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Mathematics C30 Module 1 Lesson 5 Mathematics C30 Trigonometric Equations Introduction to the Conics 299 Lesson 5 Mathematics C30 300 Lesson 5 Introduction In an earlier lesson, you were shown how to find values of x, if an equation like sin x 1 , were given. With the calculator in degree mode, using the inverse and sine functions, or the sin 1 function, the solution x 90 was found. Even though one solution to the equation has been found, you may recall that there are infinitely many other solutions. This lesson continues the study of solving equations. The use of trigonometric identities enables one to solve more complicated equations. The latter part of this lesson begins the study of conics with a study of circles and parabolas. The study of ellipses and hyperbolas will begin the next lesson in Module 2. Mathematics C30 301 Lesson 5 Mathematics C30 302 Lesson 5 Objectives After completing this lesson you will be able to Mathematics C30 solve a trigonometric equation by finding a particular solution and by finding the general solution. convert the equation of a circle from the general form to the standard form, and vice versa. sketch the graph of a circle. convert the equation of a parabola from the general form to the standard form and vice versa. sketch the graph of a parabola. 303 Lesson 5 Mathematics C30 304 Lesson 5 5.1 Solving Equations – General and Particular Solutions y 2 2 1 150° 30° 1 30° x 1 is given, one solution for is the principal angle, 30 or 2 5 radians , radians. Another solution is the principal angle, 180 30 150 or 6 6 since sine is positive in the second quadrant also. If an equation like sin All the principal angles satisfying the equation are called particular solutions to the equation. Thus, particular solutions are found from 0 to 2 or 0° to 360°. All other angles which are coterminal with the particular solutions are also solutions to 1 the equation, sin x . The general solution to the equation is the infinite set of all 2 angles which are coterminal with the particular solutions. The general solution may be written in a compact way. x 30 n 360 and x 150 n 360 , n I or x 5 n 2 and x n 2 , n I 6 6 Mathematics C30 305 Lesson 5 Example 1 Find all the particular and general solutions to 2 sin 3 . Solution: Rewrite the equation sin 3 2 Sine is negative in the third and fourth quadrants. y x – 3 2 2 – 3 Recognize that the reference angles are both 60° by recalling the 30 60 90 triangle. 2 3 60° 1 Therefore, the particular solutions are: 180 60 240 and 360 60 300 or 4 5 and 2 . 3 3 3 3 The general solution is: 240 n 360 and 300 n 360 , or Mathematics C30 306 4 5 n 2 and n 2 , n I . 3 3 Lesson 5 Example 2 Find the particular solutions to cos 2 1 . 2 Solution: Solve for cos . 1 cos Note the value when you square root both sides. 2 1 1 There are now two equations to solve, cos Recognize that the reference angles are 45° by recalling the way the sides are labelled in the 45 45 90 triangle. 2 2 and cos . 2 1 45° 1 Each solution to the equation will have a reference angle of 45°. y Therefore, the particular solutions are: for cos 2 1 2 , 45 and 315°. 1 45° 1 x 45° –1 2 y for cos 1 2 , 135 and 225°. 1 –1 In radians, the solutions are Mathematics C30 2 45° –1 x 45° 2 3 5 7 , , , . 4 4 4 4 307 Lesson 5 Example 3 Use a calculator to solve 2 + cos x = 2.0731 for all x in [ 0 , 2 ) . Solution: Rewrite the equation. 2 cos x 2 .0731 cos x 0 .0731 With the calculator in radian mode, the solution is 1.4976 radians. This solution is in the first quadrant. The other solution, in the fourth quadrant, is 2 1.4976 4.7856 . Note that since 1 .4976 is not in [0 , 2 ) , it is not a solution. The positive angle of measure 4.7856 is coterminal with 1 .4976 and is in [0 , 2 ) . Exercise 5.1 1. Find all the solutions 0 360 for each equation. a. csc 2 b. cot 1 c. sin d. sec 1 2 e. csc 1 f. tan 0 Mathematics C30 2 3 308 Lesson 5 2. 3. 4. 5. Find the general solutions, in radians, for each equation. 3 2 a. cos b. sin 1 c. tan 1 d. 2 sin 1 0 e. cot 1 0 Find all the solutions 0 2 for each equation. 3 4 a. sin 2 b. cos 2 1 c. tan d. 3 cot 2 1 0 2 3 Find all the solutions in degrees, for each equation. a. cos x 0 .6406 b. tan x 85 c. sin d. cos x 2 x 0.5869 2 5 Solve sin x 0 .6789 for Mathematics C30 3 x . 4 4 309 Lesson 5 5.2 Solving Equations of the Form sin (ax + b) = c. The technique for solving these equations is similar to that of the previous section and will simply be illustrated by the examples. 1. Replace ax b with A. 2. Solve for A. 3. If the answer is to be a general solution, find A as a general solution by adding 2 n now. 4. Replace A with ax b and solve for x. Example 1 Solve sin 1 1 x , for all possible values of x. Then find the solution, 0 x 2 . 2 2 Solution: 1 2 Step 1: Use the solution to the simpler equation, sin A . 1 2 7 A 6 Step 2: sin A (Remember sin is negative in quadrants 3 and 4.) and 7 2 n , n I 6 Step 3: A Step 4: Substitute 1 x for A. 2 1 x A. Let 2 1 7 x 2 n , n I 2 6 A 11 6 A 11 2 n , n I 6 1 11 x 2 n , n I 2 6 Multiply each expression by 2 to isolate the x. x Mathematics C30 7 4 n , n I 3 x 310 11 4 n , n I 3 Lesson 5 Find the solution, 0 x 2 . Test various values of n I to find 0 x 2 . If n = 0, x If n = 1 or 1 , x clearly falls out of the range from 0 to 2 . 7 11 and . These are not in [0 , 2 ) . 3 3 Therefore, there are no solutions for Example 1 in [ 0 , 2 ). Example 2 This example will be used as a model for Example 3. Solve tan x 1 . Solution: Since tangent is negative in the second and fourth quadrants, the two particular 3 7 solutions are the principal angles, x , and x . 4 4 y 1 1 –1 The reference angle is –1 x . 4 The general solution to the equation is: x 3 7 n 2 , and x n 2 , n I . 4 4 Mathematics C30 311 Lesson 5 Example 3 Solve tan 2 x 1 . Solution: Use the general solution to Example 2 to solve tan 2 x 1 . (Let A 2 x .) tan A 1 has the solution tan 2 x 1 3 n 2 , 4 7 A n 2 , n I 4 A 3 n 2 , 4 7 2 x n 2 , n I 4 2x has the solution Solve for x. 3 n 2 4 3 2 x n 2 4 7 2 x n 2 4 7 x n , n I 8 2x 7 n 2 4 7 2 x n 2 4 11 2x n 2 4 11 x n , n I 8 2x or Therefore, the general solution to tan 2 x 1 is: x 7 11 n , n I and x n , n I 8 8 Mathematics C30 312 Lesson 5 Exercise 5.2 1. 2. a. 1 Find the general solutions to cos x . 2 b. 1 1 Find the general solution to cos x . 2 2 a. Find the general solution to sin 3 x b. Find all solutions in 0, 2 . 3. 1 Find the general solution to cos x 1 . 3 4 4. Find all x values for which sec x is undefined. Mathematics C30 313 3 . 2 Lesson 5 5.3 Using Algebraic Techniques and Trigonometric Identities to Solve Equations Some trigonometric equations must be reduced to a simpler form before they can be solved easily. The following examples illustrate some of the techniques that can be used. Example 1 (Factoring) Solve 3 sin 2 4 sin 1 0 for the particular solutions. Solution: This is a quadratic equation and can first be solved for sin by factoring. 3 sin 1 sin 1 0 3 sin 1 0 sin 1 0 1 sin 3 sin 1 recall if a b 0 then a 0 or b 0 set each factor to 0 and isolate the trig function Using a calculator, the solution to sin 1 is 0 .3398 radians. 3 y 0.3398 0.3398 x The particular solutions are: 0.3398 3.4814 , and 2 0.3398 5.9434 . The solution to sin 1 is Mathematics C30 3 . 2 314 Lesson 5 Some trigonometric equations must first be changed to quadratic form before they can be solved. This is done by applying one of the trigonometric identities. Example 2 (Factoring) Solve sin 2 cos 2 sin . Solution: Before the quadratic formula can be applied, the trigonometric ratios must be the same. The equations must contain all sines or all cosines, but not several different functions. Substitute for cos 2 sin 2 cos 2 sin 0 sin 2 1 sin 2 sin 0 Simplify sin 2 1 sin 2 sin 0 2 sin 2 sin 1 0 2 sin 1sin 1 0 Factor Solve for as in the previous example. 2 sin 1 0 sin sin 1 0 sin 1 1 2 Alternatively, the quadratic formula may be used to solve 2 sin 2 sin 1 0 . 1 1 4 2 1 1 9 1 3 2 2 4 4 1 sin 1, 2 sin n 2 , n I . 2 1 7 If sin , then n 2 and 2 6 11 n 2 , n I . 6 If sin 1 , then Mathematics C30 315 Lesson 5 Example 3 (Squaring Both Sides) Solve the equation sin cos 1 . Solution: Square both sides of the equation. sin cos 1 2 2 sin 2 2 sin cos cos 2 1 sin 2 cos 2 2 sin cos 1 1 2 sin cos 1 2 sin cos 0 sin 2 0 Determine the general solution. Double-angle identity Let A 2 sin A 0 A n 2 n n, n I 2 In the interval [ 0 , 2 ) , the particular solutions would be 0 , 3 , , and , n I. 2 2 Each solution must be checked in the original equation since squaring both 3 sides may introduce values which are not solutions. The values and do 2 not satisfy the equation; therefore, 0 , is the solution. 2 Mathematics C30 316 Lesson 5 Example 4 (Using an Addition or Subtraction Identity) Solve sin 5 x cos 3 x sin 3 x cos 5 x 1 . Solution: Apply the subtraction identity sin A B sin A cos B sin B cos A . Substitute A = 5x and B = 3x. sin 5 x 3 x 1 sin 2 x 1 n 2 2 x n , n I 4 2x Determine the general solution. Exercise 5.3 1. Find all the particular solutions in radians for each equation in the interval 0, 2 . a. 2 cos x 1tan b. sin cos 0 c. 2 cos 2 cos 0 d. 1 sin 2 sin e. 3 sin 2 5 sin 2 0 f. tan g. cos 2 x 3 sin x 1 h. cos x sin x Mathematics C30 2 x 1 0 2 x sec x 1 0 317 Lesson 5 2. Find the general solution in degrees. 2 2 a. cos 3 x cos x sin 3 x sin x b. sin x 180 cos 90 sin 90 cos x 180 c. tan 3 x tan x 1 1 tan 3 x tan x d. sin tan e. cos 2 2 cos 2 0 3 2 5.4 Conic Sections – The Circle Conic Section Par abola Cir cle H yper bola E llipse Mathematics C30 318 Lesson 5 The conic sections are the curves generated by the intersections of a plane with one or two sections of a cone. y For a plane parallel to the top or bottom, a circle is produced. x y The closed curve produced by the intersection of a single section with an inclined plane is an ellipse. x y The curve produced by a vertical plane intersecting both sections is a hyperbola. x y The curve produced by the intersection of a single section with a plane, parallel to a side of the cone, is a parabola. Mathematics C30 319 x Lesson 5 Because of this simple geometric description, the conic sections were studied by the Greeks long before their application to planetary orbits was known. Apollonius wrote the classic ancient work on the subject entitled On Conics. Kepler was the first to notice that planetary orbits were ellipses, and Newton was then able to derive the shape of orbits mathematically using calculus, under the assumption that gravitational force goes as the inverse square of distance. Depending on the energy of the orbiting body, orbit shapes, which are any of the four types of conic sections, are possible. In this course, the equations of the four conic sections will be studied, and the properties of the graph of the conic will be determined from the equation. Since the algebraic technique of completing the square is important to the work on conics, a brief review is provided first. Completing the Square A complete or perfect square is a product of two equal factors. x b 2 , x b 2 , or 2 a 5 b 2 , etc. Some quadratic expressions can be expressed as a perfect square of a binomial. x 2 2 bx b 2 x b x 2 2 bx b 2 x b 2 2 These equations can be checked by multiplying the right hand side by the FOIL method. An expression that has one of the above forms can be written as a perfect square. Mathematics C30 320 Lesson 5 Example 1 Write each quadratic expression as a perfect square of a binomial. x2 6x 9 u 2 2u 1 16 c 2 24 ac 9 a 2 a. b. c. Solution: a. b. c. x 2 6 x 9 x 2 2 3 x 3 x 3 2 u 2 2 u 1 u 1 2 2 16 c 2 24 ac 9 a 2 4 c 2 4 c 3 a 3 a 2 4 c 3 a 2 2 It will be necessary, in the study of circles, to change a quadratic expression in such a way that a portion of it may be written as a perfect square. Example 2 Given x 2 6 x 5 , add the correct amount to each side of the equation so that the left side may be expressed as a perfect square. Solution: This equation is in the form ax 2 bx c , where a 1 . 2 b Determine . 2 Add 9 to both sides. Write as a perfect square. Mathematics C30 2 1 2 6 9 x2 6x 9 5 9 x 3 2 14 321 Lesson 5 Example 3 Complete the square, for each variable separately, in the expression x 2 2 x y 2 10 y 9 . Solution: Add 1 to the x terms and 25 to the y terms. x 2 2 x 1 y 2 10 y 25 9 1 25 Write the expression with each variable as a perfect square. x 1 2 y 5 2 35 Note that the same amount has to be added to the right side of the equation to maintain equality. 2 b In general, if is added to the left and right side of x 2 bx K , then the left 2 side can be written as a perfect square. 2 2 2 2 b b x 2 bx K 2 2 b b x K 2 2 Mathematics C30 322 Lesson 5 Example 4 Complete the square in x 2 7 x 11 . Solution: 2 2 49 1 7 2 7 2 4 Add 49 to both sides. 4 2 7 7 x 2 7 x 11 2 2 2 2 7 49 x 11 2 4 2 7 44 49 x 2 4 4 2 7 93 x 2 4 Completing the Square when the Coefficient of x 2 is not 1. To complete the square in equations where the coefficient of x 2 is not 1, and a 0 , such as in ax 2 bx K , you must first factor the a and then complete the square of the remaining factor. b a x 2 x K a 2 2 2 b b b a x x K a a 2 a 2a 2 1 b b Inside the brackets, “the square of one-half the coefficient of x” is or . 2a 2 a 2 2 b Since the new term on the left side is actually a , this is what must be added 2a to the right side to maintain equality. 2 b A common error is to add only to the right side. 2a Mathematics C30 323 Lesson 5 Example 5 Complete the square, for each variable separately, in 2 x 2 5 x 3 y 2 9 y 1 . Solution: Combine common terms. 2 x Factor 5 2 x 2 x 3 y 2 3 y 1 2 . Complete each square. 2 5x 3 y2 9 y 1 2 2 2 2 5 5 3 5 3 2 x 2 x 3 y 2 3 y 1 2 3 2 4 2 4 2 2 2 5 3 25 27 2 x 3 y 1 4 2 8 4 2 2 5 3 87 2 x 3 y 4 2 8 Example 6 Complete the square for the variable x in the following expressions. a. y x2 9x 1 b. y 3x2 7x 5 Mathematics C30 324 Lesson 5 Solution: a. y x2 9x 1 2 2 9 9 y x 2 9 x 1 2 2 2 9 Note that is added and subtracted on the same side of the equation. It could 2 have just as well been added to the y on the left side of the equation. 2 9 81 y x 1 2 4 2 9 77 y x 2 4 b. Factor the 3. Complete the square. y 3x2 7x 5 7 y 3 x 2 x 5 3 2 2 7 7 7 y 3 x 2 x 5 3 3 6 6 2 7 Note that the new term added is not but is 6 be subtracted. 2 7 3 , so the same amount must 6 2 7 11 y 3 x 6 12 The Exercises at the end of this section contain several completing the square problems, which you may do before going on to conics. Mathematics C30 325 Lesson 5 The Circle Geometrically, a circle is defined to be a curve created by the intersection of a horizontal plane with a vertical cone. Another definition of a circle is that it is the set of points in a plane that are a fixed distance from a fixed point. The fixed distance is called the radius of the circle and the fixed point is called the center of the circle. Using this definition, the equation of a circle can be developed. A Circle whose Center is at ( h, k) and whose Radius is r y P(x, y) r C(h, k) x Let P x, y be any point on a circle, whose center C is the point (h, k), and whose radius is r. The distance from C to P equals r. Use the distance formula. x h 2 y k 2 r Square both sides. x h 2 y k 2 r2 Remember that the distance formula can determine the distance between two points x, y and x1 , y1 on the coordinate plane using the formula, y y1 2 x x1 2 . The Standard Equation of a circle with radius r and center (h, k) is x h 2 y k 2 r 2 , where: Mathematics C30 • r is the radius, and • h, k is the center. 326 Lesson 5 Example 7 Find the standard equation of the circle whose center is (4, 1) and radius is 5. Solution: r=5 h=4 k=1 Write the standard form. x h 2 y k 2 r2 Substitute the values for r, h, and k. x 4 2 y 1 2 x 4 2 y 1 2 52 25 Example 8 5 and center is 7, 5 . Find the standard equation of the circle whose radius is Solution: r 5 h 7 k 5 Write the standard form. Substitute the values for r, h, and k. Mathematics C30 x h 2 y k 2 r2 x 7 2 y 5 2 x 7 2 y 5 2 5 327 5 2 Lesson 5 Example 9 A circle whose center is at 5, 2 passes through the point (2, 2). Determine its equation and sketch the graph of the circle. Solution: The radius must be found first by calculating the distance from the center to the given point on the circle. Use the distance formula. r= 5 2 2 2 2 2 r 25 r 5 The equation of the circle is x 5 y 2 25 . 2 2 Sketch the graph. y x Use a compass and sketch the circle by putting the point at the center 5, 2 with a radius of 5. Mathematics C30 328 Lesson 5 The General Form of the Equation of the Circle The standard form of a circle can be expanded. x 2 x y 2 hx 2 yk h k r 2 0 2 xh h 2 y 2 2 yk k 2 r 2 2 2 2 2 Let the constants be: C 2 h D 2 k E h2 k2 r2 Therefore, x 2 y 2 Cx Dy E 0 . The General Equation of a Circle x 2 y 2 Cx Dy E 0 where C, D, and E, are constants. The recognizing features are: • The x 2 and y 2 have coefficients 1. • x 2 and y 2 are added. Example 10 Write the general form of the circle with center (0, 1) and radius 6. Solution: Write the standard form. Expand x 0 2 y 1 2 36 x 2 y 2 2 y 1 36 0 x 2 y 2 2 y 35 0 Note that the general form does not show the x term, 0 x . Mathematics C30 329 Lesson 5 Completing the Square to Determine the Center and Radius Example 11 Find the center and radius of the circle whose general equation is x 2 6 x y 2 4 y 12 . Solution: Group the common variables. Complete the squares inside the brackets. Write in standard form. x x 2 6 x y 2 4 y 12 x 6 x 9 y 4 y 4 12 9 4 2 2 6 x y 2 4 y 12 2 x 3 2 y 2 2 25 x h 2 y k 2 r 2 h 3 k 2 r 5 The center h, k is 3, 2 . The radius, r, is 5. Example 12 Change the given general form of a circle to standard form and state the center and radius of the circle. 2 x 2 8 x 2 y 2 12 y 48 Mathematics C30 330 Lesson 5 Solution: Since the coefficients of x 2 and y 2 are 2, the entire equation may be divided by 2 to get a coefficient of 1 for x 2 and y 2 . Divide the expression by 2. Group the common variables. Complete the squares. Write in standard form x x 2 4 x y 2 6 y 24 x 4 x 4 y 6 y 9 24 4 9 2 4 x y 2 6 y 24 2 2 x 2 2 y 3 2 37 37 and center 2, 3 . This is a circle with radius Exercise 5.4 1. 2. For what value of m is each a perfect square? a. x2 6x m b. x2 2 xm 3 c. x2 1 xm 2 d. 2 x2 5x m e. 1 2 x 2x m 2 Complete the square in the variable x. a. 3 y x2 2 b. y 2 x 2 4 x 11 c. y 3x2 x 2 Mathematics C30 x 7 331 Lesson 5 3. 4. 5. 6. Complete the square in each variable separately. a. x2 x y2 y 4 b. x2 2x y2 8 y 0 c. 2 x 2 4 x 3 y 2 12 y 1 d. x2 y2 2x 6 y 7 0 e. 5x2 4 x 3 2 y2 4 y 8 0 f. x2 2 y2 4 y 0 Write the standard form of the equation of each circle. a. center (2, 0), radius 2 b. center 3, 4 , radius 5 c. center (0, 0), passing through (5, 4) d. center 3, 3 , passing through (0, 1) e. x2 y2 9 0 f. x2 y2 4 x 2 y 4 0 State why each is not an equation of a circle. a. y 2x 3 b. x2 y2 2x 6 y 1 0 c. x2 3 y2 2x 2 y 5 0 Determine if x 2 y 2 2 x 2 y 100 0 is the equation of a circle. Explain your answer. Mathematics C30 332 Lesson 5 5.5 The Parabola The geometric description of a parabola is the curve which is formed when a plane intersects a cone and the plane is parallel to a slant side of the cone. Another definition of a parabola, is the one which can be used to find the equation of the curve. Definition: The parabola is the set of all points equidistant from a fixed point F and a fixed line not containing F. In the diagram below, FP MP for every point P on the curve. The fixed point is called the focus of the parabola and the fixed line is called the directrix. The axis of symmetry is the line through the focus perpendicular to the directrix. The portion of the parabola on one side of the axis is the mirror image of the portion on the other side of the axis. The vertex is the point at which the locus intersects the axis of symmetry. The vertex is always midway between the focus and the directrix. axis of symmet r y arm P ( x, y) arm F focu s ver t ex M direct rix FP PM for every point P. By use of the distance formula, if the focus and directrix are known, then the equation of the parabola can be found. Mathematics C30 333 Lesson 5 Example 1 Find the equation of the parabola whose focus is the point (4, 5) and whose directrix is the straight line y 3 . Solution: y 5 4 3 2 1 –1 –2 F(4, 5) P( x, y) x 1 2 3 4 M (x, –3) y = –3 Dir ect r ix Let P(x, y) be any point on the parabola. Join PF. Draw PM perpendicular to the directrix y 3 . The coordinates of M are x, 3 . Use the definition of a parabola and the distance formula to determine the equation. PF PM x 4 2 y 5 2 x x 2 y 3 2 x 4 2 y 5 2 x x 2 y 3 2 x 4 2 y 5 2 y 3 2 x 4 2 y 3 2 y 5 2 x 4 2 y 2 6 y 9 y 2 10 y 25 x 4 2 16 y 16 x 4 2 16 y 1 If this equation is solved for y you get y Square both sides. 1 2 1 x x 2 , which is the familiar quadratic 16 2 equation. Mathematics C30 334 Lesson 5 From the definition of a parabola, the following points are important to note. The arms of the parabola extend indefinitely away from the directrix. The parabola does not cross the directrix but is entirely on the same side of the directrix that the focus is on. The vertex is midway between the focus and the directrix. The axis of symmetry passes through the vertex and the focus and is perpendicular to the directrix. From any point on the parabola the distance to the focus equals the perpendicular distance to the directrix. Example 2 From the information given about a parabola, determine the remaining information and then, make a sketch of the parabola including the focus and directrix. a b c d Focus (2, 0) 0, 3 2, 2 Directrix x 2 y3 Vertex x=4 (0, 0) (4, 2) Axis of Symmetry Solution: y a. The axis of symmetry is y = 0, or the x-axis. The vertex is (0, 0), which is the same distance to the focus as to the directrix. y= 0 (2, 0) x x = –2 Mathematics C30 335 Lesson 5 y b. The axis of symmetry is the y-axis or x 0 . The vertex is (0, 0), on the axes of symmetry and midway between the focus and directrix. y= 3 x (0, –3) x= 0 y c. The axis of symmetry is y = 0, or the x-axis. Since the vertex is 4 units away from the directrix, it must also be 4 units from the focus. This puts the focus at 4 , 0 on the axis of symmetry. y= 0 x x= 4 d. y The distance between the focus and vertex is 4 + 2 = 6. Both points lie on the line y = 2 so the axis of symmetry is y = 2. The directrix is x = 10 since it must be perpendicular to the axis of symmetry and 6 units away from the vertex. f(–2, 2) v(4, 2) y= 2 x x = 10 Mathematics C30 336 Lesson 5 The Standard Equation of a Parabola Suppose that the vertex of a parabola is at some point (h, k), and that the directrix is a distance p units away from the vertex. (Note that p is defined to be a distance and distance is always a positive quantity.) The focus is then also a distance p units away from the vertex. If a parabola with arms upward is considered first, the following diagram and coordinates are obtained. y P (x, y ) F p p V (h, k ) y= k –p M x The coordinates of M are x, k p . The coordinates of F are h , k p . The equation of the above parabola can now be determined using the definition that a parabola is a set of points equidistant from a fixed line and a point not on the line, just as in Example 1. This can be done using the distance formula. PF PM x h 2 y k p 2 x x 2 y k p 2 x h 2 y k p 2 y k p 2 x h 2 y k 2 2 p y k p 2 y k 2 2 p y k p 2 x h 2 4 p y k Mathematics C30 337 Lesson 5 Similarly, if the arms of the parabola are downward, the equation obtained would be x h 2 4 p y k . y x M p V (h, k ) p F P (x, y ) The coordinates of M are x, k p . The coordinates of F are h , k p . PF PM x h 2 y k p 2 x x 2 y k p 2 x h 2 y k p 2 y k p 2 x h 2 y k 2 2 p y k p 2 y k 2 2 p y k p 2 x h 2 4 p y k Mathematics C30 338 Lesson 5 A horizontal parabola as shown in the next diagram will have the equation y k 2 4 p x h . If the arms were to the left, the equation would be y k 2 4 p x h . y x M (h – p, k ) P (x, y ) 2 ( y – k ) = 4p (x – h ) p p (h, k ) F (h + p, k ) x= h –p Standard form of the Equation of a Parabola x h 2 4 p y k x h 2 4 p y k y k 2 4 p x h y k 2 4 p x h arms up arms down arms right arms left The vertex of each parabola is (h, k) and p is the distance from the vertex to the directrix or the focus. The standard equations of the vertical and horizontal parabolas must be remembered. When finding the equation of a parabola you need not always use the distance formula. You may use the standard equation and simply fill in the values for h, k, p. Mathematics C30 339 Lesson 5 Example 3 Find the standard form of the equation of a parabola for which the following information is given. a. b. c. vertex (0, 0); directrix x + 1 = 0 focus (2, 2); vertex (1, 2) focus 1, 0 ; directrix y 2 Solution: a. vertex (0, 0); directrix x 1 0 y Sketch the parabola. V (0 , 0 ) x x = –1 The arms must be to the right. Determine the equation. k 0 h 0 p 1 Substitute the values. y k 2 4 p x h V 0, 0 is the vertex. The distance from the vertex to the directrix. y k 2 y 0 2 4 p x h 4 1 x 0 y2 4 x Mathematics C30 340 Lesson 5 b. focus (2, 2); vertex (1, 2) y Sketch the parabola. Remember that the directrix and focus are on opposite sides of the vertex. F (2 , 2 ) V (1 , 2 ) x The focus is one unit away from the vertex so the directrix is the y-axis, one unit away from the vertex. Therefore, p = 1. From V(1, 2) we get h = 1, k = 2. The arms open to the right. Write the equation. Substitute the values. Simplify. c. y k 2 4 p x h y 2 2 4 1 x 1 y 2 2 4 x 1 focus 1, 0 ; directrix y 2 y Sketch the parabola. F(–1, 0) The given information shows that the arms of the parabola must be up. The vertex must be V 1, 1 . p=1 h 1 k 1 Write the equation. Substitute the values. Simplify. Mathematics C30 V x y = –2 x h 2 4 p y k x 1 2 4 1 y 1 x 1 2 4 y 1 341 Lesson 5 The General Equation of a Parabola Any quadratic equation of the form y ax 2 bx c, or x ay 2 by c with a 0 is called a general equation of a parabola. These quadratic equations are converted to the standard form, which shows the vertex and the distance p from the vertex to the focus. This is done by completing the square of the second degree variable. Example 4 The equation of a parabola is given by y of the vertex and the value of p. 1 2 x 2 x 4 . Determine the coordinates 8 Solution: Use the completing the square method to convert the equation to the standard form. y 1 2 x 2x 4 8 Multiply by 8. Complete the square in the expression which contains the quadratic variable. 8 y x 2 16 x 32 Factor 8 from the left term. 8 y 32 x 8 8 y x 8 y x 2 16 x 32 2 16 x 64 32 64 8 y x 8 32 2 2 8 y 4 x 8 x 8 2 2 8 y 4 This equation is now in the form, x h 4 p y k . 2 The vertex is V 8, 4 . The arms are up. Since 4p = 8, p = 2. Mathematics C30 342 Lesson 5 Sketch the graph of the parabola. y (8, –2) x 2 (8, –4) p= 2 Dir ect r ix y = –6 Example 5 Write the standard form of the general parabola equation, y 2 8 x 4 y 4 0 , and sketch the parabola. Solution: y 2 4 y 8 x 4 Complete the square in the y variable. y 2 4 y 4 8 x 4 4 y 2 2 8 x 8 y 2 2 8 x 1 This is in the form y k 4 p x h . 2 The vertex h , k 1, 2 . Since 4 p 8 , p 2 . The arms are to the left. Mathematics C30 343 Lesson 5 Sketch the parabola. y 2 F(–1, –2) x 2 V(1, –2) x= 3 Information about Parabolas form of equation vertex focus directrix direction of opening axis of symmetry x h 2 4 p y k (h, k) h , k p yk p up (+4p) down 4 p xh y k 2 4 p x h (h, k) h p, k xh p right (+4p) left 4 p yk Application A parabaloid is a three-dimensional object whose shape is obtained by rotating a parabola about its axis. Some examples of this are automobile headlamps, television antennas, and search lights. All of these make use of a reflecting property of the parabola. F When light rays come in parallel to the axis of the parabola, they are all reflected to the focus. Conversely, when there is a light source at the focus, the light rays will reflect off the surface parallel to the axis. Mathematics C30 344 Lesson 5 Example 6 Suppose that a parabolic TV antenna has a diameter of 3 m and a depth of 0.6 m. How far away from the vertex should the device catching the radio waves be placed? Solution: Sketch a cross section of the antennae on the x-y axis with the vertex at the origin and the arms up. y (1.5, 0.6) 0.6 m x 1.5 m The information given indicates that one point on the parabola is (1.5, 0.6) and the vertex is at (0, 0). The standard equation of such a parabola is x h 4 p y k . h 0 k 0 2 Write the equation. Substitute the values. x 0 2 4 p y 0 x 4 py 2 Solve for p by substituting (1.5, 0.6) to get p = 0.9375 m. The focus is at (0, 0.9375), and the device should be placed on the axis of the parabola, 0.9375 m from the vertex. Mathematics C30 345 Lesson 5 Exercise 5.5 1. Sketch the parabola from the given information. Label the focus, vertex, directrix and axis of symmetry in each case. a. b. c. 2. Use the definition of a parabola and the distance formula to find the standard equation of each parabola from the given information. a. b. c. 3. focus 1, 0 , directrix y = 10 vertex (5, 7), directrix x 1 vertex 4 , 4 , focus 0, 4 vertex (0, 0), focus 4 , 0 Write the standard form of the parabola equation. Determine p and the coordinates of the vertex, and focus. Write the equation of the directrix. a. b. c. d. 5. focus (0, 5), directrix y 5 vertex (0, 0), focus (3, 0) focus (2, 1), directrix x 2 Write the standard equation of each parabola with the given properties. a. b. c. d. 4. focus (6, 1), directrix, x 2 focus (6, 1), directrix y = 3 focus (0, 0), vertex 0, 3 y2 x 2 y y 2 8 x 4 y 20 0 x2 6x 8 y 7 0 x2 4 x 2 y 6 0 How far away is the focus from the vertex of a parabola with vertex (0, 0), if (5, 4) and 5, 4 are also on the curve? 6. A parabolic reflector is 40 cm in diameter and 5 cm deep. How far from the vertex should the light source be placed? Mathematics C30 346 Lesson 5 Answers to Exercises Exercise 5.1 1. a. b. c. d. e. f. 2. a. b. c. d. e. 3. a. b. c. d. 4. a. b. c. d. Mathematics C30 225°, 315° 135°, 315° 210°, 330° 150°, 210° 270° 0°, 180°, 360° 11 2 n and 2 n , n I 6 6 3 2 n , n I 2 5 2 n and 2 n , n I 4 4 5 2 n and 2 n , n I 6 6 3 7 2 n and 2 n , n I 4 4 2 4 , , , 3 3 3 0, , 2 2 4 , , , 3 3 3 2 4 , , , 3 3 3 5 3 5 3 5 3 129.8° + n(360°) and 230.2° + n(360°), n I 89.3° + n(360°) and 269.3° + n(360°), n I 50° + n(360°), 130° + n(360°), 230° + n(360°) and 310° + n(360°) n I 66.4° + n(360°) and 293.6° + n(360°), n I 347 Lesson 5 5. In decimal form, the restriction on x is 0 .7854 x 2 .3562 . By calculator, x 0 .7463 and 0.7463 2.3953 . Therefore, no solution fits the restriction. Exercise 5.2 1. 2. a. 2 n 2 , n I 3 4 x n 2 , n I 3 b. x a. 7 2 n , x 9 3 For n 1, 0 , 1, 7 13 1 x , , , 9 9 9 x 10 4 n , n I 3 14 x 4 n , n I 3 x b. 8 14 2 , , 9 9 9 16 8n 3 3. x 4. x Mathematics C30 8 2 n , n I 9 3 2 n , n I 2 348 Lesson 5 Exercise 5.3 1. a. b. c. d. 2 4 1 5 , , , 3 3 4 4 3 0, , , 2 2 3 5 , , , 2 2 3 3 5 3 , , 6 6 2 e. 3 sin 1sin 2 0 f. 1 , 3 0.3398 , 0.3398 sin 2 (no solution) 2 4 0 , , 3 3 5 , 6 6 5 , 4 4 sin g. h. 2. a. 22 .5 n 180 , 157 .5 n180 , n I b. sin x 90 c. 33.75° + n90°, 78.75° + n90°, n I d. n180 , n I e. 60° + n360°, 120° + n360°, 3 , x 30 n 360 , n I 2 240° + n360°, 300° + n360°, n I Mathematics C30 349 Lesson 5 Exercise 5.4 1. a. b. c. d. e. 9 1 9 1 16 25 4 1 2 2. a. 3 103 y x 4 16 b. y 2x 1 9 c. 1 11 y 3 x 1 6 12 a. 1 1 1 x y 4 2 2 2 b. x 1 2 y 4 2 c. 2 x 1 3 y 2 15 d. x 1 2 y 3 2 e. 2 4 2 5 x 2 y 1 3 5 5 f. x 2 2 y 1 2 2 2 2 3. 2 17 2 2 17 2 Mathematics C30 2 350 Lesson 5 4. a. b. c. x 2 2 y 2 4 x 3 2 y 4 2 25 x y 41 2 2 d. e. x 3 2 y 3 2 f. x 2 2 y 1 2 25 x y 9 2 2 9 y x (2, –1) 5. a. b. c. This is a linear equation and its graph is a straight line. There is a negative between x 2 and y 2 . The coefficient of x 2 and y 2 are not equal. 6. Completing the squares gives x 1 y 1 98 . 2 2 The radius squared cannot be a negative, so this is not the equation of a circle. Mathematics C30 351 Lesson 5 Exercise 5.5 1. a. b. c. V(2, 1), arms to the right V(6, 2), arms down directrix, y 6 , arms up 2. a. b. x 2 20 y y 2 12 x c. y 1 2 a. x 1 2 20 y 5 y 7 2 24 x 5 y 4 2 16 x 4 3. b. c. d. 4. y 2 16 x 1 3 , F , 4 4 8 x 2 , V 2, 2 , p 2, F 4 , 2 , a. y 1 b. c. y 2 2 x 3 2 8 y 2 , d. x 2 2 2 x 1, V 1, 1 , p 1 1 , x 1 4 x 0 V 3, 2 , p 2, F 3, 0 , y 4 1 1 1 2 y 1 , V 2, 1 , p , F 2, 1 , y 2 2 2 25 9 1 16 16 5. p 6. 20 cm Mathematics C30 8x 352 Lesson 5 Mathematics C30 Module 1 Assignment 5 Mathematics C30 353 Lesson 5 Mathematics C30 354 Lesson 5 Optional insert: Assignment #5 frontal sheet here. Mathematics C30 355 Lesson 5 Mathematics C30 356 Lesson 5 Assignment 5 Values (40) A. Multiple Choice: Select the best answer for each of the following and place a check () beside it. 1. 2. The particular solutions to sin ____ a. ____ b. ____ c. ____ d. Mathematics C30 4 3 6 a. b. c. d. x 3 have a reference angle ***. 4 3 2 The solutions to cos x ____ ____ ____ ____ 3. 2 1 in [ 0 , 2 ) are ***. 2 120°, 120 120°, 240° 60°, 60 30°, 60° The general solution to sin x 1 is ***. ____ a. ____ b. ____ c. ____ d. 2 n , n I 2 2 n , n I 3 2 n , 2 n , n I 4 4 2 n , n I 2 357 Lesson 5 4. The number of solutions that sin ____ ____ ____ ____ 5. 6. 7. ____ a. ____ b. ____ c. ____ d. x 2 sin x 0 has in [ 0 , 2 ) is ***. 0 1 2 4 The general solutions of cos 2 2 n , 3 4 2 n , 3 4 4 n , 3 4 8 , 3 3 1 1 x are ***. 2 2 4 2 n , n I 3 8 2 n , n I 3 8 4 n , n I 3 All the values of x for which csc 2 x is not defined are ***. ____ a. ____ b. ____ ____ c. d. n, n I 2 n, n I 4 0, n , n I 1 1 The number of solutions that the equation cos 2 x has in 2 2 [ 0 , 2 ) is ***. ____ ____ ____ ____ Mathematics C30 a. b. c. d. 2 a. b. c. d. 0 1 2 4 358 Lesson 5 8. 9. The particular solutions to tan 2 x 1 are ***. 4 ____ a. ____ b. ____ c. ____ d. If sin ____ ____ ____ ____ 10. 11. Mathematics C30 2 n , n I 2 5 , 4 4 0, 0, 2 n , x sin x 6 0 , then x ***. a. b. c. d. equals 2, 3 equals 2 , 3 equals 1, 6 has no solution If 2 cos 2 x sin x 1 0 , then sin x equals ***. ____ ____ a. b. ____ c. ____ d. 1 0, 1 1 ,1 2 1 , 1 2 The solution to sin x cos x in 0, 2 is ***. ____ a. ____ b. ____ c. ____ d. , 4 , 2 , 4 4 3 5 7 , , 4 4 4 3 2 5 4 359 Lesson 5 12. The equation cos 5 x cos 3 x sin 5 x sin 3 x 0 is equivalent to ***. ____ ____ ____ ____ 13. cos 8 x 0 cos 2 x 0 sin 8 x 0 sin 2 x 0 To complete the square in 5 x 2 6 x A B , the values of A and B are ***. ____ ____ ____ ____ 14. a. b. c. d. a. b. c. d. A =5, B = 5 A = 9, B = 9 A = 9, B = 45 A = 5, B = 25 The equation x 2 3 x 3 y 2 12 y 0 is equivalent to ***. 2 a. 3 1 2 x y 2 6 2 4 ____ b. 3 3 2 x 3 y 2 9 2 4 ____ c. ____ d. ____ 2 2 15. The circle whose equation is x 2 y 2 8 y 0 has center and radius ***. ____ ____ ____ ____ Mathematics C30 3 3 2 x 3 y 2 9 2 4 3 1 x 3 y 2 7 2 2 a. b. c. d. (0, 4), 4 0, 4 , 16 0, 4 , 4 4, 0 , 16 360 Lesson 5 16. The one equation of a circle is ***. ____ ____ ____ ____ 17. Mathematics C30 x2 2x y 4 x2 2 y2 3x 3 y 0 x2 2x 3 y2 0 2 a. b. c. d. 3 4 6 12 The equation of the directrix of a parabola with vertex at 5, 7 and focus at 3, 7 is ***. ____ ____ ____ ____ 19. x 2 y 2 100 In the parabola whose equation is x 4 12 y 2 , the distance from the focus to the vertex is ***. ____ ____ ____ ____ 18. a. b. c. d. a. b. c. d. x 13 x 13 y 13 y 7 The vertex of the parabola whose equation is x 5 y 2 2 y 1 is at ***. ____ a. ____ b. ____ c. ____ d. 1 4 , 5 5 4 1 , 5 5 3 1 , 4 5 17 1 , 20 5 361 Lesson 5 20. Mathematics C30 The general equation of the parabola with vertex at 1, 1 and focus one unit to the left is ***. ____ a. ____ b. ____ c. ____ d. x 1 2 y 1 2 1 1 1 3 x y2 y 4 2 4 2 x 1 4 y 1 y 1 2 4 x 1 362 Lesson 5 Answer Part B and Part C in the space provided. Evaluation of your solution to each problem will be based on the following: (8) B. • A correct mathematical method for solving the problem is shown. • The final answer is accurate and a check of the answer is shown where asked for by the question. • The solution is written in a style that is clear, logical, wellorganized, uses proper terms, and states a conclusion. 1. a. Determine the general solution for in degrees. 3 cot 3 0 b. Determine the general solution for all first quadrant values for in radians. 3 cos 2 2 .586 0 Mathematics C30 363 Lesson 5 (8) (8) 2. 3. Find a solution in 0, 2 for each trigonometric equation. a. 1 sin x 0 2 b. 1 csc x 1 6 4 Find the general solutions to each trigonometric equation. a. Mathematics C30 sin 2 3 cos 3 0 example 1, 5.3) 364 Determine in degrees. (Refer to Lesson 5 b. (5) 4. Mathematics C30 1 Determine in radians. 2 (Hint: use addition and subtraction identity) sin 3 x cos 2 .5 x sin 2 .5 x cos 3 x Write the standard form of the equation for each circle and sketch. a. center (3, 1) diameter 6 b. center 1, 1 passing through point 4 , 3 365 Lesson 5 (8) (5) 5. 6. Find the center and radius of each of the circles whose general equation is given. a. x2 y2 4 x 6 y 7 0 b. 3 x 2 36 x 3 y 2 0 Given the standard form of the equation for a parabola, determine the coordinates of the focus, vertex, and equation of the directrix and axis of symmetry. x 5 2 Mathematics C30 12 y 7 366 Lesson 5 (8) 7. Write the standard equation of each parabola which has the given properties. Sketch the graph, and label the vertex, focus, and directrix, with the coordinates and equation. a. focus 2, 3 ; directrix y 3 y x b. 1 focus 0 , ; directrix x = 7. 4 y x Mathematics C30 367 Lesson 5 (10) C. 1. Write a point form summary of the lesson, including essential definitions. _____ (100) Mathematics C30 368 Lesson 5