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Topics : Chapter 1 : Conic section Chapter 2 : Differential Equation Chapter 3 : Numerical Method Chapter 4 : Data descriptions Chapter 5 : Probability and Random    variables Chapter 6 : Special Distribution    Function Mathematics 102/3
ASSESSMENT 100% Total 100 2 hours Subjective Question All topic Paper 2 60% 100 2 hours Subjective Question All topic Paper 1 Final examination 20% - Throughout The semester Assessment/Quiz/ Tutorial - - Continuous Assessment 20% 100 2 Hour Subjective Question - 1 Test Percentage Marks Time Format Topic Paper Components
MAT 102/3   CHAPTER 1: CONIC SECTIONS
1.1 : Intoduction to conic    sections ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
 
 
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
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
Equations of a cirles   ,[object Object],Squaring both sides, we have This is the equation of the circle with center (a,b) and radius r units If the origin is the center of the circle, the equation becomes
[object Object],[object Object],[object Object],[object Object],[object Object],(i) (ii)
General Equation of a Circle The equation of a circle with center (a,b) and radius r units is Now substituting g = a, f = k and c=a 2 +b 2 -r 2 Conversely, the equation Where g,f and c are constant, represent a circle  This equation is called the general equation of a circle
Center and radius of a circle  x 2 +y 2 +2gx+2fy+c=0 Completing the squares for x 2 +2gx and y 2  +2fy, x 2 +y 2 +2gx+2fy+c=0 X 2 +2gx+g 2 +y 2 +2fy+f 2 =g 2 +f 2 -c (x+g) 2 +(y+f) 2 =g 2 +f 2 -c Hence, the center of a circle is (-g,-f) and the radius is
Dertermine the equation with center (h,k) Example 2 Find the center and the radius of the circle x 2 +y 2 +5x-6y-5=0 Comparing with the general equation, x 2 +y 2 +2gx+2fy+c=0 g=5/2  f=-3 c=-5 Hence,the center is (-5/2,3) and the radius is
Determine the centre and radius of a circle. ,[object Object],[object Object],[object Object],[object Object],4
The center is at ( 3 , -2), and the length of a radius is 2 units y x (3,-2) r  =2
Point of Intersection ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Substituting x=2y+4 in (1) gives (2y+4) 2 +y2-4=0 4y 2 +16y+16+y2-4=0 5y 2 +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)
Point of a circle and a straight line Example 5 Find the coordinates of the points of intersection between the circle X 2 +y 2 -6x+9=0 and the line y=7-x Solution:- Given X 2 +y 2 -6x+9=0….(1)  y=7-x….(2) By substituting (2) into (1) gives, on simplication X 2 -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)
Circle passing through three given points   ,[object Object]
[object Object],[object Object],Suppose the equation of the circle is points into this equation Substituting the coordinates of each of the three equation gives : ---------(1)   --------(2)   Circle passing through three given points 2+2g-2f+c+0 --------(3)
[object Object],[object Object],or Then from equation [1] And from equation [3] The equation of the circle which passes through (0,1), (4,3) and   is
Find the equation of a circle passing through two points with the equation of the diameter given   ,[object Object],[object Object],Solution The standard form of the circle is  Since the circle passes through   -------[1] -------[ 2]   Since the circle passes through
[object Object],Therefore,  -----[3] Solving equations [1], [2] and [3], given ,  and  The equation of the circle is
Tangent To A Circle ,[object Object],[object Object],[object Object],[object Object],y x Figure 1.2
[object Object],Solution Method 1 By using  the common tangent equation   In this case   and  y =  3 . So the tangent is   .   Example 7
Method 2 differentiating with respect to   At the point  gradient of tangent is   .
The length of the tangent to a circle   ,[object Object],[object Object],[object Object]
see figure 1.3 Figure 1.3
[object Object],Solution We see that   and By substituting this value in  the equation   d =  ,we find   =   Example 8

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Math1.1

  • 1. Topics : Chapter 1 : Conic section Chapter 2 : Differential Equation Chapter 3 : Numerical Method Chapter 4 : Data descriptions Chapter 5 : Probability and Random variables Chapter 6 : Special Distribution Function Mathematics 102/3
  • 2. ASSESSMENT 100% Total 100 2 hours Subjective Question All topic Paper 2 60% 100 2 hours Subjective Question All topic Paper 1 Final examination 20% - Throughout The semester Assessment/Quiz/ Tutorial - - Continuous Assessment 20% 100 2 Hour Subjective Question - 1 Test Percentage Marks Time Format Topic Paper Components
  • 3. MAT 102/3 CHAPTER 1: CONIC SECTIONS
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  • 7. 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
  • 8. 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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  • 11. General Equation of a Circle The equation of a circle with center (a,b) and radius r units is Now substituting g = a, f = k and c=a 2 +b 2 -r 2 Conversely, the equation Where g,f and c are constant, represent a circle This equation is called the general equation of a circle
  • 12. Center and radius of a circle x 2 +y 2 +2gx+2fy+c=0 Completing the squares for x 2 +2gx and y 2 +2fy, x 2 +y 2 +2gx+2fy+c=0 X 2 +2gx+g 2 +y 2 +2fy+f 2 =g 2 +f 2 -c (x+g) 2 +(y+f) 2 =g 2 +f 2 -c Hence, the center of a circle is (-g,-f) and the radius is
  • 13. Dertermine the equation with center (h,k) Example 2 Find the center and the radius of the circle x 2 +y 2 +5x-6y-5=0 Comparing with the general equation, x 2 +y 2 +2gx+2fy+c=0 g=5/2 f=-3 c=-5 Hence,the center is (-5/2,3) and the radius is
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  • 15. The center is at ( 3 , -2), and the length of a radius is 2 units y x (3,-2) r =2
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  • 17. Substituting x=2y+4 in (1) gives (2y+4) 2 +y2-4=0 4y 2 +16y+16+y2-4=0 5y 2 +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)
  • 18. Point of a circle and a straight line Example 5 Find the coordinates of the points of intersection between the circle X 2 +y 2 -6x+9=0 and the line y=7-x Solution:- Given X 2 +y 2 -6x+9=0….(1) y=7-x….(2) By substituting (2) into (1) gives, on simplication X 2 -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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  • 26. Method 2 differentiating with respect to At the point gradient of tangent is .
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  • 28. see figure 1.3 Figure 1.3
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