Parametric Equations
1.
A curve has parametric equations
x = 2 cot t, y = 2 sin2 t, 0 < t 
(a) Find an expression for

.
2
dy
in terms of the parameter t.
dx
(4)
(b) Find an equation of the tangent to the curve at the point where t =

.
4
(4)
(c) Find a cartesian equation of the curve in the form y = f(x). State the domain on which the curve
is defined.
(4)
2.
Figure 1
Y
P
C
R
O
Q
x
Figure 1 shows a sketch of part of the curve C with parametric equations
x = t2 + 1,
y = 3(1 + t).
The normal to C at the point P(5, 9) cuts the x-axis at the point Q, as shown in Figure 1.
(a) Find the x-coordinate of Q.
(6)
(b) Find the area of the finite region R bounded by C, the line PQ and the x-axis.
(9
3.
Figure 1
y
C
O
2
3
1
x
The curve C has parametric equations
x=
1
1
, y=
,  t < 1.
1 t
1 t
(a) Find an equation for the tangent to C at the point where t =
1
2
.
(7)
(b) Show that C satisfies the cartesian equation y =
x
.
2x  1
(3)
The finite region between the curve C and the x-axis, bounded by the lines with equations x =
2
3
and
x = 1, is shown shaded in Figure 1.
(c) Calculate the exact value of the area of this region, giving your answer in the form a + b ln c,
where a, b and c are constants.
(6)
4.
Figure 1
y
1
2
a
A
B
O
a
The curve shown in Figure 1 has parametric equations
x = a cos 3t, y = a sin t,
0t

.
6
The curve meets the axes at points A and B as shown in Figure 1.
x
The straight line shown is part of the tangent to the curve at the point A.
Find, in terms of a,
(a) an equation of the tangent at A,
(6)
(b) an exact value for the area of the finite region between the curve, the tangent at A and the
x-axis, shown shaded in Figure 1.
(9)
5.
Figure 2
y
0.5
–1
–0.5
O
0.5
1
x
The curve shown in Figure 2 has parametric equations


x = sin t, y = sin  t 

,
6


2
<t<

.
2
(a) Find an equation of the tangent to the curve at the point where t =

.
6
(6)
(b) Show that a cartesian equation of the curve is
y=
1
3
x + (1 – x2),
2
2
–1 < x < 1.
(3)
Question
Number
1.
Scheme
Marks
dx
dy
 2 cosec 2 t ,
 4sin t cos t
dt
dt
(a)
M1 A1
both
d y 2sin t cos t

dx
cosec 2 t
  2sin
3
t cos t 
M1 A1
(4)
At t  4 , x  2, y  1
(b)
B1
both x and y
Substitutes t  4 into an attempt at
dy
to obtain gradient
dx
M1
 1
 2 


Equation of tangent is y  1  
1
 x  2
2
M1 A1
Accept x  2 y  4 or any
(4)
correct equivalent
(c)
t
Uses 1  cot 2 t  cosec 2 t , or equivalent, to eliminate
M1
2
2
x
1   
y
2
A1
correctly eliminates t
y
8
4  x2
A1
cao
B1
(4)
The domain is x …0
[12]
An alternative in (c)
1
1
x
x  y 2
 y 2
sin t    ; cos t  sin t   
2
2 2
2
sin t  cos t  1 
2
2
y x2 y
  1
2 4 2
Leading to y 
8
4  x2
M1 A1
A1
2.
Question
Number
Scheme
Marks
_______ ___________________________________________________________________ __________
3. (a)
dx
1

dt
(1  t ) 2

and
B1, B1
dy
1

dt (1  t ) 2
dy (1  t )2
and at t = ½, gradient is –9

dx (1  t )2
M1 requires their dy/dt / their dx/dt
M1 A1cao
and substitution of t.
At the point of contact x =
(b)
2
3
B1
and y =2
Equation is y – 2 = -9 ( x - 23 )
M1 A1
1
1
(or both)
 1 or t  1 
x
y
Then substitute into other expression y = f(x) or x = g(y) and rearrange
1
1
(or put  1  1  and rearrange)
x
y
x
To obtain y 
*
2x 1
M1
(7)
Either obtain t in terms of x and y i,e, t 
1
x
(1  t )
Or Substitute into

2
2x 1
1
1 t
1
1
=

2  (1  t ) 1  t
= y *
(c)
1
Area =
M1
A1
(3)
M1
A1
M1
(3)
x
 2 x  1dx
B1
2
3
u  1 du
1
=  1  u1 du
2u 2
4
1
1
1
=  4 u  4 ln u 1
=

putting into a form to integrate
M1
M1 A1
3
 ( 121  14 ln 13 )
1 1
=  ln 3 or any correct equivalent.
6 4
=
1
4
M1
A1
(6)
Question
Number
4(a)
Scheme
dx
 3a sin 3t ,
dt

6
3
Gradient is 
6
dy
 a cos t
dt
therefore
Marks
dy
cos t

dx 3sin 3t
B1
When x = 0, t =
M1
Line equation is ( y  12 a )  
(b)
Area beneath curve is
3
( x  0)
6
M1 A1
 a sin t (3a sin 3t )dt
3a 2
(cos 2t  cos 4t )dt
2 
3a 2 1
1


2 sin 2t  4 sin 4t 
2
3 3a 2

Uses limits 0 and
to give
16
6
1 a
Area of triangle beneath tangent is   3a =
2 2
2
3a 3 3a 2
3a 2
Thus required area is
=
16
4
16
M1
M1
=
N.B.
M1 A1
M1 A1
M1 A1
3a 2
4
M1 A1
A1
The integration of the product of two sines is worth 3 marks (lines 2 and 3 of
scheme to part (b))
If they use parts
 sin t sin 3tdt   cos t sin 3t   3cos 3t cos tdt
  cos t sin 3t  3cos 3t sin t   9sin 3t sin tdt
8I = cost sin3t – 3 cos3t sint
M1
M1 A1