Review exercise 1
1 a z1 − z2
= 4 − 5i − pi
=4 − (5 + p )i
b z1 z2
= (4 − 5i) pi
= 4 pi − 5 pi 2
= 4 pi + 5 p
= 5 p + 4 pi
c
z1
z2
=
4 − 5i
pi
i ( 4 − 5i )
−p
4i + 5
=
−p
5 4
=
− − i
p p
=
2 a
z 3 − kz 2 + 3 z
= z ( z 2 − kz + 3)
So if there are 2 imaginary roots, the
discriminant of z 2 − kz + 3 < 0
⇒ (−k ) 2 − 12 < 0
k 2 < 12
−2 3 < k < 2 3
b z 3 − 2 z 2 + 3z =
0
0
⇒ z ( z 2 − 2 z + 3) =
2 ± −8
2
⇒ z =0, z =1 ± i 2
⇒ z= 0, z=
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1
3
5 ± 25 − 52
2
5 ± −27
=
2
5 3 3
=
±
i
2
2
5 3 3 5 3 3
+
So z1 , z2 =
i, −
i
2
2 2
2
z=
4 (2 − i) x − (1 + 3i) y − 7 =
0
⇒ (2 x − y − 7) + (− x − 3 y ) i =0
⇒ 2 x − y= 7, x + 3 y= 0
3, y =
⇒x=
−1
5 a
b
2 + 3i 5 − i 10 − 2i + 15i + 3
×
=
5+i 5−i
26
13 + 13i 1 1
=
=
+ i
26
2 2
1
=
(1 + i)
2
1
λ=
2
4
(5 + i)(5 − i) = 52 + 12 = 26 You should practise doing such
calculations mentally.
You use the result from part a to simplify the working in part b.
4
 2 + 3i   1

=

  (1 + i) 
 5+i  2

1
=
(1 + 4i + 6i 2 + 4i3 + i 4 )
16
1
=
(1 + 4i − 6 − 4i +1)
16
1
1
= × −4 = − , a real number
16
4
(1 + i) 4 is expanded using the binomial expansion
(a + b) 4 =a 4 + 4a 3b + 6a 2b 2 + 4ab3 + b 4
i3 =i 2 × i =−1× i =−i
i 4 = i 2 × i 2 = −1× −1 = 1
6 −1 + i is a root ⇒ −1 − i is also a root
⇒ ( z + 1 − i)( z + 1 + i) is a factor
⇒ z 2 + 2 z + 2 is a factor
⇒ z 3 + 5 z 2 + 8 z + 6=
⇒ z =−3, −1 ± i
(z 2 + 2 z + 2)( z + 3)
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2
7 a f(2 − 3i) =
0
⇒ (2 − 3i)3 − 6(2 − 3i) 2 + k (2 − 3i) − 26 =
0
⇒ 8 − 36i − 54 + 27i − 24 + 72i + 54 + 2k − 3ki − 26 =
0
Equating real coefficients − 42 + 2k = 0 ⇒ k = 21
b 2 + 3i must also be a factor
⇒ ( z − 2 + 3i)( z − 2 − 3i) = z 2 − 4 z + 13 is a factor
⇒ z 3 − 6 z 2 + 21z − 26 = ( z 2 − 4 z + 13)( z − 2)
⇒ z= 2, 2 + 3i are the other two factors
8 a b − 3 =−1 ⇒ b =2
4
−4c =
−16 ⇒ c =
⇒ z 4 − z 3 − 6 z 2 − 20 z − 16 = ( z 2 − 3 z − 4)( z 2 + 2 z + 4)
b z 4 − z 3 − 6 z 2 − 20 z − 16 = ( z − 4)( z + 1)( z 2 + 2 z + 4)
−2 ± 12
2
⇒ z = 4, −1, −1 ± 3i
⇒ z = 4, −1,
9 ( z − 1 − 2i)( z − 1 + 2i) must be a factor
⇒ z 2 − 2 z + 5 is a factor
⇒ z 4 − 8 z 3 + 27 z 2 − 50 z + 50
= ( z 2 − 2 z + 5)( z 2 + kz + 10)
Equating coefficients of z3
−2 + k =−8 ⇒ k =−6
0
⇒ ( z 2 − 2 z + 5)( z 2 − 6 z + 10) =
6 ± −4
2
⇒ z =1 ± 2i,3 ± i
⇒ z =1 ± 2i,
10 a Comparing constant coefficients
α×
4
α
× (α +
⇒ 4(α +
4
α
4
α
+ 1) =
12
+ 1) =
12
⇒ α 2 + 4 +α =
3α
⇒ α 2 − 2α + 4 =
0
⇒ α =1 ± 3i
So the roots are 1 ± 3i,3
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3
10 b f( z ) = ( z − 3)( z − 1 − 3i)( z − 1 − 3i)
= ( z − 3)( z 2 − 2 z + 4)
=z 3 − 5 z 2 + 10 z − 12
⇒ p=
−5, q =
10
11 a
3z − 1
4
=
2 − i 1 + 2i
8 − 4i 1 − 2i
×
3 z −=
1
1 + 2i 1 − 2i
8 − 16i − 4i − 8 −20i
=
=
= −4i
5
5
3 z = 1 − 4i
1 4
z= − i
3 3
b
You multiply both sides of the equation by 2 − i.
Then multiply the numerator and denominator by
the conjugate complex of the denominator.
You place the points in the Argand diagram which represent
conjugate complex numbers symmetrically about the real x-axis.
Label the points so it is clear which is the original number (z)
and which is the conjugate (z*).
c
2
3
 1   4  1 16 17
z =   +−  = + =
9
3  3 9 9
2
z =
17
3
tan θ = = 4 ⇒ θ ≈ 76
The diagram you have drawn in part b shows that
z is in the fourth quadrant. There is no need to
draw it again.
z is in the fourth quadrant.
It is always true z* = z and arg z* = − arg z ,
arg z = −76, to the nearest degree.
so you just write down the final answer without
further working.
4
3
1
3
=
z
=
z*
17
17
cos(−76) + i
sin(−76)
3
3
17
17
cos 76 +i
sin 76
3
3
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4
12 a
b tan θ =
17
⇒ θ = 1.084...
9
z is in the second quadrant.
You have to give your answer to 2 decimal places. To do this
accurately you must work to at least 3 decimal places. This avoids
rounding errors and errors due to premature approximation.
arg z =
π − 1.084 =
2.057 
= 2.06, in radians to 2 d.p.
c
25 + 35i 25 + 35i 25 + 35i −9 − 17i
= =
×
z
−9 + 17i −9 + 17i −9 − 17i
−225 − 425i − 315i + 595
=
(−9) 2 + 17 2
370 − 740i
=
= 1 − 2i
370
=
w
13 a
In this question, the arithmetic gets complicated.
Use a calculator to help you with this. However,
when you use a calculator, remember to show
sufficient working to make your method clear.
z 2 =(2 + i) 2 =4 − 4i + i 2
=4 − 4i − 1
= 3 − 4i, as required.
b From part a, the square roots of
3 − 4i are 2 − i and −2 + i.
Taking square roots of both sides
of the equation ( z + i) 2 =3 − 4i
z + i = 2 − i ⇒ z = 2 − 2i
z + i =−2 + i ⇒ z =−2
z1= 2 − 2i , say, and z2 = −2
The square root of any number k, real or complex, is a root
of z 2 = k . Hence, part a shows that one square root of
3 − 4i is 2 − i .
If one square root of 3 − 4i is 2 − i , then the other is
−(2 − i).
z1 and z2 could be the other way round but that
would make no difference to z1 − z2 or z1 − z2 , the
expressions you are asked about in parts d and e.
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5
13 c
z1 − z2 can be represented on the diagram you drew in part c
by the vector joining the point representing z1 to the point
representing z2 . The modulus of z1 − z2 is then just the
length of the line joining these two points and this length can
be found using coordinate geometry.
d Using the formula
d 2 = ( x1 − x2 ) 2 + ( y1 − y2 ) 2
= (2 − (−2)) 2 + (−2 − 0) 2
2
Hence z1 − z2 =
20 = 2 5
2
= 4 + 2 = 20
e
z1 + z2 =2 − 2i − 2 =−2i
The argument of any number on the negative
imaginary axis is −
π
or −90.
2
π
arg ( z1 + z2 ) =
−
2
14 a g ( x ) = x3 − x 2 − 1
g (1.4 ) = −0.216
g (1.5 ) = 0.125
There is a change of sign so there must be a root between x = 1.4 and x = 1.5
b g (1.4655 ) = −0.00025...
g (1.4665 ) = 0.00326...
Therefore 1.4655 < α < 1.4665 and so α = 1.466 to 2 d.p.
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6
15 a i
3 x 2 + 4 x − 1 =0 has roots α and β
4
3
1
The product of the roots is αβ = −
3
2
2
2
α + β = (α + β ) − 2αβ
The sum of the roots is α + β =
−
2
 4
 1
= −  − 2  − 
 3
 3
22
=
9
ii α 3 + β 3 = (α + β ) − 3α 2 β − 3αβ 2
3
=(α + β ) − 3αβ (α + β )
3
3
 4
 1  4 
= −  − 3  −   − 
 3
 3  3 
100
= −
27
b Roots are
β
α
and 2
2
α
β
Sum of the roots is:
α
β α3 + β3
+
=2 2
2
2
β
α
α β
=
α3 + β3
2
(αβ )
 100 
−

27 
=
2
 1
− 
 3
100
= −
3
Product of the roots is:
α  β  1
=
β 2  α 2  αβ
1
=
 1
− 
 3
= −3
So
100
x2 +
x −3 =
0
3
3 x 2 + 100 x − 9 =
0
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7
16 a 2 x 2 + 5 x − 4 =
0 has roots α and β
5
2
The product of the roots is αβ = −2
The sum of the roots is α + β =
−
1
b When roots are
1
and
α
β
The sum of the roots is:
1 1 α +β
+ =
α β
αβ
 5
− 
2
=
( −2 )
5
4
The product of the roots is:
11 1
=
α  β  αβ
1
=
( −2 )
=
1
2
= −
So
5
1
x2 − x − =
0
4
2
4 x2 − 5x − 2 =
0
17 a i
x 2 − 3x + 1 =
0 has roots α and β
The sum of the roots is α + β =
3
The product of the roots is αβ = 1
α 2 + β 2 = (α + β ) − 2αβ
2
( 3)
=
2
− 2 (1)
=7
ii α 3 + β 3 = (α + β ) − 3α 2 β − 3αβ 2
3
=(α + β ) − 3αβ (α + β )
3
( 3)
=
3
− 3 (1)( 3)
= 18 as required
iii α 4 + β 4 = (α 2 + β 2 ) − 2α 2 β 2
2
=(α 2 + β 2 ) − 2 (αβ ) as required
2
2
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8
17 b When roots are α 3 − β and β 3 − α
The sum of the roots is:
α 3 − β + β 3 − α = α 3 + β 3 − (α + β )
= 18 − 3
= 15
The product of the roots is:
(α 3 − β )( β 3 − α )= α 3β 3 − α 4 − β 4 + αβ
=
(αβ )
3
− (α 4 + β 4 ) + αβ
=
(αβ )
3
− (α 2 + β 2 ) − 2 (αβ ) + αβ
(
(
2
)
2
)
=(1) − ( 7 ) − 2 (1) + (1)
3
2
2
= −45
So x − 15 x − 45 =
0
2
18 P has coordinates (x, y)
Let the distance from (x, y) to (5, 0) be d
d=
( x − 5) + ( y − 0 )
2
2
Let the distance from (x, y) to the line x = −5 also be d
d=
( x + 5) + ( y − y )
2
2
= x+5
Equating both values of d gives:
( x − 5) + ( y − 0 ) =x + 5
2
2
( x − 5) + y 2 = ( x + 5)
2
2
y 2 = ( x + 5) − ( x − 5)
2
2
=x 2 + 10 x + 25 − x 2 + 10 x − 25
= 20 x
So the locus of P is of the form y2 = 4ax with a = 5
19 a y2 = 16x
A parabola of the form y2 = 4ax has the focus at (a, 0)
So y2 = 16x has the focus S at (4, 0)
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9
19 b P has coordinates (16, 16)
The gradient of PS is:
y − yS
mPS = P
xP − xS
16 − 0
=
16 − 4
4
=
3
To find the equation of PS use
4
y − y1= m ( x − x1 ) with m = at (4, 0)
3
4
y − 0=
( x − 4)
3
y 4 x − 16
3=
4 x − 3 y − 16 =
0
c 4 x − 3 y − 16 =
0 meets y2 = 16x at Q
y2
16
y2
Substituting x =
into 4 x − 3 y − 16 =
0 gives:
16
 y2 
4   − 3 y − 16 =
0
 16 
y 2= 16 x ⇒ x=
0
y 2 − 12 y − 64 =
0
( y − 16 )( y + 4 ) =
y = 16 or y = −4
When y = −4, x = 1 so Q has coordinates (1, −4)
20 a C has equations x = 3t2, y = 6t
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10
20 b y= x − 72 meets C at A and B
Substituting x = 3t2 and y = 6t into y= x − 72 gives:
6=
t 3t 2 − 72
0
t 2 − 2t − 24 =
0
( t + 4 )( t − 6 ) =
t=
6
−4 or t =
When t = −4
x = 3t 2
= 3 ( −4 )
2
= 48
y = 6t
= 6 ( −4 )
= −24
So A is the point (48, −24)
When t = 6
x = 3t 2
= 3( 6)
2
= 108
y = 6t
= 6 ( 6)
= 36
So B is the point (108, 36)
The length of AB is given by:
=
AB
(108 − 48) + ( 36 − ( −24 ) )
=
602 + 602
2
2
= 60 2
21 A parabola of the form y 2 = 4ax has the directrix at x + a = 0
So the parabola y 2 = 12 x has the directrix at x + 3 = 0
If P and Q lie on the parabola at a distance of 8 from the directrix then they both have x-coordinates
of 5.
When x = 5
y 2 = 12 x
= 12 ( 5 )
= 60
y = ±2 15
(
)
(
So P is the point 5, 2 15 and Q is the point 5, −2 15
)
The distance PQ is 4 15
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11
22 a P(2, 8) lies on y 2 = 4ax
Substituting (2, 8) into y 2 = 4ax gives:
(8)
2
= 4 ( 2) a
a =8
So y 2 = 32 x
1
2
b y = 32 x ⇒ y= 4 2 x
The tangent of a point to the curve is:
1
−
dy
= 2 2x 2
dx
2 2
=
x
At x = 2
dy 2 2
=
dx
2
2
( )
=2
To find the equation of the tangent use
y − y1= m ( x − x1 ) with m = 2 at (2, 8)
y − 8= 2 ( x − 2 )
=
y 2x + 4
c The tangent cuts the x-axis at y = 0
Substituting y = 0 into =
y 2 x + 4 gives:
( 0=) 2 x + 4
x = −2
So X has coordinates (−2, 0)
The tangent cuts the y-axis at x = 0
Substituting x = 0 into =
y 2 x + 4 gives:
=
y 2 ( 0) + 4
y=4
So Y has coordinates (0, 4)
1
Area OXY = ( 2 )( 4 )
2
=4
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12
23 a P has coordinates (3, 4) and lies on xy = 12
Q has coordinates (−2, 0)
The gradient of l is
y − yQ
ml = P
xP − xQ
=
4−0
3 − ( −2 )
4
5
To find the equation of l use
=
y − y1= m ( x − x1 ) with m =
4
( x + 2)
5
4
8
=
y
x+
5
5
4
at (−2, 0)
5
y − 0=
4
8
x + cuts xy = 12 at the point R
5
5
12
4
8
Substituting y =
into =
y
x + gives:
5
5
x
12 4
8
x+
=
5
x 5
2
60 4 x + 8 x
=
b =
y
0
x 2 + 2 x − 15 =
0
( x − 3)( x + 5) =
x = 3 or x = −5
12
5
12 

So R has coordinates  −5, − 
5

When x = −5, y = −
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13
24 a P has coordinates (12, 3) and lies on xy = 36
xy = 36 ⇒ y = 36 x −1
The gradient of the tangent to a point on the curve is:
dy
= −36 x −2
dx
36
= − 2
x
At x = 12
dy
36
= −
2
dx
(12 )
1
4
To find the equation of the tangent at (12, 3) use
1
y − y1= m ( x − x1 ) with m = − at (12, 3)
4
1
y − 3 =− ( x − 12 )
4
4 y − 12 =− x + 12
x + 4 y − 24 =
0
= −
b The tangent cuts the x-axis at M and the y-axis at N
At M, y = 0
Substituting y = 0 into x + 4 y − 24 =
0 gives:
x + 4 ( 0 ) − 24 =
0
x = 24
Therefore M has coordinates (24, 0)
At N, x = 0
Substituting x = 0 into x + 4 y − 24 =
0 gives:
0
( 0 ) + 4 y − 24 =
y=6
Therefore N has coordinates (0, 6)
The length of MN is given by:
( 24 − 0 ) + ( 0 − 6 )
2
MN=
2
= 6 17
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14
25 C has equations x = 8t, y =
16
t
1
x + 4 intersects C at A and B
4
16
1
Substituting x = 8t and y = into =
y
x + 4 gives:
t
4
 16  1
(8t ) + 4
=

 t  4
The line =
y
= 2t 2 + 4t
16
t 2 + 2t − 8 =
0
0
( t − 2 )( t + 4 ) =
t = 2 or t = −4
When t = 2
x = 8(2)
= 16
and
16
y=
( 2)
=8
So A has coordinates (16, 8)
When t = −4
x= 8(−4)
= −32
and
16
y=
( −4 )
= −4
So B has coordinates (−32, −4)
The midpoint of AB is found using:
 x A + xB y A + yB   16 + ( −32 ) 8 + ( −4 ) 
,
,


=
2  
2
2 
 2
= ( −8, 2 )
So M has coordinates (−8, 2)
26 a P ( 24t 2 , 48t ) lies on y 2 = 96 x and on xy = 144
Substituting ( 24t 2 , 48t ) into xy = 144 gives:
( 24t ) ( 48t ) = 144
2
144
1152
1
t=
2
Therefore P has coordinates (6, 24)
t3 =
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15
1
2
26 b y = 96 x ⇒ y= 4 6 x
The gradient of a tangent to a point on the parabola is:
1
−
dy
= 2 6x 2
dx
2 6
= 1
x2
At x = 6
dy 2 6
=
1
dx
6 2
2
( )
=2
To find the equation of the tangent at (6, 24) use
y − y1= m ( x − x1 ) with m = 2 at (6, 24)
y − 24 = 2 ( x − 6 )
=
y 2 x + 12
27 a P(9, 8) and Q(6, 12) lie on xy = 72
The gradient of PQ is found using:
yP − yQ
mPQ =
xP − xQ
8 − 12
9−6
4
= −
3
To find the equation of PQ:
=
y − y1= m ( x − x1 ) with m = −
4
− ( x − 6)
y − 12 =
3
−4 x + 24
3 y − 36 =
4x + 3y =
60 as required
4
at (6, 12)
3
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16
27 b The tangent at R is parallel to PQ therefore the gradient at R is −
xy = 72 ⇒ y = 72 x −1
The gradient of a tangent to a point on the curve is:
dy
= −72 x −2
dx
72
= − 2
x
Equating both values of the tangent gives:
72
4
− 2 =
−
x
3
2
4 x = 216
4
3
x = ±3 6
When
6, y 4 6
x 3=
=
When x =
−3 6, y =
−4 6
(
)
(
So the possible coordinates of R are 3 6, 4 6 and −3 6, −4 6
)
 3
28 a The point  3t ,  lies on xy = 9
 t
xy = 9 ⇒ y = 9 x −1
The gradient of a tangent to a point on the curve is:
dy
= −9 x −2
dx
9
= − 2
x
At x = 3t
dy
9
= −
2
dx
( 3t )
= −
1
t2
 3
To find the equation of the tangent at  3t ,  use:
 t
1
 3
y − y1= m ( x − x1 ) with m = − 2 at  3t , 
t
 t
3
1
y− =
− 2 ( x − 3t )
t
t
2
t y − 3t =− x + 3t
6t as required
x + t2 y =
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17
 3
28 b The tangent at  3t ,  cuts the x-axis at A and the y-axis at B
 t
At A, y = 0
Substituting y = 0 into x + t 2 y =
6t gives:
x + t 2 ( 0) =
6t
x = 6t
So A has coordinates (6t, 0)
At B, x = 0
Substituting x = 0 into x + t 2 y =
6t gives:
6t
( 0) + t 2 y =
y=
6
t
 6
So B has coordinates  0, 
 t
1
6
Area OAB = ( 6t )  
2
t
= 18
Therefore the area of triangle OAB is constant
 c
29 a The point  ct ,  lies on xy = c2
 t
2
xy = c ⇒ y = c 2 x −1
The gradient of a tangent to a point on the curve is:
dy
= −c 2 x −2
dx
c2
= − 2
x
At x = ct
dy
c2
= −
2
dx
( ct )
1
t2
1
 c
At  ct ,  the tangent has gradient − 2 so the normal has gradient t2
t
 t
 c
To find the equation of the normal at  ct ,  use:
 t
 c
y − y1= m ( x − x1 ) with m = t2 at  ct , 
 t
c
y − = t 2 ( x − ct )
t
ty − c = t 3 x − ct 4
= −
0 as required
t 3 x − ty − c ( t 4 − 1) =
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18
29 b The normal at P meets the line y = x at G
Substituting y = x into t 3 x − ty − c ( t 4 − 1) =
0 gives:
t 3 x − t ( x ) − c ( t 4 − 1) =
0
tx ( t 2 − 1)= c ( t 4 − 1)
x=
=
c ( t 4 − 1)
t ( t 2 − 1)
c ( t 2 + 1)
t
 c ( t 2 + 1) c ( t 2 + 1) 

,
and since x = y, G has coordinates 


t
t


The length of PG is found using:
PG =

c ( t 2 + 1)   c c ( t 2 + 1) 
 ct −
 + −


 t

t
t

 

=
 ct 2 − c ( t 2 + 1)   c − c ( t 2 + 1) 

 +


 

t
t

 

2
2
2
=
2
 −c   ct 
+
−

  
 t   t 
2
2
2
2
2
 −c 
PG
=   + ( −ct )
 t 
c2
= 2 + c 2t 2
t
1

= c 2  2 + t 2  as required
t

2
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19
 c
30 a The point  ct ,  lies on xy = c2
 t
2
xy = c ⇒ y = c 2 x −1
The gradient of a tangent to a point on the curve is:
dy
= −c 2 x −2
dx
c2
= − 2
x
At x = ct
dy
c2
= −
2
dx
( ct )
= −
1
t2
 c
To find the equation of the tangent at  ct ,  use:
 t
1
 c
y − y1= m ( x − x1 ) with m = − 2 at  ct , 
t
 t
c
1
y− =
− 2 ( x − ct )
t
t
2
t y − ct =− x + ct
x + t2 y =
2ct as required
b Tangents are drawn from (−3, 3) to xy = 16
Substituting (−3, 3) into x + t 2 y =
2ct gives:
2
2ct
( −3) + t ( 3) =
3t 2 − 2ct − 3 =
0
Comparing xy = 16 to xy = c2 gives c = 4 , since the formula for a rectangular hyperbola assumes c to
be a positive constant
When c = 4 the equation of the tangent is:
3t 2 − 2 ( 4 ) t − 3 =
0
3t 2 − 8t − 3 =
0
0
( 3t + 1)( t − 3) =
1
t=
− or t =
3
3
When c = 4 and t = −
1
3
 c  4

 ct ,  = − , −12 
 t  3

When c = 4 and t = 3
 c  4
 ct ,  = 12, 
 t   3
So the coordinates of the points where the tangent meets the curve are
 4

 4
 − , −12  and 12, 
 3

 3
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20
31 P ( at 2 , 2at ) , t > 0 lies on y 2 = 4ax
1
y 2= 4ax ⇒ y= 2 ax 2
The gradient of a tangent to a point on the curve is:
1
−
dy
= ax 2
dx
a
= 1
x2
At x = at2
dy
a
=
1
dx
( at 2 ) 2
1
t
To find the equation of the tangent use:
1
y − y1= m ( x − x1 ) with m = at ( at 2 , 2at )
t
1
y − 2at = ( x − at 2 )
t
2
ty − 2at =
x − at 2
=
x − ty + at 2 =
0
The tangent cuts the x-axis at T where y = 0
Substituting y = 0 into x − ty + at 2 =gives:
0
2
x − t ( 0 ) + at =
0
x = −at 2
So T has coordinates (−at2, 0)
The length of PT is found using:
PT
=
( at − ( −at ))
2
2
=
4a 2t 4 + 4a 2t 2
=
4a 2t 2 ( t 2 + 1)
= 2at
(t
2
2
+ ( 2at − 0 )
2
+ 1)
Since the gradient of the tangent at ( at 2 , 2at ) is
To find the equation of the normal use:
y − y1= m ( x − x1 ) with m = −t at ( at 2 , 2at )
1
the gradient of the normal is −t
t
y − 2at =
−t ( x − at 2 )
y − 2at =−tx + at 3
tx + y − 2at − at 3 =
0
The normal cuts the x-axis at N where y = 0
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21
Substituting y = 0 into tx + y − 2at − at 3 =gives:
0
tx + 0 − 2at − at 3 =
0
=
x a (2 + t2 )
So N has coordinates (a(2 + t2),0)
The length of PN is found using:
PN =
( at
=
4a 2 + 4a 2t 2
2
− a (2 + t 2 ) ) + ( 2at − 0 )
2
2
( t + 1)
2at ( t + 1)
=
2a ( t + 1)
= 2a
2
2
PT
PN
2
=t
32 a P ( ap 2 , 2ap ) , p > 0 lies on y 2 = 4ax
1
2
2
y = 4ax ⇒ y= 2 ax
The gradient of a tangent to a point on the curve is:
1
−
dy
= ax 2
dx
a
= 1
x2
At x = ap2
dy
a
=
1
dx
2 2
( ap )
1
p
To find the equation of the tangent use:
1
y − y1= m ( x − x1 ) with m = at ( ap 2 , 2ap )
p
1
y − 2ap = ( x − ap 2 )
p
2
py − 2ap =
x − ap 2
=
py= x + ap 2 as required
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22
32 b The tangents at P ( ap 2 , 2ap ) and Q ( aq 2 , 2aq ) meet at N
The tangent at P has equation:
x
py = x + ap 2 ⇒ y = + ap
p
The tangent at Q has equation:
x
qy = x + aq 2 ⇒ y = + aq
q
To find the coordinates of N, equate the equations:
x
x
+ ap = + aq
p
q
x x
− = aq − ap
p q
qx − px
= a (q − p)
pq
x ( q − p=
) apq ( q − p )
apq
As p ≠ x, x =
When x = apq
( apq ) + aq
=
y
q
= a ( p + q)
So N has coordinates (apq, a(p + q))
c Since N lies on y = 4a
a(p + q) = 4a
p=4−q
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23
33 a P ( at 2 , 2at ) , t > 0 lies on y 2 = 4ax
1
y 2= 4ax ⇒ y= 2 ax 2
The gradient of a tangent to a point on the curve is:
1
−
dy
= ax 2
dx
a
= 1
x2
At x = at2
dy
a
=
1
dx
( at 2 ) 2
=
1
t
At ( at 2 , 2at ) the gradient of the tangent is
1
so the gradient of the normal is −t
t
To find the equation of the normal use:
y − y1= m ( x − x1 ) with m = −t at ( at 2 , 2at )
y − 2at =
−t ( x − at 2 )
y − 2at =−tx + at 3
y + tx = 2at + at 3 as required
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24
33 b The normal meets the curve again at Q
y
y + tx =
2at + at 3 ⇒ x =
− + 2a + at 2
t
y
Substituting x =
− + 2a + at 2 into y 2 = 4ax gives:
t
y


y 2= 4a  − + 2a + at 2 
 t

4ay
− 8a 2 − 4 a 2 t 2 =
y2 +
0
t
We know that y = 2at is one solution to this equation.
So, use the fact that ( y − 2at ) is a factor to rewrite the left-hand side
( y − 2at )  y + 2at +

4a 
0
=
t 
−2at −
y=
2at or y =
When y =
−2a ( t 2 + 2 )
t
Substituting y =
4a −2a (t 2 + 2)
=
t
t
−2a ( t 2 + 2 )
 −2a ( t 2 + 2 ) 

4ax = 


t


2 2
2
4a (t + 2)
=
t2
a (t 2 + 2) 2
x=
t2
2
t
into y 2 = 4ax gives:
 a ( t 2 + 2 )2 −2a ( t 2 + 2 ) 

So Q has coordinates 
,


t2
t


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25
 c
34 a The point  ct ,  lies on xy = c2
 t
2
xy = c ⇒ y = c 2 x −1
The gradient of a tangent to a point on the curve is:
dy
c2
=
−c 2 x −2 =
− 2
dx
x
At x = ct
dy
c2
= −
2
dx
( ct )
= −
1
t2
1
so the gradient of the normal is t2
t2
 c
To find the equation of the normal at  ct ,  use:
 t
 c
y − y1= m ( x − x1 ) with m = t 2 at  ct , 
 t
c
y − = t 2 ( x − ct )
t
c
y − = t 2 x − ct 3
t
c
y = t 2 x + − ct 3 as required
t
At x = ct the gradient of the tangent is −
b The normal meets the equation again at Q
c
Substituting y = t 2 x + − ct 3 into xy = c2 gives:
t
c

 2
x  t 2 x + − ct 3  =
c
t


cx
t 2 x 2 + − ct 3 x − c 2 =
0
t
c2
c

x 2 +  3 − ct  x − 2 =
0
t
t

c
0
=
t3 

c
x = ct or x = − 3
t
c
Substituting x = − 3 into xy = c 2 gives:
t
 c
c2 ⇒ y =
−ct 3
− 3  y =
 t 
 c

So Q is the point  − 3 , −ct 3 
 t

( x − ct )  x +
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26
 c
34 c P has coordinates  ct ,  and Q has coordinates
 t
The midpoint of PQ is found using:


 c
ct +  − 3  c + ( −ct 3 ) 

+
+
x
x
y
y
 P Q P
Q 
 t , t

,

=
2
2
2
2








 c ( t 4 − 1) c (1 − t 4 ) 

=
,
 2t 3
2t 


4
4
c ( t − 1)
c (1 − t )
So X =
=
and
Y
2t 3
2t
4
c ( t − 1)
3
X
= 2t 4
Y c (1 − t )
=
 c
3
 − 3 , −ct 
 t

2t
2ct ( t 4 − 1)
2ct 3 (1 − t 4 )
= −
1
as required
t2

c
35 a The point P  cp,  lies on xy = c2
p

2
2 −1
xy = c ⇒ y = c x
The gradient of a tangent to a point on the curve is:
dy
= −c 2 x −2
dx
c2
= − 2
x
At x = cp
dy
c2
= −
2
dx
( cp )
= −
1
p2

c
To find the equation of the tangent at  cp,  use:
p


c
1
y − y1= m ( x − x1 ) with m = − 2 at  cp, 
p
p

c
1
y− =
− 2 ( x − cp )
p
p
p 2 y − pc =− x + cp
p 2 y =− x + 2cp as required
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27


c
c
35 b The tangents at P  cp,  and Q  cq,  meet at N
p
q


The tangent at P has equation:
p 2 y =− x + 2cp ⇒ x =2cp − p 2 y
The tangent at Q has equation:
q 2 y =− x + 2cq ⇒ x =2cq − q 2 y
To find the coordinates of N, equate the equations:
2cp − p 2 y = 2cq − q 2 y
y ( q 2 − p 2 ) = 2c ( q − p )
(q − p)
(q − p)
=
y 2=
c 2
2c
2
(q − p )
(q − p )(q + p )
2c
y=
as required
p+q

c
c P has coordinates  cp,  and Q has coordinates
p

The gradient of PQ is found using:
y − yQ
mPQ = P
xP − xQ

c
 cq, 
q

c c
−
p q
=
cp − cq
q− p
=
pq ( p − q )
= −
1
pq
N has y-coordinate
2c
. Substituting this into the equation=
x 2cp − p 2 y gives:
p+q
 2c 
=
x 2cp − p 2 

 p+q
2cp ( p + q ) − 2cp 2
x=
p+q
2cpq
=
p+q
 2cpq 2c 
O has coordinates (0, 0) and N has coordinates 
,

 p+q p+q
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28
The gradient of ON is found using:
y − yO
mON = N
xN − xO
2c
p+q
=
2cpq
p+q
=
2c ( p + q )
2cpq ( p + q )
1
pq
Since mPQ and mON are perpendicular
mPQ × mON =
−1
=
−
1
1
×
=
−1
pq pq
p2q2 = 1
36 a P ( ap 2 , 2ap ) , p ≠ 0 lies on y 2 = 4ax
1
2
2
y = 4ax ⇒ y= 2 ax
The gradient of a tangent to a point on the curve is:
1
−
dy
= ax 2
dx
a
= 1
x2
At x = ap2
dy
a
=
1
dx
2 2
( ap )
=
1
p
1
so the normal has gradient − p
p
To find the equation of the normal use:
y − y1= m ( x − x1 ) with m = − p at ( ap 2 , 2ap )
At x = ap2 the tangent has gradient
y − 2ap =
− p ( x − ap 2 )
y − 2ap =
− px + ap 3
y + px = 2ap + ap 3 as required
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29
36 b The normal meets the curve again at Q ( aq 2 , 2aq )
Substituting ( aq 2 , 2aq ) into y + px = 2ap + ap 3 gives:
2ap + ap 3
( 2aq ) + p ( aq 2 ) =
apq 2 + 2aq − ap ( 2 + p 2 ) =
0
2q
− ( 2 + p2 ) =
0
p
We know that q = p would give one solution to this equation.
So, use the fact that ( q − p ) is a factor to rewrite the left-hand side:
q2 +


2 
(q − p)  q +  p +   =
0
p 


2
q =p or q =− p −
p
Since q ≠ p
2
q =− p −
p
 125

c The midpoint of PQ is 
a, −3a 
 18

P has coordinates P ( ap 2 , 2ap ) and Q has coordinates P ( aq 2 , 2aq )
Find the midpoint of PQ using:
 xP + xQ yP + yQ   ap 2 + aq 2 2ap + 2aq 
,
,

=

2  
2
2

 2
 a ( p2 + q2 )

, a ( p + q)
= 


2


  2

4 
 a2p + 4+ 2 

p  2a 
 125a



Comparing 
,−
, −3a  to

2
p 
 18





−3a = a ( p + q )
p + q =−3
Since q =− p −
p=
2
3

2
2
, p +  − p −  =−3
p
p

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30
37 a A parabola of the form y 2 = 4ax has the focus at (a, 0)
So y 2 = 32 x has the focus S at (8, 0)
b A parabola of the form y 2 = 4ax has the directrix at x + a = 0
So y 2 = 32 x has the directrix at x + 8 = 0
c P has coordinates (2, 8) and Q has coordinates (32, −32)
The gradient of PQ is found using:
yP − yQ
mPQ =
xP − xQ
8 − ( −32 )
2 − 32
4
= −
3
To find the equation of PQ use:
4
y − y1= m ( x − x1 ) with m = − at (2, 8)
3
4
y − 8 =− ( x − 2 )
3
3 y − 24 =
−4 x + 8
4 x + 3 y − 32 =
0
If S lies on 4 x + 3 y − 32 =
0 then (8, 0) will satisfy 4 x + 3 y − 32 =
0
Substituting (8, 0) into 4 x + 3 y − 32 =
0 gives:
4 ( 8 ) + 3 ( 0 ) − 32 =
0
=
0=0
So S lies on PQ
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31
1
2
37 d y =
32 x ⇒ y =
±4 2 x
The gradient of a tangent to point on the curve is:
1
−
dy
= ±2 2 x 2
dx
2 2
= ± 1
x2
At P, x = 2
dy 2 2
=
1
dx
2 2
2
( )
=2
To find the equation of the tangent at P use:
y − y1= m ( x − x1 ) with m = 2 at (2, 8)
y − 8= 2 ( x − 2 )
=
y 2x + 4
At Q, x = 32
Since Q is the point (32, −32), use
dy
2 2
= − 1
dx
x2
dy
2 2
= −
1
dx
( 32 ) 2
1
2
To find the equation of the tangent at Q use:
1
y − y1= m ( x − x1 ) with m = at (32, −32)
2
1
y + 32 =
− ( x − 32 )
2
1
y=
− x − 16
2
Equate the tangents to find the point where they meet:
1
2 x + 4 =− x − 16
2
4 x + 8 =− x − 32
5 x = −40
x = −8
Therefore D lies on the directrix
= −
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32
Challenge
1 x 2 + 2ix + 5 =
0
2
2
( x + i ) − i + 5 =0
( x + i)
2
=
−6
x + i =±i 6
x= i±i 6
(
=
x i 1± 6
)
2 ax 2 + bx + c =
0 has roots α and β
α 4 + β 4 = (α 2 + β 2 ) − 2 (αβ )
2
2
79
7
and α 2 + β 2 =
−
16
4
Substituting gives:
2
79  7 
2
− =−

 − 2 (αβ )
16  4 
49 79
2
2 (αβ =
+
)
16 16
αβ = ±2
Since αβ > 0, αβ = 2
2
(α + β ) = α 2 + β 2 + 2αβ
α4 + β4 =
−
7
4
Substituting gives:
7
2
(α + β ) =− + 2 ( 2 )
4
9
=
4
3
α +β =
±
2
The equation is:
2 x 2 −=
3 x + 4 0 or 2 x 2 +=
3x + 4 0
α2 + β2 =
− and αβ = 2
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33