[VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 1

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1 COURSE CODE: MTH 112 COURSE TITLE: VECTOR AND COORDINATE GEOMETRY NUMBER OF UNITS: 3 Units COURSE DURATION: Three hours per week COURSE LECTURER: ALHASSAN CHARITY INTENDED LEARNING OUTCOMES At the end of this course, students are expected to: 1. Define the concept of vectors, and state the types of vectors 2.Understand the concept of Addition of vectors with examples 3. Understand the concept of two dimensional coordinate geometry 4. Define circle with solved examples. 5. Solve problems on parabola and ellipse. COURSE DETAILS: Week 1-2.Introduction to the concept of vectors, types of vectors Week 3-4. Concept of addition of vectors with examples. Week 5-6.Introduction to two dimensional coordinate Straight lines, angles and distance between two lines and point Week 7-8 Introduction to circles, equation of a circle and tangent to the point of a circle Week 9-10.concept of parabola, equation of a parabola, tangent to the point of a parabola. Week 11 Introduction to ellipse, equation of an ellipse Week 12.Revision [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 1

2 RESOURCES Lecturer s Office Hours: Alhassan Charity Tuesdays 8-10am Enoyoze Esosa Wednesday 12-2pm Course lecture Notes: Books Pure mathematics for advanced level 2 nd Edition by Bunday, B.D. and Mulholland,H Introduction to Fundamental Mathematics by kuhe, David Adugh 2011 Edition.ISBN: Homework ~30% of final grade Exams: Final, comprehensive(according to university schedule):~70% of final grade Assignment and grading Academic Honesty: All classwork should be done and submitted to the course representative You may take the assignment home to be taught but make sure you write the solution yourself. NO LATE HOMEWORK All homework should be done at the start of class Late assignment will not be accepted but penalized accordingly. [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 2

3 PREAMBLE A quantity that has magnitude as well as direction is called a vector. In our day to day life, we come across many queries such as What is your height? How should a football player hit the ball to give a pass to another player of his team? Observe that a possible answer to the first query may be 1.6 meters, a quantity that involves only one value (magnitude) which is a real number. Such quantities are called scalars. However, an answer to the second query is a quantity (called force) which involves muscular strength (magnitude) and direction (in which another player is positioned). Such quantities are called vectors. In mathematics, physics and engineering, we frequently come across with both types of quantities, namely, scalar quantities such as length, mass, time, distance, speed, area, volume, temperature, work, money, voltage, density, resistance etc. and vector quantities like displacement, velocity acceleration, force, weight, momentum, electric field intensity etc. Types of Vectors 1. Free vectors: These have magnitude and direction but have no particular position associated with them e.g Displacement A displacement vector 20km north is the same in Kano, lagos, or anywhere. 2. Time related vectors :These are located along a straight line e.g force acting on a rigid body always moves along their lines of action without changing its effect on the body 3. Point located vectors: This is a vector whose position in space is fixed in addition to its magnitude and direction e.g a position vector 4. Position vector: A position vector has magnitude and direction as well as a particular position associated with it.in the diagram below a vector directed from p to q shows the position of q relative to a fixed point p as the origin. PQ is the position vector of Q relative to P.A vector which is not associated with any specific position is called a free vector [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 3

4 Remark The vectors defined above are such that any of them may be subject to its parallel displacement without changing its magnitude and direction. Such vectors are called free vectors. Example 1.Classify the following measures as scalars and vectors i. 5 seconds ii. 1000cm 3 iii. 10 Newtons iv. 30km/hr v. 10g/cm 3 vi. 20m/s towards north Solution i) Time scalar ii) Volume-scalar iii) Force-vector Iv) Speed-scalar v) Density-scalar vi) 20m/s towards north Addition of vectors A vector AB A simply means the displacement from a point A to the point B.Now consider a situation that a boy moves from A to B and then from B to C (Fig 10.7).The displacement made by the boy from A to C, is given by the vector AC and expressed as AC = AB + BC.This is known as the triangle law of vector addition C A B Fig ( 10.5) [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 4

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7 Vectors one is the velocity imparted to the boat by its engine and other one is the velocity of the flow of river water. Under the simultaneous influence of these two velocities, the boat in actual starts travelling with a different velocity. To have a precise idea about the effective speed and direction (i.e., the resultant velocity) of the boat, we have the following law of vector addition. Note from fig 10.9,using the triangle law,one may note that OA + AC = OC or (sinceac = OB ) OA + OB = OC Which is parallelogram law.thus, we may say that the two laws of vector addition are equivalent to each other Properties of vector addition Property 1 For any vectors a and b a +b -b +a (commutative property) [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 7

8 Then and so a + b PQ + QR = PR b + c QR + RS = QS (a + b ) + c = PR + RS = PS [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 8

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10 Figure 1 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 10

11 Class work 1. Given the vectors, R 1 = 3i-2j+k; R 2 =2i-4j-3k, R 3 =-i+2j+2k, find the magnitude. ANS: Consider the vectors V 1 =2i-j, V 2 =i+3j,v 3 =-2i+j. find scalers,α, β such that V 3 = αv 1 + βv 2.. ANS =-1, β=0 3. If E=3i-j-2k, and F=2i+3j+k find ExF. ANS: Consider the following vectors G=i+3j-2k and H=4i-2j+4k, evaluate 3G+2H. ANS: Find the angle between the vectors C=2i+2j-k and D=6i-3j+2k. ANS:= 79 o 6. Evaluate (2i-j)(3i+k). ANS=6 7. If A= 2i-3j-k, B=I+4j-2k, evaluate the cross product BxA. ANS=-10i-3j-11k 8. Given the vectors R 1 =3i-2j+k, R 2 =2i-4j-3k, R 3 =-i+2j+2k, find the magnitudes of 2R 1-3R 2-5R 3. ANS: What is the magnitude of the vector 3i+2j-k ;ANS= Add the following vectors A=3i+4j-12k, B=i+12k,C=i-j+k ANS: =5i+3j+k 11. Find the angle between vectors M=2i+3j+4k and N= 4i-3j+2k, ANS= o Two Dimensional Co-ordinate Geometry 1 Definition: A variable is anything that varies or changes values according to a given or prevailing situation. Definition: Independent variable is a variable whose values are independent of those taken by the dependent variable e.g Teacher s competency is independent of student s performance. [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 11

12 Definition: Dependent variable is a variable whose values depend on another variable. e.g the level of students performance clearly depends on the competency of their teachers; here student s performance is a dependent variable. Definition: A function is a relationship among variable such that any given values assigned to certain variable determine the values of other variables Definition: A linear function is one which only the first power of the independent variable appears in the equation. The graph of a linear function is a straight line graph e.g. y = 3x + 4. Note: y is a dependent variable while x is the independent variable Distance between two point. Let Z(x 1 y 1 ) and P(x 2 y 2 ) be the two points, then the distance between ZP and is given as = (x 2 x 1 ) 2 + (y 2 y 1 ) 2. Example 1. Find the distance between the point Z( 1,5)and P( 2,1) Solution: Using the distance formula, the required distance between Z and P is ZP = (x 2 x 1 ) 2 + (y 2 y 1 ) 2 x 1 = 1 x 2 = 2 y 1 = 5 y 2 = 6 ZP = ( 2 1) 2 + (1 5) 2 = = 25 =5 To find the co-ordinates of the point which divides joins of two given points (x 1 y 1 ) and (x 2 y 2 ) in the ratio m 1 :m 2. Two divisions are involved which are: Internal Division [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 12

13 Let Z and P be the two given points with co-ordinates (x 1 y 1 ) and (x 2 y 2 ) respectively and let R(x, y) be the points which divides ZP in the ratio. m 1 : m 2 internally. Then x = m 1x 2 + m 2 x 1 m 1 + m 2 similarly y = m 1y 2 + m 2 y 1 m 1 + m 2 The Co-ordinate of R is ( m 1x 2 + m 2 x 1, m 1 + m 2 m 1 y 2 + m 2 y 1 m 1 + m 2 ) Corollary if R is the middle point of ZP, we have x = x 1+ x 2, 2 y = y 1+ y 2 2 Then, the co-ordinate of the middle point of the line joining the points (x 1 y 1 ) and (x 2 y 2 ) are ( x 1+ x 2, 2 External Division y 1 + y 2 ). 2 If R now divide ZP externally in the ratio m 1 : m 2 then x = m 1x 2 m 2 x 1 m 1 m 2 similarly Example 1 y = m 1y 2 m 2 y 1 m 1 m 2 Find the co-ordinates of the point which divides the line joining the point (8,9) and ( 7,4) internally in the ratio 2:3. Solution x 1 = 8 x 2 = 7 y 1 = 9, y 2 = 4 The co-ordinates of the point is obtained by substitution. x = y = 2( 7) (4) = 2 = 7 2. Find the co-ordinates of the point which divides the line joining the point (2,1) and (3,5) externally in the ratio 2:3). [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 13

14 Solution Let R (x, y) be the required point, then we have m 1 = 2 and m 2 = 3 Using the formula for external division we have x = m 1x 2 m 2 x 1 = = 0 m 1 m y = m 1y 2 m 2 y 1 m 1 m 2 = So R is the point(0, 7). The straight line Equation of a straight line 2 x = 7 The equation of a straight line is the relation between x and y which is satisfied by the co-ordinates of each and every point on the line and by those of no other point. Gradient of a straight line The gradient of a line passing through Z(x 1 y 2 ) is defined as the ratio Increase in the y co ordinate Increase in the x co ordinate in going From one point to another on the line. Let on be the gradient of the line ZP, then T S M = PR ZR = y 2 y 1 x 2 x 1 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 14

15 Angle of slope y P Z R x Line ZP makes an angle with the positive x axis is called the angle of slope of the line Gradient of the line ZP = PR ZP = tanθ. Hence the gradient of a line = tangent of the angle the line makes with the positive x-axis. Example 1 Find the gradient of the line joining (5,4) and (7,8) and the angle of slope of the line. Solution Let m be the gradient of the line. Then m = y 2 y 1 x 2 x 1 = = 4 2 = 2 Let θ be the angle of slope of the line, thus tanθ = 2 θ = tan 1 (2) θ = Angles Between two lines The angles between two lines whose inclination to the x axis are θ 1 and θ 2 respectively. [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 15

16 Let θ and x be the interior and exterior angles between the given lines, then. θ 1 = θ + θ 2 θ = θ 1 θ 2 tanθ = tan(θ 1 θ 2 ) = tanθ 1 tanθ 2 1+tanθ 1 tanθ 2 If we put m 1 = tanθ 1 and m 2 = tanθ 2 We have tanθ = m 1 m 2 1+m 1 m 2 But if we consider x to be the exterior angle between the line then tan x = tan(a θ) = tanθ = m 1 m 2 1+ m 1 m 2 y 0 x Hence the angle becomes Example tanθ = ± ( m 1m 2 1+ m 1 m 2 ) Find the angle between the lines whose slopes are 2 and 4 Solution If the angle is θ, then tanθ = 2 ( 4) 1+( 2)( 4) = = 2 9 = tan =tan = [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 16

17 Parallel and perpendicular lines Condition for parallelism If the two lines are parallel, the angle between them is zero, hence tanθ = 0 and hence m 1 = m 2. Example Investigate the parallelism of line joining BD and QR (a) B(3,1), D(4,3), Solution Q(4,6) R(5,8) Let m 1 be the gradient of the line joining B and D let me be the gradient of the line joining Q and R (a) m 1 = = 2 1 = 2, Since m 1 = m 2 BD // QR. Condition for perpendicularity m 2 = = 2 1 = 2 If the lines are perpendicular, then angle between them is a right angle that is θ = 90 and tanθ = The denominator 1 + m 1 m 2 = 0 or m 1 m 2 = 1 => m 2 = 1 m 1 Hence two lines are perpendicular if the products of their slopes is 1 that is m 1 m 2 = 1 Example 2 The perpendicularity of lines BD and QR B(5, 1) (3,2) (2,4) R(5,6). m 1 = 2 ( 1) = m 2 = = 1 Since m 1 m 2 = 1, BD QR [means BD is perpendicular to QR]. The circle [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 17

18 A circle is a closed plane curve consisting of all point at a given distance or a locus of points equidistant from a given point. The two main part of the circle is (i) the Centre (ii) the radius Equation of a circle with Centre at the origin and radius r (x h) 2 + (y k) 2 = r 2 r (h,k) y Z(h,k) r R P(x,y) y-k y k x-h x A circle Centre (h,k) radius r consider ZPR ZR = x h RP = y k Since ZPR is a right-angled 1 we have ZR 2 + RQ 2 = PQ 2 Hence (x h) 2 + (y k) 2 = r 2 Equation 1 is the equation of a circle Centre (h, k) radius r. [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 18

19 If the Centre of the circle is the origin (O, O), then eqn 1 r x R P y In OPR, OR 2 + RP 2 = OP 2 x 2 + y 2 = r 2 Examples 1. Find the equation of the circle Centre (2-1) radius 2 units Solution 1. The equation of the circle is (x h) 2 + (y k) 2 = r 2 (x 2) 2 + (y + 1) 2 = 4 x 2 4x y 2 + 2y + 1 = 4 x 2 + y 2 4x + 2y + 1 = 0 2. Find the equation of the circle origin, radius 5 units Solution x 2 + y 2 = r 2 x 2 + y 2 = 5 2 x 2 + y 2 = Find the equation of the circle whose Centre is (1, 2) and passes through( 2, 2). Solution Let r be the radius of the circle, then r 2 = ZP 2 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 19

20 z(1, -2) r (-2, 2) = (1 ( 2) 2 + ( 2 2) 2 = (1 + 2) 2 + ( 4) 2 = = r 2 = 25, r = 5 Hence the equation of the circle Centre (1, 2) and radius 5 units is (x 1) 2 + (y + 2) 2 = 5 2 (x 1)(x 1) + (y + 2)(y + 2) = 25 x 2 2x y 2 + 2y + 2y + 4 = 25 x 2 2x y 2 + 4y + 4 = 25 x 2 + y 2 2x + 4y + 5 = 25 x 2 + y 2 2x + 4y = 0 x 2 + y 2 2x + 4y 20 = 0 General equation of the circle Recall that the equation of the circle Centre (h, k) radius r is (x h) 2 + (y k) 2 = r 2 We have (x h)(x h) + (y k)(y k) = r 2 x 2 xh xh + h 2 + y 2 yk yk + k 2 = r 2 x 2 2xh + h 2 + y 2 2yk + k 2 = r 2 x 2 + y 2 2xh 2yk + h 2 + k 2 r 2 = 0 Putting h = g, k = f and c = h 2 + k 2 r 2 The above equation can be written as [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 20

21 x 2 + y 2 + 2gx + 2fy + c = 0 Thus the equation x 2 + y 2 + 2gx + 2fy + c = 0 is called the general equation of the circle. Observe that the equation of the circle 1. Have coefficients of x 2 and y 2 being equal. 2. Has no xy term 3. Is a second degree equation in x and y. So given the Centre and radius of a circle, we can easily find the general equation and also find the radius and Centre of the circle, by the method of completing the squares. Examples Find the general of a circle Centre (1,4) radius 3 units Solution The equation of a circle Centre (1,4) radius 3 units is (x 1) 2 + (y 4) 2 = 3 2 x 2 2x y 2 8y + 16 = 9 x 2 + y 2 2x 8y + 17 = 9 x 2 + y 2 2x 8y + 8 = 0 2. Find the Centre and radius of a circle whose equation is x 2 + y 2 6x + 4y 3 = 0 Solution x 2 + y 2 6x + 4y 3 = 0 x 2 6x + y 2 + 4y = 3 x 2 6x y 2 + 4y = [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 21

22 Complete the square (x 3) 2 + (y + 2) 2 = 16 (x 3) 2 + (y + 2) 2 = 4 2 Hence the Centre is (3, 2) radius = 4 units. Equation of the Tangent at the point (x 1 y 1 ) on the circle. x 2 + y 2 + 2gx + 2fy + c = 0 Let the equation of the circle be At (x 1 y 1 ) From x 2 + y 2 + 2gx + 2fy + c = 0 x y gx 1 + 2fy 1 + c = 0 C = (x y gx 1 + 2fy 1 ). (1) x 2 + y 2 + 2gx + 2fy + C = 0 2x + 2y dy dy + 2g + 2f = 0 dx dx x + y dy dy + g + f = 0 dx dx (y + f) dy dx (y + f) dy dx dy = x+g dx y+f + x + g = 0 At (x 1 y 1 ), dy = (x 1+g) dx y 1 + f = (x + g) [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 22

23 The equation of the tangent is y y 1 = (x 1+g) x x 1 y 1 + f (y y 1 )(y 1 + f) = (x 1 + g)(x x 1 ) yy 1 + yf y 1 2 y 1 f = (x 1 x x xg x 1 g) yy 1 + yf y 1 2 y 1 f = x 1 x + x 1 2 xg + x 1 g yy 1 + yf + x 1 x + xg = y y 1 f + x x 1 g xx 1 + yy 1 + xg + yf = x y x 1 g + y 1 f (2) Adding x 1 g + y 1 f to both sides of (2), we have xx 1 + yy 1 + xg + x 1 g + yf + y 1 f = x y x 1 g + 2y 1 f xx 1 + yy 1 + (x + x 1 )g + (y + y 1 )f = x y x 1 g + 2y 1 f From (1), x y x 1 g + 2y 1 f = c xx 1 + yy 1 + (x + x 1 )g + (y + y 1 )f + c = 0 Hence, the equation of the tangent to the circle x 2 + y 2 + 2gx + 2fx + c = 0 at the point (x 1 y 1 ) on the circle is xx 1 + yy 1 + g(x + x 1 ) + f(y + y 1 ) + c = 0. Example Show that the point (2,3)lies on the circle x 2 + y 2 3x + 4y 19 = 0. Hence or otherwise, determine the equation of tangent to the circle at the point (2,3). Solution Substituting the co-ordinate (2,3) into the equation x 2 + y 2 3x + 4y 19 = 0 LHS (2) + 4(3) = 0 = RHS. Hence the point (2,3) lies on the circle x 2 + y 2 3x + 4y 19 = 0 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 23

24 Equation of the tangent at (x 1, y 1 ) is xx 1 + yy 1 + (x + x 1 )g + (y + y 1 )f + c = 0 From x 2 + y 2 3x + 4y 19 = 0 g = 3 2, f = 2, c = 19 Hence the equation of the tangent at (2,3) is 2x + 3y 3 2 (x + 2) + 2(y + 3) 19 = 0 This simplifies to x + 10y 32 = 0 Hence the equation of the tangent at (x 1, y 1 ) to the circle x 2 + y 2 = r 2 on it is given by xx 1 + yy 1 = r 2 Equation of Normal to a circle To find the equation of the normal to a circle at any point (x 1 y 1 ) on the circle x 2 + y 2 = r 2. Let P be the point (x 1, y 1 ), equation of the tangent at P is xx 1 + yy 1 = r 2. Now slope of tangent = x 1 y 1 and slope of normal = y 1 x 1 Hence equation of normal at (x 1, y 1 ) is y y 1 = y 1 x 1 (x x 1 ) or yx 1 xy 1 = 0 If the equation of the circle is given to be x 2 + y 2 + 2gx + 2fy + c = 0 The equation of normal at any point on the circle will bey(x 1 + g) x(y 1 + f) + fx 1 gy 1 = 0. Example Find the equation of tangent and the normal to the circle 2x 2 + 2y 2 2x 5y + 3 = 0 at(1, 1). Solution The equation of the circle can be written as x 2 + y 2 x 5 2 y = 0 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 24

25 Here g = 1 2 f = 5 4 c = 3 2 The equation of the tangent at (1, 1) is xx 1 + yy 1 + g(x + x 1 ) + f(y + y 1 ) + C = 0 x 1 + y 1 1 (x + 1) 5 (y + 1) + 3 = 0 or 2x y 1 = The equation of normal is of the formx + 2y = K, it passes through (1, 1) if = K i. e. K = 3 Hence the equation of the normal is x + 2y = 3. Define Parabola A Parabola is a locus of points, equidistant from a given point, called the focus and from a given line called the directrix. y P(x,y) R B(-a,y) A(-a,o) V F(a,o) x Q The line AB a distance of a from the y axis is called the directrix The line AF is called the axis of symmetry. Since BP = FP BP 2 = FP 2 (x + a) 2 = (x a) 2 + (y 0) 2 x 2 + 2ax + a 2 = x 2 2ax + a 2 + y 2 2ax = 2ax + y 2 4ax = y 2 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 25

26 Hence the equation of parabola is y 2 = 4ax The line segment through the focus, and to the axis symmetry and with end points R and Q on the parabola, is called the LATUS RECTUM. The point V is called the Vertex of the parabola. Length of latus rectum= 4a. (a) y x f=(a,0) (b) y f= (-a,0 ) x (c) [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 26

27 y f=(0,b) x Directrix (d) y Directrix f=(0,-b) x If the vertex of the parabola y 2 = 4ax is translated to the point (x 1 y 1 ), the equation of the corresponding parabola becomes. (y + y 1 ) 2 = 4a(x x 1 ). The above equation is said to be in the standard or canonical form Example 1) Find the focus and directrix of the parabola y 2 = 16x Solution By comparing y 2 = 16x with y 2 = 4ax [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 27

28 16x = 4ax a = 16x 4x a = 4 Hence, the focus is (4,0) while the directirx is x = 4 2) Find the equation of the parabola whose vertex is the origin and whose focus is the point F(5,0) Solution Let the equation of parabola be y 2 = 4ax The focus F(a, 0) = (5,0) a = 5 Hence the equation of the parabola is y 2 = 20x 3) Write down the equation of the parabola y 2 4y 12x + 40 = 0 in its Solution canonical form and hence find i) The vertex ii) The focus (iii) the directrix of the parabola Given y 2 4y 12x + 40 = 0 y 2 4y x + 36 = 0 y 2 4y + 4 = 12x 36 (y 2) 2 = 12(x 3) i) Hence, the vertex is (3, 2) ii) Since 4a = 12 a = 3, The focus is 3 + 3,2) = (6,2) iii) The directrix is x = 3 3 = 0 Equation of the Tangent to y 2 = 4ax at the point (x 1 y 1 ) y 2 = 4ax By differentiating implicitly 2y dy dx = 4a dy dx = 4a 2y [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 28

29 = 2a y At the point (x 1y 1 ) dy = 2a dx y 1 y 1 (y y 1 ) = 2a(x x 1 ) y 1 y = y 1 2 = 2ax 2ax 1 Since (x 1, y 1 ) is on y 2 = 4ax, it implies that y 1 2 = 4ax 1 Thus yy 1 4ax 1 = 2ax 2ax 1 yy 1 = 2ax 2ax 1 + 4ax 1 yy 1 = 2a(x + x 1 ) Equation of Normal to y 2 = 4ax at the point (x 1 y 1 ) Recall that the normal to a curve at the point (x 1, y 1 ) is to the tangent at the point (x 1 y 1 ). Since the gradient of the tangent at (x 1 y 1 ) is 2a, the gradient of normal is y y 1 2a the equation of the normal becomes y y 1 = y 1 x x 1 2a thus 2ay 2ay 1 + x 1 y 1 is the equation of the normal to the parabola y 2 = 4ax at the point (x 1, y 1 ) on the parabola. Example 4) Find the equation of the tangent to parabola y 2 = 12x at the point (3,6) Solution By comparing y 2 = 12x with y 2 = 4ax, a = 3 The equation of the tangent at (x 1, y 1 ) is yy 1 = 2a(x + x 1 ), since (x 1, y 1 ) = (3,6) Equation of tangent is 6y = 2 3(x + 3) 6y = 6(x + 3) Thus 6y 6x 18 = 0 or y x 3 = 0 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 29

30 5 The equation of the normal to parabola y 2 = 16x at the point Comparing y 2 = 16x with y 2 = 4ax a = 4 The equation of the normal at the point (x 1, y 1 ) is 2ay + xy 1 = xy 1 + x 1 y 1 Hence the equation of the normal at (1, 4) is 2(4)y 4x = 2(4)( 4) + (1) ( 4) 8y 4x = y 4x = 36 8y 4x + 36 = 0 2y x + 9 = 0 6 The tangent and the normal to the parabola y 2 = 20x at the point meet the x-axis at the point A and B respectively. Show AB 2 = AT 2 + TB 2 y T(5,10) O B x Comparing y 2 = 20x with y 2 = 4ax a = 5 Equation of the tangent to the parabola y 2 = 4ax at (x 1, y 1 ) is yy 1 = 2a(x + x 1 ) Hence, the equation of the tangent to y 2 = 20x at (5,10) is 10y = 10(x + 5) y = x + 5 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 30

31 When y = 0, x = 5. Hence the point A has coordinates ( 5,0) Equation of the normal to y 2 = 4ax at the point (x 1, y 1 ) is 2ay + xy 1 = 2ay 1 + x 1 y 1 Hence the equation of the normal at (5,10) is 10y + 10x = 10(10) y + 10x = 150 When y = 0, x = 15 Hence the points B has coordinates (15,0) AB 2 = (15 + 5) = 20 2 = 400 AT 2 = (5 + 5) 2 + (10 0) = 200 TB 2 = (15 5) 2 + (0 10) = 200 AT 2 + TB 2 = = 400 Hence AB 2 = AT 2 + TB 2 The Ellipse An eclipse is the locus of a point P, moving in a plane such that the sum of it distance from two fixed point F 1 and F 2 called Foci, is a constant. [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 31

32 y x The point v 1, v 2, v 3 and v 4 are called the vertices of the ellipse. The line segment v 1, v 2 is called the major axis while the line segment v 3 v 4 is called the minor axis. The point 0 is called the Centre of the ellipse From the figure above, v 1, v 2 = 2a and v 3, v 4 = 2b Also, From the locus condition we want f 2 P + Pf 1 = constant Then P is on v 1 f 2 P + Pf 1 = f 2 v 1 + v 1 f 1 = f 2 v 1 + v 2 f 2 = v 1 f 1 + f 2 v 2 = v 1 v 2 = 2a [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 32

33 Thus f 2 P + Pf 1 = 2a (x + c) 2 + y 2 + (x c) 2 + y 2 =2a (x + c) 2 + y 2 = 2a (x c) 2 + y 2 (x + c) 2 + y 2 = (2a (x c) 2 + y 2 ) 2 = 4a 2 4a (x c) 2 + y 2 + (x + c) 2 + y 2 x 2 + 2cx + c 2 + y 2 = 4a 2 4a (x c) 2 + y 2 + x 2 2cx + c 2 + y 2 2cx = 4a 2 4a (x c) 2 + y 2 2cx 4a (x c) 2 + y 2 = 4a 2 4cx a (x c) 2 + y 2 = a 2 cx (x c) 2 + y 2 = a cx a (x c) 2 + y 2 = a 2 2cx + c2 a 2 x2 x 2 2cx + c 2 + y 2 = a 2 2cx + c2 a 2 x2 x 2 + y 2 + c 2 = a 2 + c2 a 2 x2 x 2 c2 a 2 x2 + y 2 = a 2 c 2 ( a2 c 2 a 2 ) x2 + y 2 = a 2 c 2 Dividing through by a 2 c 2, we have x 2 + y 2 = (1) a 2 a 2 c 2 From OPF 1 in the above figure, a 2 = b 2 + c 2 b 2 = a 2 c (2) Substituting b 2 and a 2 c 2 in equation (1), we have [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 33

34 [ x2 + y2 a 2 b 2 = 1] (a > b) (3) Equation (3) is the canonical or standard equation of an ellipse with major axis horizontal. Similarly, if the major axis is vertical, it can be shown that x 2 + y2 = 1 Gives the equation of the ellipse. a 2 b2 Example 1) Find four vertices and foci of the ellipse x2 4 + y2 16 = 1 Solution By comparing x2 + y2 x2 = 1 with + y2 = a 2 b 2 We have, b 2 = 4, b = ±2 a 2 = 16, a = ±4 Hence the four vertices are v 1 (0,4), v 2 (0, 4), v 3 (2,0), v 4 ( 2,0). Since c 2 = a 2 b 2 = 16 4 = 12 c = ±3.46 Hence the foci are f 1 (0,3.46) and f 2 (0, 3.46) Change or Origin Recall that, the canonical form of the equation of an ellipse Centre origin is x 2 + y2 = 1 (with major axis on x-axis) a 2 b2 If the Centre of the ellipse is transferred to the point (x, y ) in such a way that the axes are parallel to the x and y axes, then the new canonical form of equation of the ellipse becomes. (x x ) 2 + (y y ) 2 = 1 (With major axis horizontal) a 2 b 2 (x x ) 2 + (y y ) 2 = 1 (With major axis vertical) b 2 a 2 Example [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 34

35 2) Find the equation of the ellipse 25x 2 + 4y 2 50x 16y 59 = 0 in the canonical form and hence, Determine i. The coordinates of the Centre of the ellipse; ii. The four vertices of the ellipse Solution iii. The two foci of the ellipse 25x 2 + 4y 2 50x 16y 59 = 0 25x x + 4y 2 16y = 59 25(x 2 2x) + 4(y 2 4y) = 59 25(x 2 2x + 1 1) + 4(y 2 4y + 4 4) = 59 25(x 2 2x + 1) + 4(y 2 4y + 4) = 59 25(x 2 2x + 1) + 4(y 2 4y + 4) 41 = 59 25(x 1) 2 + 4(y 2) 2 = 100 (x 1)2 4 Hence + (y 2)2 25 = 1 i. The coordinates of the Centre (1,2) ii. Comparing (x 1) (y 2)2 25 ±2, a 2 = 25, a = ±5 with Hence, the vertices on the vertical axes are (x x ) 2 b 2 + (y y ) 2 a 2 = 1, we have b 2 = 4, b = V 1 (0 + x, a + y ) = V 1 (1,7) and V 2 (0 + x, a + y ) = V 2 (1, 3) The vertices on the horizontal axes are V 3 (b + x, 0 + y ) = V 3 (3,2) and V 4 ( b + x, 0 + y ) = V 4 ( 1,2) Since c 2 = a 2 b = 21 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 35

36 iii. The two foci are F 1 (0 + x, c + y ) = F 1 (1, )and F 2 (0 + x, c + y ) = F 2 (1, ) Equation of the tangent at (x 1, y 1 ) to the Ellipse x 2 + y2 a 2 b2 = 1 Differentiating implicitly 2x + 2y dy = 0 a 2 b 2 dx 2y dy = 2x b 2 dx a 2 dy = 2x b2 dx a 2 2y = b2 a 2 x y At the point (x 1, y 1 ); dy dx = b2 a 2 x 1 y 1 The equation of the tangent becomes y y 1 = b2 x 1 x x 1 a 2 y 1 a 2 y 1 (y y 1 ) = b 2 x 1 (x x 1 ) a 2 yy 1 a 2 y 1 = b 2 x 1 x + b 2 x 1 2 a 2 yy 1 + b 2 xx 1 a 2 y 1 b 2 x 1 2 = (1) As the point (x 1, y 1 ) is on the ellipse x 1 2 a 2 + y2 b 2 = 1 or b 2 x a 2 y 1 2 a 2 b (2) From equation (1) a 2 yy 1 + b 2 xx 1 (a 2 y b 2 x 1 2 ) = (3) Substituting a 2 b 2 in (2) for a 2 y b 2 x 1 2 in (3) a 2 yy 1 + b 2 xx 1 a 2 b 2 = 0 a 2 yy 1 + b 2 xx 1 a 2 b (4) Dividing equation (4) through by a 2 b 2 yy 1 + xx 1 = 1 or b 2 a 2 xx 1 a 2 + yy 1 b 2 = 1 [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 36

37 Hence, the equation of the tangent to the ellipse x2 a 2 + y2 xx 1 a 2 + yy 1 b 2 = 1 PRACTICAL WORK ACTIVITY ONE: b 2 = 1 at point (x 1y 1 ) on the ellipse is 1) Find the equation of the tangent to parabola y 2 = 12x at the point (3,6) ACTIVITY TWO: 2) Find the vector parallel to the resultant of the vectors P 1 = 2i + 4j 5k, P 2 = i + 2j + 3k... 3) If A=2i-3j-k and B=i+4j-2k, Evaluate the cross product AxB: ACTIVITY THREE 4) Find the equation of the circle whose Centre is (1, 2) and passes through( 2, 2). 5) Find the resultant of the vectors BC, DC and FD. ANS: BF 6) If the cosine of the angle between U = i + 2j + 2k and V = i 4j + Pk is 1 3. Find the value of the parameter p [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 37

38 VECTOR AND COORDINATE GEOMETRY by ALHASSAN CHARITY is licensed under a creative common Attribution-Noncommercial-share Alike 4.0 International License [VECTOR AND COORDINATE GEOMETRY, EDO UNIVERSITY, IYAMHO] P a g e 38