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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 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. The curves formed depends on the angel at which the plane intersects the cone.

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 From the definition of the circle, a point P(x,y) lies on the circle if and only if PC = r, that is 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

Example 1 Find the equation of the circle with :- Center at origin and radius 3 units (ii) Center (2,-3) and radius 5 units Solutions : (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. 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 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 Example 4 Find the coordinates of the point of intersection of the circles x 2 +y 2 -4=0 and x 2 +y 2 -2x+4y+4=0 x 2 +y 2 -4=0 ……….(1) x 2 +y 2 -2x+4y+4=0…..(2) Solving the equation simultaneously for the point of intersection, – (2) 2x-4y-8=0 x=2y+4

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 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 .

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 Substituting the coordinates of each of the three equation gives : ---------(1) --------(2) Circle passing through three given points 2+2g-2f+c+0 --------(3)

Multiplying equation [3] by 4 and then subtracting from equation [2], gives Multiplying equation [1] by 3 and adding to equation [4] gives 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 Example 6 Find the equation of the circle passing through the points (1,1) and (3,2) and with diameter Solution The standard form of the circle is Since the circle passes through -------[1] -------[ 2] Since the circle passes through

The center of the circle must passes through the diameter Therefore, -----[3] Solving equations [1], [2] and [3], given , and The equation of the circle is

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 y x Figure 1.2

Find the equation of the tangent to a circle at the point 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 Teorem The length of the tangent from a fixed point to a circle with equation (denote by d ), is given by

Find the length of the tangent from the point to the circle Solution We see that and By substituting this value in the equation d = ,we find = Example 8

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