Download CO-ORDINATE GEOMETRY IN THE (x,y) PLANE

C4 Chapter 2
CO-ORDINATE GEOMETRY IN THE (x,y) PLANE
PARAMETRIC EQUATIONS
Coordinates x and y are expressed as functions of a parameter.
x = f(t) and y = g(t)
Example
Draw the curve given by the parametric equations:
2
x = t + 3,
y = t
for
– 3  t  3
Draw up a table of values
t
x
y
-3
0
9
-2
1
4
-1
2
1
0
3
0
1
4
1
2
5
4
3
6
9
y
10
8
6
4
2
-2
JMcC
0
2
4
6
x
1
C4 Chapter 2
To find the Cartesian equation of the curve, you need to
eliminate the parameter between the two equations
Example 1
Find the Cartesian equation of the curve given by the
parametric equations:
2
x = t + 3,
y = t
Make t the subject  t = x – 3
2
Substitute into the y equation  y = (x – 3)
Check: this is the parabola y = x2 translated 3 units to the
right (see previous example)
Example 2
Find the Cartesian equation of the curve given by the
parametric equations:
1
2
x =
,y =
, t  – 3, t  5
t + 3
t – 5
Make t the subject:
x =
1
t + 3
 t + 3 =
1
x
 t =
1
x
– 3 =
1 – 3x
x
Substitute into the y equation:
t – 5 =
1 – 3x
x
y = 2 
– 5 =
1 – 8x
x
1 – 8x
x
 y =
2x
1 – 8x
Ex 2A p 10
JMcC
2
C4 Chapter 2
USING PARAMETRIC EQUATIONS TO SOLVE PROBLEMS
Example 1
Find the coordinates of the point where the curve with
parametric equations: x = √t, y = 4t – 25 meets the x-axis
At the x – axis, y = 0
x =
t = 
25
4
= 
 4t – 25 = 0  t =
25
4
5
2
 5

Points of intersection are  , 0 
 2

and

5

–
,
0


2


Example 2
Find the coordinates of the points of intersection of the curve
with parametric equations: x = 2t, y = t2 and the straight line
y=x+3
Substitute the parameters into the equation of the line:
2
2
t = 2t + 3  t – 2t – 3 = 0
 (t – 3)(t + 1) = 0  t = 3, – 1
 when t = 3  x = 6 and y = 9
 when t = – 1  x = – 2 and y = 1
Points of intersection are (6, 9) and ( – 2, 1)
Ex 2B p 13
JMcC
3
C4 Chapter 2
PARAMETRIC EQUATIONS INVOLVING TRIGONOMETRIC
PARAMETERS
You will need to use some of the Trigonometric Identities from
C3
Example 1
Find the Cartesian equation of the curve given by:
x = cos t, y = sin 2t
y = sin 2t = 2 sin t cos t = 2x sin t
 we need to find sin t in terms of x or y
2
2
sin t = 1 – cos t  sin t =
2
2
1 – cos t =
1 – x
2
y = 2x 1 – x
Example 2
Find the Cartesian equation of the curve given by:
x = sec t, y = 4 sin t
y = 4 sin t and x = sec t =
sin t =
2
1 – cos t =
1
cos t
2
 cos t =
1 – (1  x ) =
1
x
2
x – 1
2
x
2
=
(x – 1)
x
2
 y =
4 (x – 1)
x
Ex 2C p 15
JMcC
4
C4 Chapter 2
AREA UNDER A CURVE GIVEN BY PARAMETRIC
EQUATIONS
Since Area =


y dx, Chain Rule 
y dx =

y
dx
dt
dt
Example
The diagram shows the curve with parametric equations:
2
x = t , y = t(5 – t)
The region R is bounded by the curve and the x-axis.
Find the area of R.
y
6
4
R
2
0
5
10
15
20
25 x
Curve crosses x – axis at (25, 0)  t(5 – t) = 0  t = 5
Area =
=

5
0

25
y dx =
0

5
y
0
dx
dt
dt =
3
 10t
(10t – 2t ) dt = 
 3
2
3

5
t(5 – t)  2t dt
0
4
5
2t 
–
= 104·17

4  o
Ex 2D p 18 & Mixed Ex 2E p 19
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