Parametric equations define a curve where x and y are defined in terms of a third variable called a parameter. The graph of parametric equations consists of all points (x,y) obtained by allowing the parameter to vary over its domain. Eliminating the parameter between the two equations yields a non-parametric equation of the curve. Examples are provided of eliminating parameters between various parametric equations to obtain the curve and sketching the resulting graphs. Exercises are given to further practice eliminating parameters and sketching curves.

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
t=0
●
t=-1● ● t=1
x
t=-2 ● ● t=2
t=-3 ● ● t=3

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
2
1
0 1 2
x
-2 -1
-1
-2

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
2
1
0 1 2
x
-2 -1
-1
-2

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
cos t = 1 − sin2 t
Substitute into x = sin t + cos t , then
x = sin t + 1 − sin2 t
But y = sin t and y2 = sin2 t
Therefore x=y+ 1 − y2
x–y= 1 − y2
Squaring both sides
(x – y)2 = 1 – y2

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