Ace Ahead Mathematic T Exam Practise Chapter 5

10
29 –6 Exam Practice 5 1. 9x 2 + y 2 = 36 x 2 —– 4 + y 2 —– 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 —————– m 2 + 1 = 4m + 12 ———— m 2 + 1 If the line is a tangent to the circle centre C, then, 4m + 12 ———— m 2 + 1 = 10 4(m + 3) = 10 m 2 + 1 2(m + 3) = 5 m 2 + 1 Squaring, 4(m + 3) 2 = 25(m 2 + 1) 4m 2 + 24m + 36 = 25m 2 + 25 21m 2 – 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 4x 2 + y 2 = 36. Hence, 4α 2 + 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 + 4y 2 = 36 (2x – 1) 2 + y 2 = 9 y 2 = 9 – (2x – 1) 2 = 9 – 4x 2 + 4x – 1 = 4(2 + x x 2 ) = 4(2 – x)(1 + x) This equation is the locus of the midpoint of AP. 7. Coordinates of midpoint of PQ are at 2 —– 2 , 2at Let x = at 2 —– 2 ......(1) y = 2at ......(2) From (2), t = y —– 2a Substitute t = y —– 2a into (1), x = a 2 y 2 —– 4a 2 y 2 = 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(at 2 , 2at) O y 2 = 4ax Ace Ahead Mathematics S & T Volume 1 ACE STPM Math (Text Ans) 3rd.indd 29 ACE STPM Math (Text Ans) 3rd.indd 29 3/27/2008 4:13:54 PM 3/27/2008 4:13:54 PM

Transcript of Ace Ahead Mathematic T Exam Practise Chapter 5

Page 1: 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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32

(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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