Math1.1

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Topics :Topics :Chapter 1 : Conic sectionChapter 1 : Conic sectionChapter 2 : Differential Chapter 2 : Differential EquationEquationChapter 3 : Numerical MethodChapter 3 : Numerical MethodChapter 4 : Data descriptionsChapter 4 : Data descriptionsChapter 5 : Probability and Chapter 5 : Probability and Random Random

variablesvariablesChapter 6 : Special Distribution Chapter 6 : Special Distribution

FunctionFunction

Mathematics 102/3

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Components Paper Topic Format Time Marks Percentage

Test 1 - SubjectiveQuestion

2 Hour 100 20%

ContinuousAssessment

- - Assessment/Quiz/

Tutorial

ThroughoutThe

semester

- 20%

Final examination

Paper 1 All topic

SubjectiveQuestion

2 hours 100 60%

Paper 2 All topic

SubjectiveQuestion

2 hours 100

Total 100%

ASSESSMENT

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MAT 102/3MAT 102/3 MAT 102/3MAT 102/3

CHAPTER 1:CHAPTER 1:

CONIC SECTIONSCONIC SECTIONS

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1.1 : Intoduction to conic sections

o Circles, parabola, ellipses and hyperbolas are called conic sections because they are curves obtained by the intersection of a right circular cone and a plane.o The curves formed depends on the angel at which the plane intersects the cone.

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1.2 : Circles

Definition:

A circle is defined as the set of all points P in a plane that are at a constant distance from a fixed point.

This fixed point is called the centre and the fixed distance is called the radius

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),( yxP

y

x

P( x, y )

C ( h, k )

r

Figure shows a circle with center (h,k) and radius r

r X - h

Y - k

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Equations of a cirles From the definition of the circle, a point

P(x,y) lies on the circle if and only if PC = r, that is

rkyhx 22 )()(Squaring both sides, we have

222 )()( rkyhx This is the equation of the circle with center (a,b) and radius r unitsIf the origin is the center of the circle, the equation becomes

222 ryx

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Example 1

Find the equation of the circle with :-

(i) Center at origin and radius 3 units

(ii) Center (2,-3) and radius 5 units

Solutions :

9

322

222

222

yx

yx

ryx(i)

01264

5)3()2(

)()(

22

222

222

yxyx

yx

rkyhx(ii)

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General Equation of a Circle

The equation of a circle with center (a,b) and radius r units is

022

)()(22222

222

rbabyaxyx

rbyax

Now substituting g = a, f = k and c=a2+b2-r2

Conversely, the equation

02222 cfygxyxWhere g,f and c are constant, represent a circle

This equation is called the general equation of a circle

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Center and radius of a circle x2+y2+2gx+2fy+c=0Completing the squares for x2+2gx and y2

+2fy,x2+y2+2gx+2fy+c=0

X2+2gx+g2+y2+2fy+f2=g2+f2-c(x+g)2+(y+f)2=g2+f2-c

Hence, the center of a circle is (-g,-f) and the radius is

cfgr 22

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Dertermine the equation with center (h,k)Example 2

Find the center and the radius of the circle x2+y2+5x-6y-5=0

Comparing with the general equation, x2+y2+2gx+2fy+c=0

g=5/2 f=-3 c=-5

Hence,the center is (-5/2,3) and the radius is

2

9)5()3()

2

5( 22

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Determine the centre and radius of a circle.• Example 3

Graph

• Solution We can change the given equation into the standard

form of the circle by completing the square on x and y as follow

094622 yxyx

094622 yxyx

9)4()6( 22 yyxx 99)44()96( 22 yyxx

222 2)2()3( yx222 2))2(()3( yx

4

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The center is at ( 3 , -2), and the length of a radius is 2 units

(3,-2) r =2

y

x

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Point of IntersectionExample 4

Find the coordinates of the point of intersection of the circles x2+y2-4=0 and x2+y2-2x+4y+4=0

x2+y2-4=0 ……….(1)

x2+y2-2x+4y+4=0…..(2)

Solving the equation simultaneously for the point of intersection,

(1)– (2) 2x-4y-8=0

x=2y+4

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Substituting x=2y+4 in (1) gives

(2y+4)2+y2-4=0

4y2+16y+16+y2-4=0

5y2+16y+12=0

(5y+6)(y+2)=0

Y=-6/5 or 2

When y=-6/5, x=12/5+4=8/5

When y=-2 x=-4+4=0

Therefore, the points of intersection

are (8/5,-6/5) and (0,-2)

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Point of a circle and a straight lineExample 5

Find the coordinates of the points of intersection between the circle X2+y2-6x+9=0 and the line y=7-x

Solution:-

Given X2+y2-6x+9=0….(1) y=7-x….(2)

By substituting (2) into (1) gives, on simplication

X2-8x+15=0

(x-5)(x-3)=0

x=5, y=2

x=3, y=4

So, intersection point are (5,2) and (3,4)

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Circle passing through three given

points

If we are given the coordinates of three

points on the circumference of a circle,

we can substitute these values of x and

y into the equation of the circle and

obtain three equations which can be

solved simultaneously to find the

constants g, f and c.

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• Find the equation of the circle passing through the points (0,1). (4,3), and (1,-1).

• Solution:Suppose the equation of the circle is points into this equation 02222 cfygxyx

Substituting the coordinates of each of the three equation gives:

021 cf06825 cfg

---------(1)

--------(2)

Circle passing through three given points

2+2g-2f+c+0 --------(3)

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• Multiplying equation [3] by 4 and then subtracting from equation [2], gives

Multiplying equation [1] by 3 and adding to equation [4] gives

031417 cff

02020 f 1for

Then from equation [1] 1cAnd from equation [3]

212g

The equation of the circle which passes through (0,1), (4,3) and

)1,1( is 012522 yxyx

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Find the equation of a circle passing through two points with the equation of the diameter

given • Example 6

Find the equation of the circle passing through the points (1,1) and (3,2) and with diameter 073 xy

SolutionThe standard form of the circle is 02222 cfygxyxSince the circle passes through 1,1

0222 cfg -------[1] 2,3

01346 cfg -------[ 2]

Since the circle passes through

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• The center of the circle must passes through the diameter

fg ,

073 xy

Therefore, 073 gf -----[3]

Solving equations [1], [2] and [3], given

2

5g

,

2

1f and 4c

The equation of the circle is 04522 yxyx

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Tangent To A Circle• Teorem 1

Suppose we have a standard equation, so the equation of a tangent for the circle at the point of is given by

see figure 1.2

222 ryx

),( 11 yxP 211 ryyxx

yxP ,

y

x

222 ryx 2

11 ryyxx

Figure 1.2

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• Find the equation of the tangent to a circle at the point 1322 yx 3,2T

SolutionMethod 1By using the common tangent equation

211 ryyxx

In this case 21 x and y1= 3 . So the tangent is

1332 yx

.

Example 7

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Method 2

1322 yx differentiating with respect to x

022 dx

dyyx

y

x

dx

dy

2

2

y

x

At the point 3,2

3

2

dx

dy

gradient of tangent is

3

2

.

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The length of the tangent to a circle

• TeoremThe length of the tangent from a fixed point to a circle with equation

(denote by d), is given by

baN ,

02222 cfygxyx

cfbgabad 2222

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see figure 1.3

Figure 1.3

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• Find the length of the tangent from the point to the circle 6,4K 62422 yxyx

Solution

We see that 4a and 6b By substituting this value in the equation

d = cfbgaba 2222

,we find 661242264 22 d

= 42

Example 8