Ace Ahead Mathematic T Exam Practise Chapter 5
Transcript of Ace Ahead Mathematic T Exam Practise Chapter 5
29
–6
Exam Practice 5
1. 9x2 + y2 = 36
x 2
—–4
+ y2
—–36
= 1
Coordinates are (0, ± 6) and (±2, 0).
2. (x – 5)2 + (y – 7)2 = 25 Coordinates of C = (5, 7) Gradient of PC = 7 – 3——–
5 – 2 = 4—
3
Hence, gradient of tangent = – 3—4
Equation of tangent at P:
y – 3 = – 3—4
(x – 2)
4y – 12 = –3x + 6 3x + 4y = 18
3. Equation of line with gradient m and passing through P(0, 18):
y – 18 = mx mx – y + 18 = 0 Perpendicular distance from C(4, 6) to this
line: 4m – 6 + 18—————–m2 + 1
= 4m + 12————m2 + 1
If the line is a tangent to the circle centre C, then,
4m + 12————
m2 + 1 = 10
4(m + 3) = 10 m2 + 1 2(m + 3) = 5 m2 + 1
Squaring, 4(m + 3)2 = 25(m2 + 1) 4m2 + 24m + 36 = 25m2 + 25 21m2 – 24m – 11 = 0 [shown]
4. Gradient of the line
3y – x + 2 = 0 ......(1) is 1—3
.
Gradient of PQ = –3. Equation of PQ: y + 6 = –3 (x – 2) y = –3x Substitute y = –3x into (1), –9x – x + 2 = 0 10x = 2
x = 1—5
When x = 1—5
, y = – 3—5
.
Coordinates of the foot of the perpendicular is
� 1—5
, – 3—5 �.
5. Let Q be the point (α, β). Q lies on the curve xy = 12.
Hence, αβ = 12
Coordinates of mid-point of OQ = � α—2
, β—2 �
Let x = α—2
......(1)
y = β—2
......(2)
Eliminating α and β from these equations, α = 2x, β = 2y αβ = 4xy 4xy = 12 xy = 3 xy = 3 is the locus of the midpoint of OQ.
6. Let P be the point (α, β). P lies on the curve 4x2 + y2 = 36. Hence, 4α2 + β 2 = 36 ......(1)
Coordinates of midpoint of AP = �α + 1——–2
, β—2 �
Let x = α + 1——–2
⇒ α = 2x – 1
y = β—2
⇒ β = 2y
Substitute α = 2x – 1 and β = 2y into (1), 4(2x – 1)2 + 4y2 = 36 (2x – 1)2 + y2 = 9 y2 = 9 – (2x – 1)2
= 9 – 4x2 + 4x – 1 = 4(2 + x – x2) = 4(2 – x)(1 + x) This equation is the locus of the midpoint of AP.
7.
Coordinates of midpoint of PQ are � at2
—–2
, 2at� Let x = at2
—–2
......(1)
y = 2at ......(2)
From (2), t = y—–
2a
Substitute t = y—–
2a into (1),
x = a—
2 � y2
—–4a2 �
y2 = 8ax which is the locus of the midpoint of PQ.
8. Let P be the point (α, β). P lies on the hyperbola xy = 4. Hence, αβ = 4 ......(1) Let the coordinates of Q be (x, y). x = 3α + 2——–—
5 ⇒ α = 5x – 2——––
3
x
y
–2 2
P
Q
(2, –6)
3y – x + 2 = 0
6
x
y
Q P(at2, 2at)
O
y2 = 4ax
Ace Ahead Mathematics S & T Volume 1
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y = 3β + 0——–—5
⇒ β = 5y—–3
Substitute α and β into (1), (5x – 2)——–—
3 (5y)—––
3 = 4
5y(5x – 2) = 36 The locus of the point Q is 5y (5x – 2) = 36.
9.
Gradient of PQ =
c—p
– c—q
————cp – cq
= c(q – p)
————–cpq(p – q)
= – 1—–pq
Equation of PQ is
y – c—p
= – 1—–pq
(x – cp)
pqy – cq = –x + cp x + pqy = c(p + q) At A, y = 0, x = c(p + q) Hence, A(c(p + q), 0)
At B, x = 0, y = c(p + q)———–
pq
Hence, B�0, c(p + q)———–
pq � Coordinates of midpoint of PQ
= � cp + cq———–2
,
c—p
+ c—q
————2 �
= � c(p + q)———–2
, c(p + q)———–2pq �
Coordinates of midpoint of AB
= � c(p + q) + 0———–——
2,
0 + c(p + q)———–——
2pq � = � c(p + q)
———–2,
c(p + q)———–2pq �
Therefore, the midpoint of PQ = the midpoint of AB
10.
Let P be the point (x, y). Equation of the line is y – 1 = m(x – 1).
At A, y = 0, x = 1 – 1—m
Hence, A�1 – 1—m
, 0� At B, x = 0, y – 1 = –m y = 1 – m
Hence, B(0, 1 – m) Coordinates of P are
x = 2�1 – 1—
m �————
3 ......(1)
y = (1 – m)———–3
......(2)
From (2), m = 1 – 3y Substitute m = 1 – 3y into (1),
3x = 2�1 – 1
———1 – 3y �
3x(1 – 3y) = 2(1 – 3y – 1) 3x – 9xy = –6y x – 3xy = –2y 3xy – x – 2y = 0 [shown]
11.
2x – 3y = 15 ......(1) 3x + 2y = 3 ......(2) Solving (1) and (2) for the point D, (1) × 2, 4x – 6y = 30 ......(3) (2) × 3, 9x + 6y = 9 ......(4) (3) + (4), 13x = 39 x = 3 From (2), 9 + 2y = 3 y = –3 Hence, D(3, –3) Let the centre of the circle C be (x, y). ABCD is a rectangle, ⇒ midpoint of CD = midpoint of AB
� x + 3——–2
, y – 3——–2 � = � 6 + 1——–
2, –1 + 0———
2 � Hence, x = 4 and y = 2 Therefore, C(4, 2) Radius of circle = BC = 32 + 22 = 13 Equation of circle is (x – 4)2 + (y – 2)2 = 13 x2 + y2 – 8x – 4y + 7 = 0
12.
The diagonals of a rhombus bisect at right angles.
Gradient of BD = – 4—3
Gradient of AC = 3—4
A�1 – 1—m, 0�
B(0, 1 – m)
P(x, y)
1
2
A(6, –1) 2x – 3y = 15 D
B(1, 0)
3x + 2y = 3
C
P�cp, c—p �x
y
A B
Q�cq, c—q �O
x
y
B (1, 1)
P(x, y)
AO
D C 4x + 3y – 48 = 0
M
A(–7, –8) B(18, –8)
P(α, β)
2
Q(x, y) 3
A(1, 0)
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Equation of AC is
y + 8 = 3—4
(x + 7)
4y + 32 = 3x + 21 3x – 4y – 11 = 0 ......(1) BD: 4x + 3y – 48 = 0 ......(2)
Solving (1) and (2) for the point M, (1) × 3, 9x – 12y – 33 = 0 ......(3) (2) × 4, 16x + 12y – 192 = 0 ......(4) (3) + (4), 25x – 225 = 0 x = 9
From (1), 27 – 4y – 11 = 0 y = 4 Hence, M(9, 4)
Let the coordinates of C be (x, y).
x – 7——–2
= 9 and y – 8——–2
= 4
x = 25, y = 16
Therefore, C(25, 16) [shown]
Let the coordinates of D be (x, y).
x + 18———2
= 9 and y – 8——–2
= 4
x = 0, y = 16 Therefore, D(0, 16)
13.
Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 Since (0, 8) lies on the circle, 64 + 16f + c = 0 ......(1) Gradient of CP =
8 + f——–g
Gradient of tangent = 4—3
Hence, 8 + f——–
g = – 3—4
32 + 4f = –3g 3g + 4f + 32 = 0 ......(2) C(–g, –f) lies on the line 3x – y = 7. –3g + f – 7 = 0 ......(3)
(2) + (3), 5f + 25 = 0 f = –5 From (2), 3g – 20 + 32 = 0 g = –4 From (1), 64 – 80 + c = 0 c = 16 Equation of the circle is x2 + y2 – 8x – 10y + 16 = 0
14.
Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 Since (2, 5) lies on the circle, 4 + 25 + 4g + 10f + c = 0 29 + 4g + 10f + c = 0 ......(1) Since (4, 3) lies on the circle, 16 + 9 + 8g + 6f + c = 0 25 + 8g + 6f + c = 0 ......(2) The centre (–g, –f) lies on the line x + y = 3. –g – f – 3 = 0 ......(3) (2) – (1), –4 + 4g – 4f = 0 –1 + g – f = 0 ......(4) (3) + (4), –4 – 2f = 0 f = –2 From (3), g = –1 From (2), 25 – 8 – 12 + c = 0 c = –5
Equation of the circle is x2 + y2 – 2x – 4y – 5 = 0
Centre is (1, 2).
Radius is (4 – 1)2 + (3 – 2)2 = 10
15.
(a) If m = 0, the line is parallel to the x-axis, and cuts the curve at only one point. Hence, m ≠ 0. [shown]
y2 = 8x ......(1) y = mx – 4 ......(2) Solve (1) and (2) for points of
intersection. Eliminating y, (mx – 4)2 = 8x m2x2 – 8mx + 16 = 8x m2x2 – 8(m + 1)x + 16 = 0 ......(3)
For real and distinct roots, b2 – 4ac � 0 64(m + 1)2 – 64m2 � 0 m2 + 2m + 1 – m2 � 0 2m � –1
m � – 1—2
[shown]
P(0, 8) 3y – 4x – 24 = 0
C(–g, –f)
3x – y = 7
(2, 5)
(–g, –f)
(4, 3)
x + y = 3C
x
y
y2 = 8x A(x1, y1)
C
O
B(x2, y2)
y = mx – 4
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(b) Roots of equation (3) are x1 and x2.
Sum of roots, x1 + x2 = 8(m + 1)——–—–m2
[shown]
(c) Eliminating x from (1) and (2),
y2 = 8(y + 4)——–—–m
my2 – 8y – 32 = 0 ......(4)
Roots of equation (4) are y1 and y2.
Sum of roots, y1 + y2 = 8—m
[shown]
Let the coordinates of C be (x, y).
Midpoint of OC = midpoint of AB
� x—2
, y—2 � = � 8(m + 1)——–—–
2m2, 8—–
2m � x = 8(m + 1)——–—–
m2 ......(5)
y = 8—m ⇒ m = 8—
y
Substitute m = 8—y into (5),
x =
8� 8—y + 1�
————64—–y2
64x—–y2
= 64—–y
+ 8
64x = 64y + 8y2
y2 + 8y = 8x is the locus of C. [shown]
16.
Coordinates of M are
� 5 – 3——–2
, 4 – 4——–2 � = (1, 0)
Gradient of AC = 4 + 4——–5 + 3
= 1
Gradient of BD = –1 Equation of BD is y – 0 = –1(x – 1) x + y = 1 ......(1) Equation of BC is y – 4 = 2(x – 5) 2x – y = 6 ......(2) Solving (1) and (2) for the point B,
(2) + (1), 3x = 7 ⇒ x = 7—3
From (1), y = – 4—3
Hence, B� 7—3
, – 4—3 �
Let the coordinates of D be (x, y).
Midpoint of BD = midpoint of AC
� x + 7—3———
2, y – 4—
3———2 � = (1, 0)
x = – 1—3
, y = 4—3
Hence, D�– 1—3
, 4—3 �
17.
Gradient of DC = 4—3
Gradient of AB = 26 – 2———3 + 15
= 24—–
18 = 4—
3 Therefore, DC is parallel to AB [proven] Equation of AP is
y – 26 = – 24—–7
(x – 3)
7y – 182 = –24x + 72 24x + 7y = 254 ......(1) Equation of BC is y = 2 ......(2)
Solving (1) and (2) for the point P, 24x + 14 = 254 24x = 240 x = 10 Therefore, P(10, 2) Length of CP = 10 Length of BP = 25 �PDC is similar to �PAB.
Area �PDC——————Area �PAB
= 102
—–252
= 100—––625
= 4—–25
[proven]
18.
x2
—–a2 +
y2
—–b2 = 1 ......(1)
y = mx + c ......(2) Solving (1) and (2) for point of intersection,
x2
—–a2 +
(mx + c)2
————b2
= 1
b2x2 + a2 (m2x2 + 2mcx + c2) = a2b2
(b2 + a2m2)x2 + 2mca2x + a2(c2 – b2) = 0
A(–3, –4) B
C(5, 4)D
M m = 2
A(3, 26)
m = – 24—–7
x
y
B(–15, 2)C(0, 2)
D
P 4x –
3y + 6 =
0
x
y
O
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If the line is a tangent, roots are equal. b2 – 4ac = 0 4m2c2a4 – 4(b2 + a2m2)a2(c2 – b2) = 0 m2c2a2 – b2c2 + b4 – a2m2c2 + a2b2m2 = 0 b2c2 = b4 + a2b2m2
c2 = a2m2 + b2 [shown]
Substitute x = a2 + b2 , y = 0 into y = mx + c,
0 = m a2 + b2 + c c2 = m2(a2 + b2) a2m2 + b2 = m2a2 + m2b2
b2 = m2b2
m2 = 1 ⇒ m = ±1 Gradient of tangents = ±1 Since the x-intercepts and y-intercepts are
equal, and by symmetry, the coordinates of the vertices of the square whose sides touch the
curve are (± a2 + b2 , 0) and (0, ± a2 + b2 ).
19.
(a) OP2 = a2 cos2 θ + b2 sin2 θ OQ2 = a2 sin2 θ + b2 cos2 θ OP2 + OQ2 = a2(cos2 θ + sin2 θ) + b2(sin2 θ + cos2 θ) = a2 + b2 [shown]
0 a cos θ –a sin θ 0(b) � � 0 b sin θ b cos θ 0
Area of �OPQ = 1—2
|ab cos2 θ + ab sin2 θ|
= 1—2
ab (cos2 θ + sin2 θ)
= 1—2
ab [shown]
(c) Coordinates of mid-point of PQ
= � a cos θ – a sin θ———————–2
, b sin θ + b cos θ———————–2 �
x = a—2
(cos θ – sin θ) ......(1)
y = b—2
(sin θ + cos θ) ......(2)
(1) + (2), 2x—–a +
2y—–b
= 2 cos θ
x—a
+ y—b
= cos θ ......(3)
(2) – (1), 2y—–b
– 2x—–a = 2 sin θ
y—b
– x—a = sin θ ......(4)
(3)2 + (4)2, � x—a + y—b �2
+ � y—b
– x—a �2
= sin2 θ + cos2 θ
x2
—–a2
+ 2xy——ab
+ y2
—–b2
+ y2
—–b2
– 2xy——ab
+ x2
—–a2
= 1
2x2
——a2
+ 2y2
——b2
= 1
which is the locus of the midpoint of PQ. [shown]
20.
Let the gradient of l be m. Equation of l is y – 2 = m(x – 3) At P, x = 0, y = 2 – 3m ⇒ P(0, 2 – 3m)
At Q, y = 0, x = 3 – 2—m ⇒ Q(3 – 2—
m , 0)
Coordinates of M are
x = 1—2 �3 – 2—
m � ......(1) y = 1—
2(2 – 3m) ......(2)
From (2), m = 2 – 2y———
3
Substitute m = 2(1 – y)————
3 into (1),
2x = 3 – 3
———1 – y
2x(1 – y) = 3(1 – y) – 3 2xy = 2x + 3y [shown]
21. (a)
Let the coordinates of M be (x, y).
x = 2(0) + 1(3)—————
3 = 1
y =
2(2) + 1(5)—————3
= 3
Hence, M(1, 3)
(b) Gradient of AC = –1 – 2———5 – 0
= – 3—
5
Equation of AC is y – 2 = – 3—
5 (x – 0)
3x + 5y – 10 = 0
Perpendicular distance from B to AC
= � 3(3) + 5(5) – 10———————32 + 52 � = 24——
34 units
x
y
O
P(a cos θ, b sin θ) Q(–a sin θ, b cos θ)
x
y
A(3, 2)
M P
Q O
A(0, 2) M(x, y) B(3, 5)
1 2
A(0, 2)
B(3, 5)
C(5, –1)
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22. Let the coordinates of the point P be (x, y).
PA = 2 13 (x – 1)2 + (y – 1)2 = 52 ......(1)
PA is perpendicular to AB,
� y – 1———x – 1 ��– 3—
2 � = –1
y – 1———x – 1
= 2—3
⇒ x = 3y – 1———2
Substitute x = 3y – 1———2
into (1),
� 3y – 1———2
– 1�2 + (y – 1)2 = 52
9—4
(y – 1)2 + (y – 1)2 = 52
13—–4
(y – 1)2 = 52
(y – 1)2 = 16 y – 1 = ±4 ⇒ y = 5 or –3
When y = 5, x = 7 and when y = –3, x = –5 Coordinates of the two points are (7, 5) and
(–5, –3).
23.
Coordinates of M, the mid-point of BC are (1, 2).
Gradient of BC = 4 – 0———3 + 1
= 1
Equation of the perpendicular bisector of BC is y – 2 = –1(x – 1) x + y = 3 ......(1) 2y – 3x = 16 ......(2) Solving (1) and (2) for the point A, (1) × 3, 3x + 3y = 9 ......(3) (2) + (3), 5y = 25 y = 5 From (1), x = –2 Therefore, A(–2, 5)
24. 4x + 3y – 6 + p(2x – 5y – 16) = 0(a) If the line passes through the origin, x = 0, y = 0 ⇒ –6 – 16p = 0
p = – 3—8
(b) If the line is parallel to the line 5x – 6y – 11 = 0, the gradients of the two lines are equal.
–(4 + 2p)————
3 – 5p = 5—
6 –24 – 12p = 15 – 25p 13p = 39 p = 3
(c) If the line is perpendicular to the line x + 4y = 0, the product of the gradients of the two lines is equal to –1,
–(4 + 2p)————
3 – 5p × �– 1—
4 � = –1
4 + 2p = –12 + 20p 18p = 16
p = 8—9
25.
Equation of AB is
y – 4 = 1—8
(x – 1)
8y – x = 31 ......(1)
Equation of BC is
y + 2 = 7—4
(x – 5)
4y – 7x = –43 ......(2)
Solving (1) and (2) for the point B, (2) × 2, 8y – 14x = –86 ......(3) (1) – (3), 13x = 117 x = 9 From (1), y = 5 Hence, B(9, 5) Let the point D be (x, y). Midpoint of BD = midpoint of AC
x + 9——–2
= 5 + 1——–2
⇒ x = –3
y + 5——–
2 = 4 – 2——–
2 ⇒ y = –3
Hence, D(–3, –3) Gradient of AC = – 6—
4 = – 3—
2
Gradient of BD = 8—12
= 2—3
Product of gradient = �– 3—2 �� 2—
3 � = –1
Hence, the diagonals meet at right angles. Therefore, ABCD is a rhombus. [proven]
26.
Gradient of AB = 12 + 2———20 – 4
= 7—8
Gradient of CD = 7—
8
B(3, –2)
2 13 2 13
P(x, y) A(1, 1) P(x, y)
B(–1, 0) M C(3, 4)
A2y – 3x = 16
D C(5, –2)
4y – 7x = 3
B A(1, 4) 8y – x = 0
D(–3, 6) C(x, y)
x + 15y = 200
A(4, –2) B(20, 12)
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Equation of CD is
y – 6 = 7—8
(x + 3)
8y – 7x = 69 ......(1) x + 15y = 200 ......(2)
Solving (1) and (2) for the point C, (2) × 7, 7x + 105y = 1400 ......(3) (1) + (3), 113y = 1469 y = 13 From (2), x + 195 = 200 x = 5 Hence, C(5, 13)
Gradient of AD = 8—––7
Product of gradients of AD and AB
= �– 8—7 �� 7—
8 � = –1
Therefore, A is a right angle [proven]
AB = 162 + 142 = 452 = 2 113
CD = 82 + 72 = 113
Therefore, AB = 2DC [shown]
27.
AB: 3x – y + 3 = 0 ......(1) AC: 2x + y – 8 = 0 ......(2)
Solving (1) and (2) for the point A, (1) + (2), 5x – 5 = 0 x = 1 From (2), y = 6 Hence, A(1, 6)
BC: x = 0 ......(3) AB: 3x – y + 3 = 0 ......(4) Solving (3) and (4) for the point B, x = 0 and y = 3 Hence, B(0, 3) BC: x = 0 AC: 2x + y = 8 Solving the equations for the point C, x = 0, y = 8 Hence, C(0, 8) Equation of the altitude AD is y = 6 ......(5) Gradient of AC = –2
Gradient of BE = 1—2
Equation of altitude BE is
y – 3 = 1—2
(x – 0)
x – 2y + 6 = 0 ......(6) Gradient of AB = 3
Gradient of CF = – 1—3
Equation of altitude CF is
y – 8 = – 1—3
(x – 0)
x + 3y – 24 = 0 ......(7)
Solving (5) and (6) for point of intersection, y = 6 and x = 6 The point (6, 6) satisfi es equation (7). Hence, the altitude meet at the point (6, 6).
28.
Gradient of line = –2
Gradient of PQ = 1—2
Equation of PQ is
y – 1 = 1—2
(x + 4)
x – 2y + 6 = 0 ......(1) 2x + y – 8 = 0 ......(2)
Solving (1) and (2) for the point M, 5x – 10 = 0 x = 2 and y = 4 Hence, M(2, 4) Let the coordinates of Q be (x, y). M is the midpoint of PQ.
x – 4——–
2 = 2 ⇒ x = 8
y + 1——–
2 = 4 ⇒ y = 7
Therefore, Q(8, 7)
29. A(0, 4), B(0, –4), C(6, 3) Let the coordinates of P be (x, y). PA2 = (x – 0)2 + (y – 4)2 = x2 + y2 – 8y + 16 PB2 = (x – 0)2 + (y + 4)2 = x2 + y2 + 8y + 16 PC2 = (x – 6)2 + (y – 3)2 = x2 + y2 – 12x – 6y + 45 The condition is PA2 + PB2 + PC 2 = 362 3x2 + 3y2 – 12x – 6y + 77 = 362 x2 + y2 – 4x – 2y – 95 = 0 which is the equation of the locus of P. This equation represents a circle,
centre = (2, 1), radius = 22 + 12 + 95 = 10 Substitute x = 8, y = 9 into the equation of the
circle. LHS = 64 + 81 – 32 – 18 – 95 = 0 (8, 9) satisfi es the equation of the circle and
hence, it lies on the circle [shown]
A
B
C
D
E
F
2x + y = 8
x =
0
3x –
y + 3
= 0
P(–4, 1)
M 2x + y – 8 = 0
Q(x, y)
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30. Let P(x, y), A(–3, 0) and B(3, 0) Condition is PA = kPB PA2 = k2PB2
(x + 3)2 + y2 = k2 [(x – 3)2 + y2] x2 + 6x + 9 + y2 = k2 (x2 – 6x + 9 + y2) (k2 – 1)x2 + (k2 – 1)y2 – 6(k2 + 1)x + 9(k2 – 1)
= 0 x2 + y2 –
6(k2 + 1)————k2 – 1
x + 9 = 0
This equation is the locus of P which represents a circle [shown]
Radius = 9(k2 + 1)2
————(k2 – 1)2 – 9
= 3———
(k2 – 1)(k2 + 1)2 – (k2 – 1)2
= 6k———
k2 – 1 [shown]
31. Let A(a, 0), B(0, b) and the midpoint of AB be
M. Hence M� a—2
, b—2 �
(a) Condition is AB = k a2 + b2 = k2 ......(1) For M, x = a—
2 ⇒ a = 2x
y = b—2
⇒ b = 2y
Substitute a and b into (1), Locus of M is 4x2 + 4y2 = k2
(b) Condition is 1—
2ab = k ⇒ ab = 2k ......(2)
Substitute a and b into (2), (2x)(2y) = 2k Locus of M is 2xy = k
(c) Condition is P(α, β) lies on AB. Gradient of PA = gradient of PB
β – 0——–α – a
= β – b——–α – 0
αβ = (α – a)(β – b) bα + aβ = ab ......(3) Substitute a = 2x, b = 2y into (3),
2αy + 2βx = 4xy Locus of M is αy + βx = 2xy
32. (a) (b)
(c) (d)
(e) (f)
33. Length of perpendicular from P(x1, y1) to ax1 + by1 + c ax + by + c = 0 is �—————–�. a2 + b2
Let P be the point (x, y). Equation of locus of P is
� y – 2x + 1————–
(12 + 22) � = x2 + y2
(y – 2x + 1)2 = 5(x2 + y2) y2 + 4x2 + 1 – 4xy + 2y – 4x = 5x2 + 5y2
x2 + 4xy + 4y2 + 4x – 2y – 1 = 0 ......(1) y = 2x ......(2)
Solving (1) and (2) for points of intersection, x2 + 8x2 + 16x2 + 4x – 4x – 1 = 0 25x2 – 1 = 0
x = ± 1—5
, y = ± 2—5
Hence A� 1—5
, 2—5 � and B�– 1—
5, – 2—
5 � Midpoint of AB = (0, 0) which is the origin. [shown]
34.
Gradient of BC = 1—4
Equation of BC is
y – 2 = 1—4
(x – 2)
4y – x = 6 ......(1)
Gradient of OA = 3
Equation of OA is y = 3x ......(2)
Solving (1) and (2) for the point D, 12x – x = 6 11x = 6
x = 6—–11
y = 18—–11
B(0, b)
M(x, y)
A(a, 0) x
y
O
P(α, β)
x
y
–2
–4x
y
1O
x
y
O (1, 0)
x
y
O
(0, –1)
–4
y
x –4 O
x
y
2O
x
y A(1, 3)
B(–2, 1)
C(2, 2) m D
O
n
O
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Hence, D� 6—–11
, 18—–11 �
Let the ratio of BD : DC = m : n Using the ratio formula,
2m – 2n————m + n
= 6—–11
22m – 22n = 6m + 6n 16m = 28n
m—n
= 7—4
Ratio of BD : DC = 7 : 4
35. Coordinates of midpoint of AB
= � x1 + x2———2
, y1 + y2———
2 � Gradient of AB =
y1 – y2———x1 – x2
Gradient of perpendicular bisector of AB
= –x1 – x2———y1 – y2
Equation of perpendicular bisector of AB is
y – y1 + y2———
2 = –
x1 – x2———y1 – y2
�x – x1 + x2———
2 � (y1 – y2)[2y – (y1 + y2)]
+ (x1 – x2)[2x – (x1 + x2)] = 0 [shown]
Let A(2, 1), B(1, 5). Take x1 = 2, y1 = 1, x2 = 1, y2 = 5
Equation of perpendicular bisector of AB is (2 – 1)[2x – (2 + 1)] + (1 – 5)[2y – (1 + 5)] = 0 2x – 3 – 8y + 24 = 0 2x – 8y + 21 = 0 The point on the line 3x – 4y + 3 = 0 which is equidistant from the points (2, 1) and
(1, 5) is given by the point of intersection of the lines
2x – 8y + 21 = 0 and 3x – 4y + 3 = 0 Solving the equations simultaneously, 4x – 15 = 0
x = 15—–4
y = 57—–16
Coordinates of the point are �3 3—4
, 3 9—–16 �.
36. Equation of the line is x—a
+ y—b
= 1 ......(1)
Let P(x, y) be the foot of the perpendicular
from O to the line x—a
+ y—b
= 1.
Gradient of given line = – b—a
Gradient of OP = a—b
Equation of OP is y = a—
b x ......(2)
Solving (1) and (2) for the point P.
x—a
+ ax—–b2 = 1
(b2 + a2)x = ab2
x = ab2
—–——a2 + b2 ......(3)
y = a2b—–——a2 + b2 ......(4)
1—–a2 + 1—–
b2 = 1—–c2
c2 = a2b2
—–——a2 + b2
From (3), x2 = a2b4
—–——–(a2 + b2)2
From (4), y2 = a4b2
—–——–(a2 + b2)2
x2 + y2 = a2b4
—–——–(a2 + b2)2 + a4b2
—–——–(a2 + b2)2
= a2b2(b2 + a2)—–——–—–(a2 + b2)2
= a2b2
—–——a2 + b2
= c2 [shown]
37. Let the coordinates of B be (x, y). AB = 10 (x + 1)2 + (y – 2)2 = 100 x2 + y2 + 2x – 4y = 95 ......(1)
AB is perpendicular to BC,
(y – 2)———(x + 1)
× (y – 7)———(x – 9)
= –1
(x + 1)(x – 9) + (y – 2)(y – 7) = 0 x2 + y2 – 8x – 9y = –5 ......(2)
Solving (1) and (2) simultaneously, (1) – (2), 10x + 5y = 100 y = 20 – 2x
Substitute y = 20 – 2x into (1), x2 + (20 – 2x)2 + 2x – 4(20 – 2x) = 95 5x2 – 70x + 225 = 0 x2 – 14x + 45 = 0 (x – 5)(x – 9) = 0 x = 5 or 9 When x = 5, y = 10 When x = 9, y = 2
x
y
O a
b P(x, y)
A(–1, 2) B(x, y)
C(9, 7) D
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Hence, B(5, 10) or B(9, 2) Let the coordinates of D be (x, y). Taking B(5, 10),
x + 5———2
= 9 – 1———2
⇒ x = 3
y + 10———
2 = 7 + 2———
2 ⇒ y = –1
Hence, D(3, –1) Take B(9, 2),
x + 9———2
= 9 – 1———2
⇒ x = –1
y + 2
———2
= 7 + 2———2
⇒ y = 7
Hence, D(–1, 7)
38. x = t(t – 2) ......(1) y = 2(t – 1) ......(2)
(a) Eliminating t from (1) and (2),
From (2), t = y—2
+ 1
Substitute t = y—2
+ 1 into (1),
x = � y—2
+ 1�� y—2
– 1� x = y2
—4
– 1
4x = y2 – 4 y2 = 4(x + 1)(b)
39.
Let A(5, –2) and the equation of BD be 3x – 2y – 6 = 0 ......(1)
Gradient of BD = 3—2
Gradient of AC = – 2—3
Equation of AC is
y + 2 = – 2—3
(x – 5)
3y + 2x = 4 ......(2)
Solving (1) and (2) for the point M, 13y = 0 y = 0 x = 2 Hence, M(2, 0) Let the coordinates of C be (x, y). Since M is the midpoint of AC,
x + 5———
2 = 2 ⇒ x = –1
y – 2———
2 = 0 ⇒ y = 2
Hence, C(–1, 2) Let B be the point (x, y) AB = BC (x – 5)2 + (y + 2)2 = (x + 1)2 + (y – 2)2
x2 – 10x + 25 + y2 + 4y + 4 = x2 + 2x + 1 + y2 – 4y + 4 12x – 8y – 24 = 0
y = 3x – 6———2
......(3)
AB is perpendicular to BC,
y + 2
———x – 5
× y – 2
———x + 1
= –1
(x – 5)(x + 1) + (y + 2)(y – 2) = 0 x2 + y2 – 4x – 9 = 0 ......(4)
Solving (3) and (4) for the point B, x2 + (3x – 6)2
———–4
– 4x – 9 = 0
4x2 + 9x2 – 36x + 36 – 16x – 36 = 0 13x2 – 52x = 0 13x (x – 4) = 0 x = 0 or 4
When x = 0, y = –3 B(0, –3)
When x = 4, y = 3 D(4, 3)
The coordinates of the other vertices are (4, 3), (–1, 2) and (0, –3)
40.
Let the coordinates of P be (t2, t3) and the midpoint of AP be M(x, y).
Since M is the midpoint of AP,
x = t2 + 2——— 2
......(1)
y = t3
—2
......(2)
From (1), t2 = 2x – 2 ⇒ t = 2x – 2
Substitute the value of t into (2), 2y = (2x – 2) 2x – 2 4y2 = (2x – 2)2(2x – 2) = 8(x – 1)2(x – 1) y2 = 2(x – 1)3
This equation represents the locus of the midpoint of AP. [proven]
x
y
2
–1–2
A(5, –2) B
C D
M
O
3x – 2y – 6 = 0
x
y
O A(2, 0)
M(x, y)
P(t2, t3)
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