
t=-3 t=3
t=0
t=1

5. Example 2: Eliminate the parameter between x = t + 1
and y = t2
+ 3t + 2 and sketch the graph.
Solution:
Solving x = t + 1 for t, we have t = x – 1.
Substitute into y = t2
+ 3t + 2, then
y = (x – 1)2
+ 3(x – 1) + 2
y = x2
– 2x + 1 + 3x – 3 + 2
y = x2
+ x
Reducing to the standard form,
y + ¼ = x2
+ x + ¼
y + ¼ = (x + ½)2
, a parabola with V(-½,-¼)
opening upward

6. y
x
-1-2 210
-2
-1
2
1

7. Example 3: Eliminate the parameter between x = sin t
and y = cos t and sketch the graph.
Solution:
Squaring both sides of the parametric equations, we
have
x2
= sin2
t and y2
= cos2
t
And adding the two equations will give us
x2
+ y2
= sin2
t + cos2
t
But
sin2
t + cos2
t = 1
Therefore
x2
+ y2
= 1 , a circle with C(0, 0) and r = 1

8. y
x
-1-2 210
-2
-1
2
1

9. Example 4: Find the parametric representation for the
line through (1, 5) and (-2, 3).
Solution:
Letting (1, 5) and (-2, 3) be the first and second points,
respectively, of
x = x1 + r(x2 – x1)
and
y = y1 + r(y2 – y1)
We then have
x = 1 + r(-2 – 1) and y = 5 + r(3
– 5)
x = 1 – 3r y = 5 – 2r

10. Example 5: Eliminate the parameter between
x = sin t + cos t and y = sin t.
Solution:
Solving sin2
t + cos2
t = 1 for cos t, we have
Substitute into x = sin t + cos t , then
x = sin t +
But y = sin t and y2
= sin2
t
Therefore x = y +
x – y =
Squaring both sides
(x – y)2
= 1 – y2
tsin1tcos 2
−=
tsin1 2
−
2
y1−
2
y1−

11. Exercises:
Eliminate the parameter and sketch the curve.
• x = t2
+ 1, y = t + 1
• x = t2
+ t – 2 , y = t + 2
• x = cos θ , y = cos2
θ + 8 cos θ
• x = 4 cos θ , y = 7 sin θ
• x = cos θ , y = sin 2θ
• x = 1 + cos 2θ , y = 1 – sin θ