text stringlengths 235 3.08k |
|---|
οΏ½ Q & Aβ¦ Can we use the same method to find the determinant of a larger matrix? No, this method only works for 2 Γ 2 and 3 Γ 3 matrices. For larger matrices it is best to use a graphing utility or computer software. Using Cramerβs Rule to Solve a System of Three equations in Three Variables Now that we can find the determinant of a 3 Γ 3 matrix, we can apply Cramerβs Rule to solve a system of three equations in three variables. Cramerβs Rule is straightforward, following a pattern consistent with Cramerβs Rule for 2 Γ 2 matrices. As the order of the matrix increases to 3 Γ 3, however, there are many more calculations required. When we calculate the determinant to be zero, Cramerβs Rule gives no indication as to whether the system has no solution or an infinite number of solutions. To find out, we have to perform elimination on the system. SECTION 9.8 solving systems with cramer's rule 847 Consider a 3 Γ 3 system of equations where β£ If we are writing the determinant D x, we replace the x column with the constant column. If we are writing the determinant D y, we replace the y column with the constant column. If we are writing the determinant D z, we replace the z column with the constant column. Always check the answer. Example 4 Solving a 3 Γ 3 System Using Cramerβs Rule Find the solution to the given 3 Γ 3 system using Cramerβs Rule. x + y β z = 6 3x β 2y + z = β5 x + 3y β 2z = 14 Solution Use Cramerβs Rule. D = β£ 1 1 β1 3 β2 1 1 3 β1 β5 β2 1 14 3 β1 3 β5 1 1 14 β2 β5 1 3 14 β£ Then3 ___ β3 _ D D y β9 ___ β3 _ D D z = 6 ___ β The solution is (1, 3, β2). Try It #3 Use Cramerβs Rule to solve the 3 Γ 3 matrix. x β 3y + 7z = 13 x + y + z = 1 x β 2y + 3z = 4 Example 5 Using Cramerβs Rule to Solve an Inconsistent System Solve the system of equations using Cramerβs Rule. ( |
2) Solution We begin by finding the determinants D, D x, and D y. 6x β 4y = 0 3x β 2y = 4 (1) D = β£ 3 β2 6 β4 β£ = 3(β4) β 6(β2) = 0 848 CHAPTER 9 systems oF eQuations and ineQualities We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables. 1. Multiply equation (1) by β2. 2. Add the result to equation (2). β6x + 4y = β8 6x β 4y = 0 0 = β8 We obtain the equation 0 = β8, which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines. See Figure 1. y 5 4 3 2 1 β1 β1 β2 β3 β4 β5 β5 β4 β3 β 21 3 4 5 Figure 1 Example 6 Use Cramerβs Rule to Solve a Dependent System Solve the system with an infinite number of solutions. x β 2y + 3z = 0 3x + y β 2z = 0 2x β 4y + 6z = 0 (1) (2) (3) Solution Letβs find the determinant first. Set up a matrix augmented by the first two columns. Then, β£ 1 β2 3 3 1 β2 2 β4 6 1 β2 3 1 2 β4 β£ β£ 1(1)(6) + (β2)(β2)(2) + 3(3)(β4) β 2(1)(3) β (β4)(β2)(1) β 6(3)(β2) = 0 As the determinant equals zero, there is either no solution or an infinite number of solutions. We have to perform elimination to find out. 1. Multiply equation (1) by β2 and add the result to equation (3): β2x + 4y β 6x = 0 2x β 4y + 6z = 0 0 = 0 2. Obtaining an answer of 0 = 0, a statement that is always true, means that the system has an infinite number of solutions. Graphing the system, we can see that two of |
the planes are the same and they both intersect the third plane on a line. See Figure 2. SECTION 9.8 solving systems with cramer's rule 849 x β 2y + 3z = 0 2x β 4y + 6z = 0 3x + y + 2z = 0 Figure 2 Understanding Properties of Determinants There are many properties of determinants. Listed here are some properties that may be helpful in calculating the determinant of a matrix. properties of determinants 1. If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. 2. When two rows are interchanged, the determinant changes sign. 3. If either two rows or two columns are identical, the determinant equals zero. 4. If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. 5. The determinant of an inverse matrix A β1 is the reciprocal of the determinant of the matrix A. 6. If any row or column is multiplied by a constant, the determinant is multiplied by the same factor. Example 7 Illustrating Properties of Determinants Illustrate each of the properties of determinants. Solution Property 1 states that if the matrix is in upper triangular form, the determinant is the product of the entries down the main diagonal1 ξ² Augment A with the first two columns1 1 2 0 2 0 0 ξ² β£ Then det(A) = 1(2)(β1) + 2(1)(0) + 3(0)(0) β 0(2)(3) β 0(1)(1) + 1(0)(2) = β2 Property 2 states that interchanging rows changes the sign. Given A = ξ° B = ξ° β1 5 4 β3 4 β3 β1 5 ξ², det(A) = (β1)(β3) β (4)(5) = 3 β 20 = β17 ξ², det(B) = (4)(5) β (β1)(β3) = 20 β 3 = 17 850 CHAPTER 9 systems oF eQuations and ineQualities Property 3 states that if two rows or two columns are identical, the determinant equals zero1 2 ξ² β1 2 2 β£ det(A) = 1(2)(2) + 2(2)(β1) + 2(2)(2) + 1 |
(2)(2) β 2(2)(1) β 2(2)(2 Property 4 states that if a row or column equals zero, the determinant equals zero. Thus, 1 2 0 0 ξ², det(A) = 1(0) β 2(0) = 0 A = ξ° Property 5 states that the determinant of an inverse matrix A β1 is the reciprocal of the determinant A. Thus, A = ξ° 1 2 3 4 ξ², det(A) = 1(4) β 3(2) = β2 A β1 = ξ° β2 1 β 1 _ 2 3 _ 2 ξ², det( A β1 ) = β2 ξ’ β 1 __ 2 ξͺ β ξ’ 3 __ 2 ξͺ (1) = β 1 __ 2 Property 6 states that if any row or column of a matrix is multiplied by a constant, the determinant is multiplied by the same factor. Thus, A = ξ° 1 2 3 4 ξ², det(A) = 1(4) β 2(3) = β2 B = ξ° 2(1) 3 2(2) 4 ξ², det(B) = 2(4) β 3(4) = β4 Example 8 Using Cramerβs Rule and Determinant Properties to Solve a System Find the solution to the given 3 Γ 3 system. 2x + 4y + 4z = 2 3x + 7y + 7z = β5 x + 2y + 2z = 4 (1) (2) (3) Solution Using Cramerβs Rule, we have β£ Notice that the second and third columns are identical. According to Property 3, the determinant will be zero, so there is either no solution or an infinite number of solutions. We have to perform elimination to find out. 1. Multiply equation (3) by β2 and add the result to equation (1). β2x β 4y β 4x = β8 2x + 4y + 4z = 2 0 = β6 Obtaining a statement that is a contradiction means that the system has no solution. Access these online resources for additional instruction and practice with Cramerβs Rule. β’ Solve a System of Two equations Using Cramer's Rule (http://openstaxcollege.org/l/system2c |
ramer) β’ Solve a Systems of Three equations using Cramer's Rule (http://openstaxcollege.org/l/system3cramer) SECTION 9.8 section exercises 851 9.8 SeCTIOn exeRCISeS VeRBAl 1. Explain why we can always evaluate the determinant 2. Examining Cramerβs Rule, explain why there is no of a square matrix. 3. Explain what it means in terms of an inverse for a matrix to have a 0 determinant. AlGeBRAIC For the following exercises, find the determinant. unique solution to the system when the determinant of your matrix is 0. For simplicity, use a 2 Γ 2 matrix. 4. The determinant of 2 Γ 2 matrix A is 3. If you switch the rows and multiply the first row by 6 and the second row by 2, explain how to find the determinant and provide the answer4 β£ β2 3.1 β3 4,000 β£. β£ 9. β£ 13. β£ 17. β£ 21. β£ 5 1 β1 2 3 1 3 β6 β3 β£ β1 2 3 β4 10 20 0 β10 β£ β£ β1.1 0.6 7.2 β0.5 2 β3 1 3 β4 1 β5 6 1 β£ β£ 6. β£ 10. β£ 14. β£ 18. β£ 22. β£ 2 β5 β1 6 10 0.2 5 0.1 β£ β£ β1 0 0 0 1 0 0 0 β3 β£ β2 1 4 β4 2 β8 2 β8 β3 β£ 7. β£ 11. β£ 15. β£ 19. β£ 23. β£ 1.1 β4 4.1 2 0 β0.4 β1 0 2.5 β£ 2 1.1 β9.3 β1.6 3.1 3 β8 0 2 β£ 8. β£ 12. β£ 16. β£ 20. β£ 24. β£ β8 4 β1 5 6 β3 4 8 β£ β£ β1 4 0 0 2 3 0 0 β3 6 β1 2 β4 β3 5 1 9 β For the following exercises, solve the system of linear equations using Cramerβs Rule. 25. 2x β 3y = β1 4x + 5y = 9 |
29. 4x + 3y = 23 2x β y = β1 26. 5x β 4y = 2 β4x + 7y = 6 27. 6x β 3y = 2 β8x + 9y = β1 28. 2x + 6y = 12 5x β 2y = 13 30. 10x β 6y = 2 β5x + 8y = β1 31. 4x β 3y = β3 2x + 6y = β4 32. 4x β 5y = 7 β3x + 9y = 0 33. 4x + 10y = 180 β3x β 5y = β105 34. 8x β 2y = β3 β4x + 6y = 4 For the following exercises, solve the system of linear equations using Cramerβs Rule. 35. x + 2y β 4z = β 1 7x + 3y + 5z = 26 β2x β 6y + 7z = β 6 36. β5x + 2y β 4z = β 47 4x β 3y β z = β 94 3x β 3y + 2z = 94 37. 4x + 5y β z = β7 β2x β 9y + 2z = 8 5y + 7z = 21 38. 4x β 3y + 4z = 10 5x β 2z = β 2 3x + 2y β 5z = β 9 39. 4x β 2y + 3z = 6 40. 5x + 2y β z = 1 β 6x + y = β 2 2x + 7y + 8z = 24 β7x β 8y + 3z = 1.5 6x β 12y + z = 7 41. 13x β 17y + 16z = 73 β11x + 15y + 17z = 61 46x + 10y β 30z = β 18 42. β4x β 3y β 8z = β 7 2x β 9y + 5z = 0.5 5x β 6y β 5z = β 2 852 CHAPTER 9 systems oF eQuations and ineQualities 43. 4x β 6y + 8z = 10 β2x + 3y β 4z = β 5 x + y + z = 1 44. 4x β 6y + 8z = 10 β2x + 3y β 4z = β 5 12x |
+ 18y β 24z = β 30 TeCHnOlOGY For the following exercises, use the determinant function on a graphing utility. 45 β£ β£ 469 1 3 β1 β2 0 β2 1 1 β£ 47. 1 _ 2 100,000 0 0 0 2 48 β£ β£ β£ β£ ReAl-WORlD APPlICATIOnS For the following exercises, create a system of linear equations to describe the behavior. Then, calculate the determinant. Will there be a unique solution? If so, find the unique solution. 49. Two numbers add up to 56. One number is 20 less 50. Two numbers add up to 104. If you add two times than the other. the first number plus two times the second number, your total is 208 51. Three numbers add up to 106. The first number is 3 less than the second number. The third number is 4 more than the first number. 52. Three numbers add to 216. The sum of the first two numbers is 112. The third number is 8 less than the first two numbers combined. For the following exercises, create a system of linear equations to describe the behavior. Then, solve the system for all solutions using Cramerβs Rule. 53. You invest $10,000 into two accounts, which receive 8% interest and 5% interest. At the end of a year, you had $10,710 in your combined accounts. How much was invested in each account? 55. A movie theater needs to know how many adult tickets and children tickets were sold out of the 1,200 total tickets. If childrenβs tickets are $5.95, adult tickets are $11.15, and the total amount of revenue was $12,756, how many childrenβs tickets and adult tickets were sold? 57. You decide to paint your kitchen green. You create the color of paint by mixing yellow and blue paints. You cannot remember how many gallons of each color went into your mix, but you know there were 10 gal total. Additionally, you kept your receipt, and know the total amount spent was $29.50. If each gallon of yellow costs $2.59, and each gallon of blue costs $3.19, how many gallons of each color go into your green mix? 54. You invest $80,000 into two accounts, $22,000 in one account, and $58,000 in the other account. |
At the end of one year, assuming simple interest, you have earned $2,470 in interest. The second account receives half a percent less than twice the interest on the first account. What are the interest rates for your accounts? 56. A concert venue sells single tickets for $40 each and coupleβs tickets for $65. If the total revenue was $18,090 and the 321 tickets were sold, how many single tickets and how many coupleβs tickets were sold? 58. You sold two types of scarves at a farmersβ market and would like to know which one was more popular. The total number of scarves sold was 56, the yellow scarf cost $10, and the purple scarf cost $11. If you had total revenue of $583, how many yellow scarves and how many purple scarves were sold? SECTION 9.8 section exercises 853 59. Your garden produced two types of tomatoes, one green and one red. The red weigh 10 oz, and the green weigh 4 oz. You have 30 tomatoes, and a total weight of 13 lb, 14 oz. How many of each type of tomato do you have? 60. At a market, the three most popular vegetables make up 53% of vegetable sales. Corn has 4% higher sales than broccoli, which has 5% more sales than onions. What percentage does each vegetable have in the market share? 61. At the same market, the three most popular fruits 62. Three bands performed at a concert venue. The make up 37% of the total fruit sold. Strawberries sell twice as much as oranges, and kiwis sell one more percentage point than oranges. For each fruit, find the percentage of total fruit sold. first band charged $15 per ticket, the second band charged $45 per ticket, and the final band charged $22 per ticket. There were 510 tickets sold, for a total of $12,700. If the first band had 40 more audience members than the second band, how many tickets were sold for each band? 63. A movie theatre sold tickets to three movies. The 64. Men aged 20β29, 30β39, and 40β49 made up 78% tickets to the first movie were $5, the tickets to the second movie were $11, and the third movie was $12. 100 tickets were sold to the first movie. The total number of tickets sold was 642, for a total revenue of $6,774. How many tickets for each movie |
were sold? of the population at a prison last year. This year, the same age groups made up 82.08% of the population. The 20β29 age group increased by 20%, the 30β39 age group increased by 2%, and the 40β49 age group decreased to 3 __ of their previous population. 4 Originally, the 30β39 age group had 2% more prisoners than the 20β29 age group. Determine the prison population percentage for each age group last year. 65. At a womenβs prison down the road, the total number of inmates aged 20β49 totaled 5,525. This year, the 20β29 age group increased by 10%, the 30β39 age group decreased by 20%, and the 40β49 age group doubled. There are now 6,040 prisoners. Originally, there were 500 more in the 30β39 age group than the 20β29 age group. Determine the prison population for each age group last year. For the following exercises, use this scenario: A health-conscious company decides to make a trail mix out of almonds, dried cranberries, and chocolate-covered cashews. The nutritional information for these items is shown in Table 1. Almonds (10) Cranberries (10) Cashews (10) Fat (g) 6 0.02 7 Protein (g) 2 0 3.5 Carbohydrates (g) 3 8 5.5 Table 1 67. For the βhikingβ mix, there are 1,000 pieces in the mix, containing 390.8 g of fat, and 165 g of protein. If there is the same amount of almonds as cashews, how many of each item is in the trail mix? 66. For the special βlow-carbβ trail mix, there are 1,000 pieces of mix. The total number of carbohydrates is 425 g, and the total amount of fat is 570.2 g. If there are 200 more pieces of cashews than cranberries, how many of each item is in the trail mix? 68. For the βenergy-boosterβ mix, there are 1,000 pieces in the mix, containing 145 g of protein and 625 g of carbohydrates. If the number of almonds and cashews summed together is equivalent to the amount of cranberries, how many of each item is in the trail mix? 854 CHAPTER 9 systems oF eQuations and ineQualities CHAPTeR 9 ReVIe |
W Key Terms addition method an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable augmented matrix a coefficient matrix adjoined with the constant column separated by a vertical line within the matrix brackets break-even point the point at which a cost function intersects a revenue function; where profit is zero coefficient matrix a matrix that contains only the coefficients from a system of equations column a set of numbers aligned vertically in a matrix consistent system a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system cost function the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable costs Cramerβs Rule a method for solving systems of equations that have the same number of equations as variables using determinants dependent system a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system determinant a number calculated using the entries of a square matrix that determines such information as whether there is a solution to a system of equations entry an element, coefficient, or constant in a matrix feasible region the solution to a system of nonlinear inequalities that is the region of the graph where the shaded regions of each inequality intersect Gaussian elimination using elementary row operations to obtain a matrix in row-echelon form identity matrix a square matrix containing ones down the main diagonal and zeros everywhere else; it acts as a 1 in matrix algebra inconsistent system a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common independent system a system of linear equations with exactly one solution pair (x, y) main diagonal entries from the upper left corner diagonally to the lower right corner of a square matrix matrix a rectangular array of numbers multiplicative inverse of a matrix a matrix that, when multiplied by the original, equals the identity matrix nonlinear inequality an inequality containing a nonlinear expression partial fraction decomposition the process of returning a simplified rational expression to its original form, a sum or difference of simpler rational expressions partial fractions the individual fractions that make up the sum or difference of a rational expression before combining them into a simplified rational expression profit function the profit function is written as P(x) = R(x) β C( |
x), revenue minus cost revenue function the function that is used to calculate revenue, simply written as R = xp, where x = quantity and p = price row a set of numbers aligned horizontally in a matrix row operations adding one row to another row, multiplying a row by a constant, interchanging rows, and so on, with the goal of achieving row-echelon form row-echelon form after performing row operations, the matrix form that contains ones down the main diagonal and zeros at every space below the diagonal row-equivalent two matrices A and B are row-equivalent if one can be obtained from the other by performing basic row operations scalar multiple an entry of a matrix that has been multiplied by a scalar solution set the set of all ordered pairs or triples that satisfy all equations in a system of equations substitution method an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable CHAPTER 9 review 855 system of linear equations a set of two or more equations in two or more variables that must be considered simultaneously. system of nonlinear equations a system of equations containing at least one equation that is of degree larger than one system of nonlinear inequalities a system of two or more inequalities in two or more variables containing at least one inequality that is not linear Key equations Identity matrix for a 2 Γ 2 matrix Identity matrix for a 3 Γ 3 matrix Multiplicative inverse of a 2 Γ 2 matrix Key Concepts 9.1 Systems of Linear Equations: Two Variables 1 = d βb 1 ξ° _______ ad β bc βc a ξ², where ad β bc β 0 β’ A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously. β’ The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. See Example 1. β’ Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution. β’ One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes. See Example 2. β’ Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation. See Example 3. |
β’ A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See Example 4. β’ It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together. See Example 5, Example 6, and Example 7. β’ Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect. See Example 8. β’ The solution to a system of dependent equations will always be true because both equations describe the same line. See Example 9. β’ Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit. See Example 10 and Example 11. 9.2 Systems of Linear Equations: Three Variables β’ A solution set is an ordered triple {(x, y, z)} that represents the intersection of three planes in space. See Example 1. β’ A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. See Example 2. β’ Systems of three equations in three variables are useful for solving many different types of real-world problems. See Example 3. β’ A system of equations in three variables is inconsistent if no solution exists. After performing elimination operations, the result is a contradiction. See Example 4. β’ Systems of equations in three variables that are inconsistent could result from three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. 856 CHAPTER 9 systems oF eQuations and ineQualities β’ A system of equations in three variables is dependent if it has an infinite number of solutions. After performing elimination operations, the result is an identity. See Example 5. β’ Systems of equations in three variables that are dependent could result from three identical planes, three planes intersecting at a line, or two identical planes that intersect the third on a line. 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables β’ There are three possible types of solutions to a system of equations representing a line and a parabola: (1) no solution, the line does not intersect the par |
abola; (2) one solution, the line is tangent to the parabola; and (3) two solutions, the line intersects the parabola in two points. See Example 1. β’ There are three possible types of solutions to a system of equations representing a circle and a line: (1) no solution, the line does not intersect the circle; (2) one solution, the line is tangent to the parabola; (3) two solutions, the line intersects the circle in two points. See Example 2. β’ There are five possible types of solutions to the system of nonlinear equations representing an ellipse and a circle: (1) no solution, the circle and the ellipse do not intersect; (2) one solution, the circle and the ellipse are tangent to each other; (3) two solutions, the circle and the ellipse intersect in two points; (4) three solutions, the circle and ellipse intersect in three places; (5) four solutions, the circle and the ellipse intersect in four points. See Example 3. β’ An inequality is graphed in much the same way as an equation, except for > or <, we draw a dashed line and shade the region containing the solution set. See Example 4. β’ Inequalities are solved the same way as equalities, but solutions to systems of inequalities must satisfy both inequalities. See Example 5. 9.4 Partial Fractions P(x) ____ Q(x) β’ Decompose by writing the partial fractions as A ________ + a 1 x + b 1 the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. See Example 1.. Solve by clearing the fractions, expanding B ________ a 2 x + b 2 β’ The decomposition of with repeated linear factors must account for the factors of the denominator in P(x) ____ Q(x) increasing powers. See Example 2. with a nonrepeated irreducible quadratic factor needs a linear numerator over the β’ The decomposition of P(x) ____ Q(x) quadratic factor, as in A __ x +. See Example 3. Bx + C ____________ (a x 2 + bx + c) P(x) ____ Q(x) β’ In the decomposition of, where Q(x) has |
a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers as Ax + B __ + (a x 2 + bx + c) A 2 x + B 2 __ (a x 2 + bx + c __ n (a x 2 + bx + c). See Example 4. 9.5 Matrices and Matrix Operations β’ A matrix is a rectangular array of numbers. Entries are arranged in rows and columns. β’ The dimensions of a matrix refer to the number of rows and the number of columns. A 3 Γ 2 matrix has three rows and two columns. See Example 1. β’ We add and subtract matrices of equal dimensions by adding and subtracting corresponding entries of each matrix. See Example 2, Example 3, Example 4, and Example 5. β’ Scalar multiplication involves multiplying each entry in a matrix by a constant. See Example 6. β’ Scalar multiplication is often required before addition or subtraction can occur. See Example 7. β’ Multiplying matrices is possible when inner dimensions are the sameβthe number of columns in the first matrix must match the number of rows in the second. β’ The product of two matrices, A and B, is obtained by multiplying each entry in row 1 of A by each entry in column 1 of B; then multiply each entry of row 1 of A by each entry in columns 2 of B, and so on. See Example 8 and Example 9. CHAPTER 9 review 857 β’ Many real-world problems can often be solved using matrices. See Example 10. β’ We can use a calculator to perform matrix operations after saving each matrix as a matrix variable. See Example 11. 9.6 Solving Systems with Gaussian Elimination β’ An augmented matrix is one that contains the coefficients and constants of a system of equations. See Example 1. β’ A matrix augmented with the constant column can be represented as the original system of equations. See Example 2. β’ Row operations include multiplying a row by a constant, adding one row to another row, and interchanging rows. β’ We can use Gaussian elimination to solve a system of equations. See Example 3, Example 4, and Example 5. β’ Row operations are performed on matrices to obtain row-echelon form. See Example 6. β’ To solve a system of equations, write it in augmented matrix form. Perform row operations to obtain row-echelon form. |
Back-substitute to find the solutions. See Example 7 and Example 8. β’ A calculator can be used to solve systems of equations using matrices. See Example 9. β’ Many real-world problems can be solved using augmented matrices. See Example 10 and Example 11. 9.7 Solving Systems with Inverses β’ An identity matrix has the property AI = IA = A. See Example 1. β’ An invertible matrix has the property A A β1 = A β1 A = I. See Example 2. β’ Use matrix multiplication and the identity to find the inverse of a 2 Γ 2 matrix. See Example 3. β’ The multiplicative inverse can be found using a formula. See Example 4. β’ Another method of finding the inverse is by augmenting with the identity. See Example 5. β’ We can augment a 3 Γ 3 matrix with the identity on the right and use row operations to turn the original matrix into the identity, and the matrix on the right becomes the inverse. See Example 6. β’ Write the system of equations as AX = B, and multiply both sides by the inverse of A: A β1 AX = A β1 B. See Example 7 and Example 8. β’ We can also use a calculator to solve a system of equations with matrix inverses. See Example 9. 9.8 Solving Systems with Cramer's Rule ξ² is ad β bc. See Example 1. β’ The determinant for ξ° β’ Cramerβs Rule replaces a variable column with the constant column. Solutions are =. See Example 2. β’ To find the determinant of a 3 Γ 3 matrix, augment with the first two columns. Add the three diagonal entries (upper left to lower right) and subtract the three diagonal entries (lower left to upper right). See Example 3. β’ To solve a system of three equations in three variables using Cramerβs Rule, replace a variable column with the constant D z _ D β’ Cramerβs Rule is also useful for finding the solution of a system of equations with no solution or infinite solutions. See column for each desired solution: x =. See Example 4 = Example 5 and Example 6. β’ Certain properties of determinants are useful for solving problems. For example: β If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. β When two rows are interchanged, the determinant changes sign. β If either two rows or two |
columns are identical, the determinant equals zero. β If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. β The determinant of an inverse matrix A β1 is the reciprocal of the determinant of the matrix A. β If any row or column is multiplied by a constant, the determinant is multiplied by the same factor. See Example 7 and Example 8. 858 CHAPTER 9 systems oF eQuations and ineQualities CHAPTeR 9 ReVIeW exeRCISeS SYSTeMS OF lIneAR eQUATIOnS: TWO VARIABleS For the following exercises, determine whether the ordered pair is a solution to the system of equations. 1. 3x β y = 4 x + 4y = β 3 and ( β 1, 1) 2. 6x β 2y = 24 β3x + 3y = 18 and (9, 15) For the following exercises, use substitution to solve the system of equations. 3. 10x + 5y = β5 3x β 2y = β12 4. 4 __ 7 5 __ 6 y = 43 ___ x + 1 __ 5 70 y = β 2 __ x β 1 __ 3 3 5. 5x + 6y = 14 4x + 8y = 8 For the following exercises, use addition to solve the system of equations. 6. 3x + 2y = β7 2x + 4y = 6 7. 3x + 4y = 2 9x + 12y = 3 8. 8x + 4y = 2 6x β 5y = 0.7 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 9. A factory has a cost of production C(x) = 150x + 15,000 and a revenue function R(x) = 200x. What is the break-even point? 10. A performer charges C(x) = 50x + 10,000, where x is the total number of attendees at a show. The venue charges $75 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point? SYSTeMS OF lIneAR eQUATIOnS: THRee VARIABleS For the following exercises, solve the system of three equations using substitution or addition |
. 11. 0.5x β 0.5y = 10 β 0.2y + 0.2x = 4 0.1x + 0.1z = 2 14. 2x β 3y + z = β1 x + y + z = β4 4x + 2y β 3z = 33 17. 2x β 3y + z = 0 2x + 4y β 3z = 0 6x β 2y β z = 0 12. 5x + 3y β z = 5 3x β 2y + 4z = 13 4x + 3y + 5z = 22 15. 3x + 2y β z = β10 x β y + 2z = 7 βx + 3y + z = β2 18. 6x β 4y β 2z = 2 3x + 2y β 5z = 4 6y β 7z = 5 13. x + y + z = 1 2x + 2y + 2z = 1 3x + 3y = 2 16. 3x + 4z = β11 x β 2y = 5 4y β z = β10 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 19. Three odd numbers sum up to 61. The smaller is 20. A local theatre sells out for their show. They sell all one-third the larger and the middle number is 16 less than the larger. What are the three numbers? 500 tickets for a total purse of $8,070.00. The tickets were priced at $15 for students, $12 for children, and $18 for adults. If the band sold three times as many adult tickets as childrenβs tickets, how many of each type was sold? CHAPTER 9 review 859 SYSTeMS OF nOnlIneAR eQUATIOnS AnD IneQUAlITIeS: TWO VARIABleS For the following exercises, solve the system of nonlinear equations. 21. y = x 2 β 7 y = 5x β 13 22. y = x 2 β 4 y = 5x + 10 23. x 2 + y 2 = 16 24. x 2 + y 2 = 25 25 For the following exercises, graph the inequality. 26. y > x 2 β 1 27. 1 __ 4 x 2 + y 2 < 4 For the following exercises, graph the system of inequalities. 28. x 2 |
+ y 2 + 2x < 3 y > β x 2 β 3 29. x 2 β 2x + y 2 β 4x < 4 y < β x + 4 30 PARTIAl FRACTIOnS For the following exercises, decompose into partial fractions. 31. β2x + 6 __________ x 2 + 3x + 2 34. x β 18 ____________ x 2 β 12x + 36 32. 10x + 2 ___________ 4 x 2 + 4x + 1 35. β x 2 + 36x + 70 _____________ x 3 β 125 33. 7x + 20 ____________ x 2 + 10x + 25 36. β5 x 2 + 6x β 2 ____________ x 3 + 27 37. x 3 β 4 x 2 + 3x + 11 ________________ ( x 2 β 2) 2 38. 4 x 4 β 2 x 3 + 22 x 2 β 6x + 48 _______________________ x( x 2 + 4) 2 MATRICeS AnD MATRIx OPeRATIOnS For the following exercises, perform the requested operations on the given matrices. A = ξ° 4 β3 11 β2 4 ξ², C = ξ° 6 7 11 β2 14 0 ξ², D = ξ° 1 β4 9 10 5 β14 3 2 β1 3 0 1 9 ξ² 39. β4A 45. CB 40. 10D β 6E 46. DE 41. B + C 47. ED 42. AB 48. EC 43. BA 49. CE 44. BC 50. A 3 SOlVInG SYSTeMS WITH GAUSSIAn elIMInATIOn For the following exercises, write the system of linear equations from the augmented matrix. Indicate whether there will be a unique solution. 51. ξ° 1 0 β3 7 ξ² β5 0 1 2 0 0 0 0 β£ 52. ξ° β9 1 0 5 0 1 β2 4 3 0 0 0 ξ² β£ For the following exercises, write the augmented matrix from the system of linear equations. 53. β2x + 2y + z = 7 2x β 8y + 5z = 0 19x β 10y + 22z = 3 54. 4x + 2y β 3z = 14 β12x + 3y + z = 100 9x β 6y + 2z |
= 31 55. x + 3z = 12 βx + 4y = 0 y + 2z = β 7 860 CHAPTER 9 systems oF eQuations and ineQualities For the following exercises, solve the system of linear equations using Gaussian elimination. 56. 3x β 4y = β 7 β6x + 8y = 14 57. 3x β 4y = 1 β6x + 8y = 6 58. β1.1x β 2.3y = 6.2 β5.2x β 4.1y = 4.3 59. 2x + 3y + 2z = 1 β4x β 6y β 4z = β 2 10x + 15y + 10z = 0 60. βx + 2y β 4z = 8 3y + 8z = β 4 β7x + y + 2z = 1 SOlVInG SYSTeMS WITH InVeRSeS For the following exercises, find the inverse of the matrix. 61. ξ° β0.2 1.4 ξ² 1.2 β0.4 62. ξ° 1 __ 2 β 1 __ 4 β 1 __ 2 3 __ 4 ξ² 63. ξ° 12 9 β6 β1 3 2 β4 β3 2 ξ² 64 ξ² For the following exercises, find the solutions by computing the inverse of the matrix. 65. 0.3x β 0.1y = β10 β0.1x + 0.3y = 14 66. 0.4x β 0.2y = β0.6 β0.1x + 0.05y = 0.3 67. 4x + 3y β 3z = β4.3 5x β 4y β z = β6.1 x + z = β0.7 68. β2x β 3y + 2z = 3 βx + 2y + 4z = β5 β2y + 5z = β3 For the following exercises, write a system of equations to solve each problem. Solve the system of equations. 69. Students were asked to bring their favorite fruit to class. 90% of the fruits consisted of banana, apple, and oranges. If oranges were half as popular as bananas and apples were 5% more popular than bananas, what are the percentages of each individual fruit? 70. A sorority held a bake sale to raise money and sold |
brownies and chocolate chip cookies. They priced the brownies at $2 and the chocolate chip cookies at $1. They raised $250 and sold 175 items. How many brownies and how many cookies were sold? SOlVInG SYSTeMS WITH CRAMeR'S RUle For the following exercises, find the determinant. 71. ξ° 100 0 ξ² 0 0 72. ξ° 0.2 β0.6 0.7 β1.1 ξ² 73. ξ° β1 4 3 0 2 3 0 0 β3 ξ² β 74 β For the following exercises, use Cramerβs Rule to solve the linear systems of equations. 75. 4x β 2y = 23 β5x β 10y = β35 76. 0.2x β 0.1y = 0 β0.3x + 0.3y = 2.5 77. β0.5x + 0.1y = 0.3 β0.25x + 0.05y = 0.15 78. x + 6y + 3z = 4 2x + y + 2z = 3 3x β 2y + z = 0 79. 4x β 3y + 5z = β 5 __ 2 7x β 9y β 3z = 3 __ 2 x β 5y β 5z = 5 __ 2 80. 3 ___ 10 1 ___ 10 2 __ 5 x β 1 __ 5 x β 1 ___ 10 x β 1 __ 2 y β 3 ___ 10 y β 1 __ 2 y β 3 __ 5 z = β 1 ___ 50 z = β 9 ___ 50 z = β 1 __ 5 CHAPTER 9 practice test 861 CHAPTeR 9 PRACTICe TeST Is the following ordered pair a solution to the system of equations? 1. β5x β y = 12 x + 4y = 9 with ( β 3, 3) For the following exercises, solve the systems of linear and nonlinear equations using substitution or elimination. Indicate if no solution exists. y = 4 x β 1 __ 3 2. 1 __ 2 3 __ x β y = 0 2 3. β 1 __ 2 x β 4y = 4 2x + 16y = 2 4. 5x β y = 1 β10x + 2y = β 2 6. x + z = 20 x + y + z = 20 x + 2y + z = 10 7. |
5x β 4y β 3z = 0 2x + y + 2z = 0 x β 6y β 7z = 0 8. y = x 2 + 2x β 3 y = x β 1 For the following exercises, graph the following inequalities. 10. y < x 2 + 9 11. 4x β 6y β 2z = 1 ___ 10 x β 7y + 5z = β 1 __ 4 3x + 6y β 9z = 6 __ 5 9. y 2 + x 2 = 25 y 2 β 2 x 2 = 1 For the following exercises, write the partial fraction decomposition. 12. β8x β 30 ____________ x 2 + 10x + 25 13. 13x + 2 ________ (3x + 1) 2 14. x 4 β x 3 + 2x β 1 ______________ x ( x 2 + 1) 2 For the following exercises, perform the given matrix operations. 15. 5 ξ° 4 9 β2 3 β6 12 ξ² + 1 __ ξ² ξ° 2 4 β8 18. det β£ 0 0 400 4,000 β£ 16. ξ° 1 4 β7 β2 9 5 12 0 β4 ξ² ξ° 3 β4 1 3 5 10 ξ² 17. ξ° 1 __ 2 1 __ 4 1 __ 2 β 1 __ 2 0 19. det β£ β 1 __ 2 0 1 __ 2 0 1 __ 2 0 β£ β1 1 __ 3 1 __ 5 ξ² 20. If det(A) = β6, what would be the determinant if you switched rows 1 and 3, multiplied the second row by 12, and took the inverse? 21. Rewrite the system of linear equations as an augmented matrix. 14x β 2y + 13z = 140 β2x + 3y β 6z = β1 x β 5y + 12z = 11 22. Rewrite the augmented matrix as a system of linear equations. 1 0 3 β2 4 9 β6 1 2 ξ° 12 ξ² β5 8 β£ 862 CHAPTER 9 systems oF eQuations and ineQualities For the following exercises, use Gaussian elimination to solve the systems of equations. 23. x β 6y = 4 2x β 12y = 0 24. 2x + y + z = β3 x β 2y + 3z = 6 x β y β |
z = 6 For the following exercises, use the inverse of a matrix to solve the systems of equations. 25. 4x β 5y = β50 βx + 2y = 80 z = β49 26. 1 ___ 100 3 ___ 100 9 ___ 100 x β 3 ___ 100 x β 7 ___ 100 x β 9 ___ 100 y + 1 ___ 20 y β 1 ___ 100 y β 9 ___ 100 z = 13 z = 99 For the following exercises, use Cramerβs Rule to solve the systems of equations. 27. 200x β 300y = 2 400x + 715y = 4 28. 0.1x + 0.1y β 0.1z = β1.2 0.1x β 0.2y + 0.4z = β1.2 0.5x β 0.3y + 0.8z = β5.9 For the following exercises, solve using a system of linear equations. 29. A factory producing cell phones has the following 30. A small fair charges $1.50 for students, $1 for cost and revenue functions: C(x) = x 2 + 75x + 2,688 and R(x) = x 2 + 160x. What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit. children, and $2 for adults. In one day, three times as many children as adults attended. A total of 800 tickets were sold for a total revenue of $1,050. How many of each type of ticket was sold? Analytic Geometry 10 Figure 1 (a) Greek philosopher Aristotle (384β322 BCe) (b) German mathematician and astronomer johannes Kepler (1571β1630) a b CHAPTeR OUTlIne 10.1 The ellipse 10.2 The Hyperbola 10.3 The Parabola 10.4 Rotation of Axes 10.5 Conic Sections in Polar Coordinates Introduction The Greek mathematician Menaechmus (c. 380βc. 320 BCE) is generally credited with discovering the shapes formed by the intersection of a plane and a right circular cone. Depending on how he tilted the plane when it intersected the cone, he formed different shapes at the intersectionβbeautiful shapes with near-perfect symmetry. It was also said that Aristotle may have had an intuitive understanding of these shapes, as he observed the orbit of the planet to be |
circular. He presumed that the planets moved in circular orbits around Earth, and for nearly 2,000 years this was the commonly held belief. It was not until the Renaissance movement that Johannes Kepler noticed that the orbits of the planet were not circular in nature. His published law of planetary motion in the 1600s changed our view of the solar system forever. He claimed that the sun was at one end of the orbits, and the planets revolved around the sun in an oval-shaped path. In this chapter, we will investigate the two-dimensional figures that are formed when a right circular cone is intersected by a plane. We will begin by studying each of three figures created in this manner. We will develop defining equations for each figure and then learn how to use these equations to solve a variety of problems. 863 864 CHAPTER 10 analytic geometry leARnInG OBjeCTIVeS In this section, you will: β’ Write equations of ellipses in standard form. β’ Graph ellipses centered at the origin. β’ Graph ellipses not centered at the origin. β’ Solve applied problems involving ellipses. 10.1 THe ellIPSe Figure 1 The national Statuary Hall in Washington, D.C. (credit: Greg Palmer, Flickr) Can you imagine standing at one end of a large room and still being able to hear a whisper from a person standing at the other end? The National Statuary Hall in Washington, D.C., shown in Figure 1, is such a room.[33] It is an ovalshaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can stand and still hear each other whisper. Writing equations of ellipses in Standard Form A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown in Figure 2. Ellipse Hyperbola Figure 2 Parabola Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form |
of the 33. Architect of the Capitol. http://www.aoc.gov. Accessed April 15, 2014. SECTION 10.1 the ellipse 865 equation for the ellipse. An ellipse is the set of all points (x, y) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci). We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. See Figure 3. Foci Figure 3 Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. See Figure 4. y Co-vertex Minor Axis Vertex Focus Major Axis Center Focus Vertex x Co-vertex Figure 4 In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the x- and y-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane. To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs. |
866 CHAPTER 10 analytic geometry Deriving the Equation of an Ellipse Centered at the Origin To derive the equation of an ellipse centered at the origin, we begin with the foci (βc, 0) and (c, 0). The ellipse is the set of all points (x, y) such that the sum of the distances from (x, y) to the foci is constant, as shown in Figure 5. (x, y) d1 y d2 (βc, 0) (c, 0) (a, 0) x (βa, 0) Figure 5 If (a, 0) is a vertex of the ellipse, the distance from (βc, 0) to (a, 0) is a β ( βc) = a + c. The distance from (c, 0) to (a, 0) is a β c. The sum of the distances from the foci to the vertex is (a + c) + (a β c) = 2a If (x, y) is a point on the ellipse, then we can define the following variables: d1 = the distance from (βc, 0) to (x, y) d2 = the distance from (c, 0) to (x, y) By the definition of an ellipse, d1 + d2 is constant for any point (x, y) on the ellipse. We know that the sum of these distances is 2a for the vertex (a, 0). It follows that d1 + d2 = 2a for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic. d1 + d2 = β ββ (x β ( β c))2 + (y β 0)2 + β β (x β c)2 + (y β 0)2 = 2a Distance formula β (x + c)2 + y 2 + β β (x β c)2 + y 2 = 2a β β (x + c)2 + y 2 = 2a β β β (x β c)2 + y 2 β (x + c)2 + y 2 = [2a β β x 2 + 2cx + c2 + y 2 = 4a2 β 4a β β (x β c)2 |
+ y 2 ] β 2 x 2 + 2cx + c2 + y 2 = 4a2 β 4a β (x β c)2 + y 2 + (x β c)2 + y 2 (x β c)2 + y 2 + x 2 β 2cx + c2 + y 2 β 2cx = 4a2 β 4a β (x β c)2 + y 2 β 2cx β 4cx β 4a2 = β4a β β β (x β c)2 + y 2 (x β c)2 + y 2 2 (x β c)2 + y 2 ξ² cx β a2 = β a β = a2 ξ° β c2 x 2 β 2a2 cx + a4 = a2 ξ’ x 2 β 2cx + c2 + y 2 ξͺ 2 ξ° cx β a2 ξ² β c2 x 2 β 2a2 cx + a4 = a2 x 2 β 2a2 cx + a2 c2 + a2 y 2 a2 x 2 β c2 x 2 + a2 y 2 = a4 β a2 c2 x 2(a2 β c2) + a2 y 2 = a2(a2 β c2) x 2 b2 + a2 y 2 = a2 b2 a2b2 ____ a2b2 x 2b2 a2y 2 ____ ____ a2b2 + a2b2 = y 2 x 2 __ __ b2 = 1 a2 + Simplify expressions. Move radical to opposite side. Square both sides. Expand the squares. Expand remaining squares. Combine like terms. Isolate the radical. Divide by 4. Square both sides. Expand the squares. Distribute a2. Rewrite. Factor common terms. Set b2 = a2 β c2. Divide both sides by a2b2. Simplify. Thus, the standard equation of an ellipse is y 2 x 2 _ __ b2 = 1. This equation defines an ellipse centered at the origin. a2 + If a > b, the ellipse is stretched further in the horizontal direction, and if b > a, the ellipse is stretched further in the vertical direction. SECTION 10.1 the ellipse 867 Writing Equations of Ellipses Centered at the Origin |
in Standard Form Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations weβve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the relationship between algebraic and geometric representations of mathematical phenomena. The key features of the ellipse are its center, vertices, co-vertices, foci, and lengths and positions of the major and minor axes. Just as with other equations, we can identify all of these features just by looking at the standard form of the equation. There are four variations of the standard form of the ellipse. These variations are categorized first by the location of the center (the origin or not the origin), and then by the position (horizontal or vertical). Each is presented along with a description of how the parts of the equation relate to the graph. Interpreting these parts allows us to form a mental picture of the ellipse. standard forms of the equation of an ellipse with center (0, 0) The standard form of the equation of an ellipse with center (0, 0) and major axis on the x-axis is where y 2 x 2 _ _ b2 = 1 a2 + β’ a > b β’ the length of the major axis is 2a β’ the coordinates of the vertices are (Β±a, 0) β’ the length of the minor axis is 2b β’ the coordinates of the co-vertices are (0, Β±b) β’ the coordinates of the foci are (Β±c, 0), where c2 = a2 β b2. See Figure 6a. The standard form of the equation of an ellipse with center (0, 0) and major axis on the y-axis is where y 2 x 2 _ _ a2 = 1 b2 + β’ a > b β’ the length of the major axis is 2a β’ the coordinates of the vertices are (0, Β± a) β’ the length of the minor axis is 2b β’ the coordinates of the co-vertices are (Β±b, 0) β’ the coordinates of the foci are (0, Β± c), where c2 = a2 β b2. See Figure 6b. Note that the vertices, co-vertices, and foci are related by the equation c2 = a2 |
β b2. When we are given the coordinates of the foci and vertices of an ellipse, we can use this relationship to find the equation of the ellipse in standard form. y (0, b) Minor Axis (βa, 0) (βc, 0) Major Axis (0, 0) (c, 0) (a, 0) x (βb, 0) Minor Axis (0, βb) (a) Figure 6 (a) Horizontal ellipse with center (0, 0) (b) Vertical ellipse with center (0, 0) y (0, a) (0, c) Major Axis (0, 0) (0, βc) (0, βa) (b) (b, 0) x 868 CHAPTER 10 analytic geometry How Toβ¦ Given the vertices and foci of an ellipse centered at the origin, write its equation in standard form. 1. Determine whether the major axis lies on the x- or y-axis. a. If the given coordinates of the vertices and foci have the form (Β±a, 0) and ( Β±c, 0) respectively, then the major axis is the x-axis. Use the standard form axis is the y-axis. Use the standard form y 2 x 2 _ _ b2 = 1. a2 + y 2 x 2 _ _ a2 = 1. b2 + b. If the given coordinates of the vertices and foci have the form (0, Β±a) and ( Β±c, 0), respectively, then the major 2. Use the equation c2 = a2 β b2, along with the given coordinates of the vertices and foci, to solve for b2. 3. Substitute the values for a2 and b2 into the standard form of the equation determined in Step 1. Example 1 Writing the Equation of an Ellipse Centered at the Origin in Standard Form What is the standard form equation of the ellipse that has vertices (Β±8, 0) and foci (Β±5, 0)? Solution The foci are on the x-axis, so the major axis is the x-axis. Thus, the equation will have the form y 2 x 2 _ __ b2 = 1 a2 + The vertices are (Β±8, 0), so a = 8 and a2 = 64. The foci are (Β±5, 0), |
so c = 5 and c2 = 25. We know that the vertices and foci are related by the equation c 2 = a 2 β b 2. Solving for b 2, we have: c 2 = a 2 β b2 25 = 64 β b2 b 2 = 39 Substitute for c 2 and a 2. Solve for b2. Now we need only substitute a2 = 64 and b2 = 39 into the standard form of the equation. The equation of the ellipse is + = 1. x 2 _ 64 y 2 _ 39 Try It #1 What is the standard form equation of the ellipse that has vertices (0, Β± 4) and foci (0, Β± β β 15 )? Q & Aβ¦ Can we write the equation of an ellipse centered at the origin given coordinates of just one focus and vertex? Yes. Ellipses are symmetrical, so the coordinates of the vertices of an ellipse centered around the origin will always have the form (Β±a, 0) or (0, Β± a). Similarly, the coordinates of the foci will always have the form (Β±c, 0) or (0, Β± c). Knowing this, we can use a and c from the given points, along with the equation c2 = a2 β b2, to find b2. Writing Equations of Ellipses Not Centered at the Origin Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated h units horizontally and k units vertically, the center of the ellipse will be (h, k). This translation results in the standard form of the equation we saw previously, with x replaced by (x β h) and y replaced by (y β k). SECTION 10.1 the ellipse 869 standard forms of the equation of an ellipse with center (h, k) The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the x-axis is where (x β h)2 _ a2 + (y β k)2 _ b2 = 1 β’ a > b β’ the length of the major axis is 2a β’ the coordinates of the vertices are (h Β± a, k) β’ the length of the minor axis is 2b β’ the coordinates of the co-vertices are (h, k Β± b) β’ the coordinates |
of the foci are (h Β± c, k), where c2 = a2 β b2. See Figure 7a. The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the y-axis is where (x β h)2 _ b2 + (y β k)2 _ a2 = 1 β’ a > b β’ the length of the major axis is 2a β’ the coordinates of the vertices are (h, k Β± a) β’ the length of the minor axis is 2b β’ the coordinates of the co-vertices are (h Β± b, k) β’ the coordinates of the foci are (h, k Β± c), where c2 = a2 β b2. See Figure 7b. Just as with ellipses centered at the origin, ellipses that are centered at a point (h, k) have vertices, co-vertices, and foci that are related by the equation c2 = a2 β b2. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given. y (h, k + b) y (h, k + a) (h β a, k) Minor Axis (h + c1, k) (h β c1, k) (h, k) Major Axis (h, k β b) (a) (h + a, k) x (h β b, k) Major Axis (h, k) Minor Axis (h, k + c) (h + b, k) x (h, k β c) (h, k β a) (b) Figure 7 (a) Horizontal ellipse with center (h, k) (b) Vertical ellipse with center (h, k) How Toβ¦ Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form. 1. Determine whether the major axis is parallel to the x- or y-axis. a. If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the x-axis. Use the standard form (x β h)2 _ a2 + (y β k)2 _ b2 = 1. b. If the x-coordinates of the given |
vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form (x β h)2 _ b2 + (y β k)2 _ a2 = 1. 870 CHAPTER 10 analytic geometry 2. Identify the center of the ellipse (h, k) using the midpoint formula and the given coordinates for the vertices. 3. Find a2 by solving for the length of the major axis, 2a, which is the distance between the given vertices. 4. Find c2 using h and k, found in Step 2, along with the given coordinates for the foci. 5. Solve for b2 using the equation c2 = a2 β b2. 6. Substitute the values for h, k, a2, and b2 into the standard form of the equation determined in Step 1. Example 2 Writing the Equation of an Ellipse Centered at a Point Other Than the Origin What is the standard form equation of the ellipse that has vertices (β2, β8) and (β2, 2) and foci (β2, β7) and (β2, 1)? Solution The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form (y β k)2 _ a2 First, we identify the center, (h, k). The center is halfway between the vertices, (β2, β 8) and (β2, 2). (x β h)2 _ b2 = 1 + Applying the midpoint formula, we have: (h, k) = ξ’ β2 + (β2) _________, 2 β8 + 2 ξͺ _______ 2 Next, we find a2. The length of the major axis, 2a, is bounded by the vertices. We solve for a by finding the distance between the y-coordinates of the vertices. = (β2, β3) 2a = 2 β (β8) 2a = 10 a = 5 So a2 = 25. Now we find c2. The foci are given by (h, k Β± c). So, (h, k β c) = (β2, β7) and (h, k + c) = (β2, 1). We substitute k = β3 using either |
of these points to solve for c. k + c = 1 β3 + c = 1 c = 4 So c2 = 16. Next, we solve for b2 using the equation c2 = a2 β b2. c2 = a2 β b2 16 = 25 β b2 b2 = 9 Finally, we substitute the values found for h, k, a2, and b2 into the standard form equation for an ellipse: (y + 3)2 _ 25 (x + 2)2 _ 9 = 1 + Try It #2 What is the standard form equation of the ellipse that has vertices (β3, 3) and (5, 3) and foci (1 β 2 β (1 + 2 β 3, 3)? β β 3, 3) and Graphing ellipses Centered at the Origin Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph a2 = 1, b2 + b2 = 1, a > b for horizontal ellipses and a2 + ellipses centered at the origin, we use the standard form a > b for vertical ellipses. SECTION 10.1 the ellipse 871 How Toβ¦ Given the standard form of an equation for an ellipse centered at (0, 0), sketch the graph. 1. Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci. a. If the equation is in the form y 2 x 2 _ _ b2 = 1, where a > b, then a2 + β’ the major axis is the x-axis β’ the coordinates of the vertices are (Β±a, 0) β’ the coordinates of the co-vertices are (0, Β± b) β’ the coordinates of the foci are (Β±c, 0) y 2 x 2 _ _ a2 = 1, where a > b, then b2 + b. If the equation is in the form β’ the major axis is the y-axis β’ the coordinates of the vertices are (0, Β± a) β’ the coordinates of the co-vertices are (Β±b, 0) β’ the coordinates of the foci are (0, Β± c) 2. Solve for c using the equation c2 = a2 β b2. 3. Plot the |
center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse. Example 3 Graphing an Ellipse Centered at the Origin x 2 _ 9 + y 2 _ 25 Graph the ellipse given by the equation, = 1. Identify and label the center, vertices, co-vertices, and foci. Solution First, we determine the position of the major axis. Because 25 > 9, the major axis is on the y-axis. Therefore, where b 2 = 9 and a2 = 25. It follows that: b 2 + the equation is in the form β’ the center of the ellipse is (0, 0) β’ the coordinates of the vertices are (0, Β± a) = (0, Β± β β’ the coordinates of the co-vertices are (Β±b, 0) = (Β± β β’ the coordinates of the foci are (0, Β± c), where c 2 = a 2 β b 2 Solving for c, we have: 25 ) = (0, Β± 5) 9, 0) = (Β±3, 0) β β β β a2 β b2 25 β 9 16 β Therefore, the coordinates of the foci are (0, Β± 4). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure 8. y (0, 5) (0, 4) (β3, 0) (0, 0) (3, 0) x (0, β4) (0, β5) Figure 8 872 CHAPTER 10 analytic geometry Try It #3 Graph the ellipse given by the equation x 2 __ 36 y 2 _ 4 + = 1. Identify and label the center, vertices, co-vertices, and foci. Example 4 Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form Graph the ellipse given by the equation 4x 2 + 25y 2 = 100. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci. Solution First, use algebra to rewrite the equation in standard form. + 4x 2 + 25y 2 = 100 25y 2 4x 2 100 ____ _ ___ 100 100 100 y 2 |
_ 4 x 2 _ 25 = 1 = + Next, we determine the position of the major axis. Because 25 > 4, the major axis is on the x-axis. Therefore, the equation is in the form y 2 x 2 _ _ a2 + b2 = 1, where a2 = 25 and b2 = 4. It follows that: β’ the center of the ellipse is (0, 0) β’ the coordinates of the vertices are (Β±a, 0) = (Β± β β’ the coordinates of the co-vertices are (0, Β± b) = (0, Β± β β’ the coordinates of the foci are (Β±c, 0), where c 2 = a 2 β b 2. Solving for c, we have: 25, 0) = (Β±5, 0) 4 ) = (0, Β± 2) β β c = Β± β = Β± β = Β± β β β a2 β b2 25 β 4 21 β Therefore the coordinates of the foci are (Β± β Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure 9. 21, 0). β (β5, 0) y (0, 2) (0, 0) (0, β2) Figure 9 (5, 0) x Try It #4 Graph the ellipse given by the equation 49x 2 + 16y 2 = 784. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci. Graphing ellipses not Centered at the Origin When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, (h, k), we use the standard forms (y β k)2 _ a2 (x β h)2 _ b 2 horizontal ellipses and + = 1, a > b for vertical ellipses. From these standard equations, we can (x β h)2 _ a 2 + (y β k)2 _ b 2 = 1, a > b for easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes. SECTION 10.1 the ellipse 873 How To |
β¦ Given the standard form of an equation for an ellipse centered at (h, k), sketch the graph. 1. Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci. a. If the equation is in the form = 1, where a > b, then (x β h)2 _ a2 + (y β k)2 _ b2 β’ the center is (h, k) β’ the major axis is parallel to the x-axis β’ the coordinates of the vertices are (h Β± a, k) β’ the coordinates of the co-vertices are (h, k Β± b) β’ the coordinates of the foci are (h Β± c, k) b. If the equation is in the form = 1, where a > b, then (x β h)2 _ b2 + (y β k)2 _ a2 β’ the center is (h, k) β’ the major axis is parallel to the y-axis β’ the coordinates of the vertices are (h, k Β± a) β’ the coordinates of the co-vertices are (h Β± b, k) β’ the coordinates of the foci are (h, k Β± c) 2. Solve for c using the equation c 2 = a2 β b2. 3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse. Example 5 Graphing an Ellipse Centered at (h, k) Graph the ellipse given by the equation, and foci. (x + 2)2 _ 4 + (y β 5)2 _ 9 = 1. Identify and label the center, vertices, co-vertices, Solution First, we determine the position of the major axis. Because 9 > 4, the major axis is parallel to the y-axis. Therefore, the equation is in the form = 1, where b2 = 4 and a2 = 9. It follows that: (x β h)2 _ b2 + (y β k)2 _ a2 β’ the center of the ellipse is (h, k) = (β2, 5) β’ the coordinates of the vertices are (h, k Β± a) = (β2, 5 Β± β β’ the coordinates of the co-vert |
ices are (h Β± b, k) = (β2 Β± β β’ the coordinates of the foci are (h, k Β± c), where c 2 = a 2 β b 2. Solving for c, we have: 9 ) = (β2, 5 Β± 3), or (β2, 2) and (β2, 8) 4, 5) = (β2 Β± 2, 5), or (β4, 5) and (0, 5) β β c = Β± β = Β± β = Β± β β β a2 β b2 9 β 4 5 β Therefore, the coordinates of the foci are (β2, 5 β β β 5 ) and (β2, 5 + β β 5 ). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. y (β2, 8) (β2, 5 + β5) (β4, 5) (0, 5) (β2, 5) (β2, 5 ββ5) (β2, 2) Figure 10 x 874 CHAPTER 10 analytic geometry Try It #5 Graph the ellipse given by the equation and foci. (x β 4)2 _ 36 + (y β 2)2 _ 20 = 1. Identify and label the center, vertices, co-vertices, How Toβ¦ Given the general form of an equation for an ellipse centered at (h, k), express the equation in standard form. 1. Recognize that an ellipse described by an equation in the form ax 2 + by 2 + cx + dy + e = 0 is in general form. 2. Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation. 3. Factor out the coefficients of the x 2 and y 2 terms in preparation for completing the square. 4. Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, m1 (x β h)2 + m2(y β k)2 = m3, where m1, m2, and m3 are constants. 5. Divide both sides of the equation by the constant term to express the equation in standard form. Example 6 Graphing an Ellipse Centered at ( h, k) by First Writing It in Standard |
Form Graph the ellipse given by the equation 4x 2 + 9y 2 β 40x + 36y + 100 = 0. Identify and label the center, vertices, co-vertices, and foci. Solution We must begin by rewriting the equation in standard form. 4x 2 + 9y 2 β 40x + 36y + 100 = 0 Group terms that contain the same variable, and move the constant to the opposite side of the equation. (4x 2 β 40x) + (9y 2 + 36y) = β100 Factor out the coefficients of the squared terms. 4(x 2 β 10x)+ 9(y 2 + 4y) = β100 Complete the square twice. Remember to balance the equation by adding the same constants to each side. 4(x 2 β 10x + 25)+ 9(y 2 + 4y + 4) = β100 + 100 + 36 Rewrite as perfect squares. Divide both sides by the constant term to place the equation in standard form. 4(x β 5)2 + 9(y + 2)2 = 36 Now that the equation is in standard form, we can determine the position of the major axis. Because 9 > 4, the major (x β 5)2 _ 9 + (y + 2)2 _ 4 = 1 axis is parallel to the x-axis. Therefore, the equation is in the form = 1, where a2 = 9 and b2 = 4. (x β h)2 _ a2 + (y β k)2 _ b2 It follows that: β’ the center of the ellipse is (h, k) = (5, β2) β’ the coordinates of the vertices are (h Β± a, k) = (5 Β± β β’ the coordinates of the co-vertices are (h, k Β± b) = (5, β2 Β± β β’ the coordinates of the foci are (h Β± c, k), where c2 = a2 β b2. Solving for c, we have: 9, β2) = (5 Β± 3, β2), or (2, β2) and (8, β2) 4 )= (5, β2 Β± 2), or (5, β4) and (5, 0) β β = Β± β 5, β2) and (5 + β Therefore, the coordinates of the foci are (5 β β |
Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in Figure 11. β β c = Β± β = Β± β β β a2 β b2 9 β 4 5 5, β2). β SECTION 10.1 the ellipse 875 β1 0 β1 β2 β3 β4 1 (5 ββ5, β2) 1 2 3 4 (5 +β5, β2) 6 7 8 9 x (5, 0) 5 (2, β2) (5, β2) (8, β2) (5, β4) Figure 11 Try It #6 Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse. 4x 2 + y 2 β 24x + 2y + 21 = 0 Solving Applied Problems Involving ellipses Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus. See Figure 12. In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the fociβabout 43 feet apartβcan hear each other whisper. Figure 12 Sound waves are reflected between foci in an elliptical room, called a whispering chamber. Example 7 Locating the Foci of a Whispering Chamber The Statuary Hall in the Capitol Building in Washington, D.C. is a whispering chamber. Its dimensions are 46 feet wide by 96 feet long as shown in Figure 13. a. What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point (0, 0). b. If |
two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot. 46 feet 96 feet Figure 13 876 Solution CHAPTER 10 analytic geometry a. We are assuming a horizontal ellipse with center (0, 0), so we need to find an equation of the form y 2 x 2 _ _ b2 = 1, where a > b. We know that the length of the major axis, 2a, is longer than the length of the a2 + minor axis, 2b. So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis. β’ Solving for a, we have 2a = 96, so a = 48, and a2 = 2304. β’ Solving for b, we have 2b = 46, so b = 23, and b2 = 529. Therefore, the equation of the ellipse is y 2 _ 529 b. To find the distance between the senators, we must find the distance between the foci, (Β±c, 0), where x 2 _ 2304 = 1. + c2 = a2 β b2. Solving for c, we have: c2 = a2 β b2 c2 = 2304 β 529 Substitute using the values found in part (a). β c = Β± β c = Β± β 2304 β 529 1775 β Subtract. Take the square root of both sides. The points (Β±42, 0) represent the foci. Thus, the distance between the senators is 2(42) = 84 feet. c β Β± 42 Round to the nearest foot. Try It #7 Suppose a whispering chamber is 480 feet long and 320 feet wide. a. What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point (0, 0). b. If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot. Access these online resources for additional instruction and practice with ellipses. β’ Conic Sections: The ellipse (http://openstaxcollege.org/l/conicellipse) β’ Graph an ellipse with Center at the Origin (http://openst |
axcollege.org/l/grphellorigin) β’ Graph an ellipse with Center not at the Origin (http://openstaxcollege.org/l/grphellnot) SECTION 10.1 section exercises 877 10.1 SeCTIOn exeRCISeS VeRBAl 1. Define an ellipse in terms of its foci. 3. What special case of the ellipse do we have when the 2. Where must the foci of an ellipse lie? 4. For the special case mentioned in the previous major and minor axis are of the same length? question, what would be true about the foci of that ellipse? 5. What can be said about the symmetry of the graph of an ellipse with center at the origin and foci along the y-axis? AlGeBRAIC For the following exercises, determine whether the given equations represent ellipses. If yes, write in standard form. 6. 2x 2 + y = 4 9. 4x 2 + 9y 2 = 1 7. 4x 2 + 9y 2 = 36 8. 4x 2 β y 2 = 4 10. 4x 2 β 8x + 9y 2 β 72y + 112 = 0 For the following exercises, write the equation of an ellipse in standard form, and identify the end points of the major and minor axes as well as the foci. 11 49 14. 4x 2 + 16y 2 = 1 17. (x + 5)2 _ 4 + (y β 7)2 _ 9 = 1 20. 9x 2 β 54x + 9y 2 β 54y + 81 = 0 22. 4x 2 + 24x + 16y 2 β 128y + 228 = 0 24. x 2 + 2x + 100y 2 β 1000y + 2401 = 0 26. 9x 2 + 72x + 16y 2 + 16y + 4 = 0 12. x 2 _ 100 + y 2 _ 64 = 1 15. (x β 2)2 _______ 49 + (y β 4)2 _______ 25 = 1 13. x 2 + 9y 2 = 1 16. (x β 2)2 _ 81 + (y + 1)2 _ 16 = 1 18. (x β 7)2 _______ 49 + = 1 (y β 7)2 _______ 49 21. 4x 2 β 24x + 36y 2 |
β 360y + 864 = 0 23. 4x 2 + 40x + 25y 2 β 100y + 100 = 0 25. 4x 2 + 24x + 25y 2 + 200y + 336 = 0 19. 4x 2 β 8x + 9y 2 β 72y + 112 = 0 For the following exercises, find the foci for the given ellipses. 27. (x + 3)2 _ 25 + (y + 1)2 _ 36 = 1 28. (x + 1)2 _ 100 + (y β 2)2 _ 4 = 1 29. x 2 + y 2 = 1 30. x 2 + 4y 2 + 4x + 8y = 1 31. 10x 2 + y 2 + 200x = 0 GRAPHICAl For the following exercises, graph the given ellipses, noting center, vertices, and foci. 32. x 2 _ 25 + y 2 _ 36 = 1 35. 81x 2 + 49y 2 = 1 38. + x 2 _ 2 (y + 1)2 _ 5 = 1 33. 36 16 (x β 2)2 _ 64 + 34. 4x 2 + 9y 2 = 1 = 1 (y β 4)2 _ 16 39. 4x 2 β 8x + 16y 2 β 32y β 44 = 0 (x + 3)2 _ 9 + 37. (y β 3)2 _ 9 = 1 40. x 2 β 8x + 25y 2 β 100y + 91 = 0 42. 64x 2 + 128x + 9y 2 β 72y β 368 = 0 44. 100x 2 + 1000x + y 2 β 10y + 2425 = 0 41. x 2 + 8x + 4y 2 β 40y + 112 = 0 43. 16x 2 + 64x + 4y 2 β 8y + 4 = 0 45. 4x 2 + 16x + 4y 2 + 16y + 16 = 0 For the following exercises, use the given information about the graph of each ellipse to determine its equation. 46. Center at the origin, symmetric with respect to the x- and y-axes, focus at (4, 0), and point on graph (0, 3). 47. Center at the origin, symmetric with respect to the x- and y-axes, focus at (0, β2), and point on graph (5, 0). |
878 CHAPTER 10 analytic geometry 48. Center at the origin, symmetric with respect to the x- and y-axes, focus at (3, 0), and major axis is twice as long as minor axis. 49. Center (4, 2); vertex (9, 2); one focus: (4 + 2 β 6, 2). β 50. Center (3, 5); vertex (3, 11); one focus: (3, 5 + 4 β For the following exercises, given the graph of the ellipse, determine its equation. 2 ) β 51. Center (β3, 4); vertex (1, 4); one focus: (β3 + 2 β β 3, 4) y β8 β6 β4 β2 0 β2 β4 β6 β5 β6 β4 β2 52. 55 53. y 6 4 2 β10 β8 β6 β4 β2 0 β2 2 4 6 8 10 x 56. β4 β6 y 5 4 3 2 1 54. β6 β4 β2 y β8 β6 β4 β2 0 β2 β4 β6 β5 x 2 4 6 β5 β4 β3 β2 β1 x 1 0 β1 β3 β2 β1 0 β1 1 2 3 4 5 x exTenSIOnS For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area = a β
b β
Ο. 57. (x β 3)2 _ 9 + (y β 3)2 _ 16 = 1 58. (x + 6)2 _ 16 + (y β 6)2 _ 36 = 1 59. (x + 1)2 _ 4 + (y β 2)2 _ 5 = 1 60. 4x 2 β 8x + 9y 2 β 72y + 112 = 0 61. 9x 2 β 54x + 9y 2 β 54y + 81 = 0 ReAl-WORlD APPlICATIOnS 62. Find the equation of the ellipse that will just fit 63. Find the equation of the ellipse that will just fit inside a box that is 8 units wide and 4 units high. 64. An arch has the shape of a semi-ellipse (the top half of an ellipse). The arch has a height of 8 feet and a span of 20 feet. Find |
an equation for the ellipse, and use that to find the height to the nearest 0.01 foot of the arch at a distance of 4 feet from the center. 66. A bridge is to be built in the shape of a semi- elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center. 68. A person is standing 8 feet from the nearest wall in a whispering gallery. If that person is at one focus, and the other focus is 80 feet away, what is the length and height at the center of the gallery? inside a box that is four times as wide as it is high. Express in terms of h, the height. 65. An arch has the shape of a semi-ellipse. The arch has a height of 12 feet and a span of 40 feet. Find an equation for the ellipse, and use that to find the distance from the center to a point at which the height is 6 feet. Round to the nearest hundredth. 67. A person in a whispering gallery standing at one focus of the ellipse can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet, and the foci are located 30 feet from the center, find the height of the ceiling at the center. SECTION 10.2 the hyperBola 879 leARnInG OBjeCTIVeS In this section, you will: β’ Locate a hyperbolaβs vertices and foci. β’ Write equations of hyperbolas in standard form. β’ Graph hyperbolas centered at the origin. β’ Graph hyperbolas not centered at the origin. β’ Solve applied problems involving hyperbolas. 10. 2 THe HYPeRBOlA What do paths of comets, supersonic booms, ancient Grecian pillars, and natural draft cooling towers have in common? They can all be modeled by the same type of conic. For instance, when something moves faster than the speed of sound, a shock wave in the form of a cone is created. A portion of a conic is formed when the wave intersects the ground, resulting in a sonic boom. See Figure 1. Wake created from shock wave Portion of a hyperbola Figure |
1 A shock wave intersecting the ground forms a portion of a conic and results in a sonic boom. Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the tip is exceeding the speed of sound. The bullets shot from many firearms also break the sound barrier, although the bang of the gun usually supersedes the sound of the sonic boom. locating the Vertices and Foci of a Hyperbola In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other. See Figure 2. Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points (x, y) in a plane such that the difference of the distances between (x, y) and the foci is a positive constant. Figure 2 A hyperbola 880 CHAPTER 10 analytic geometry Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances. As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the |
diagonals of the central rectangle. See Figure 3. y Conjugate axis Transverse axis Co-vertex Focus Vertex Vertex x Focus Co-vertex Center Asymptote Asymptote Figure 3 Key features of the hyperbola In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x- and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin. Deriving the Equation of an Ellipse Centered at the Origin Let (βc, 0) and (c, 0) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points (x, y) such that the difference of the distances from (x, y) to the foci is constant. See Figure 4. y (x, y) d2 d1 (c, 0) x (βc, 0) (βa, 0) (a, 0) Figure 4 If (a, 0) is a vertex of the hyperbola, the distance from (βc, 0) to (a, 0) is a β (βc) = a + c. The distance from (c, 0) to (a, 0) is c β a. The sum of the distances from the foci to the vertex is (a + c) β (c β a) = 2a SECTION 10.2 the hyperBola 881 If (x, y) is a point on the hyperbola, we can define the following variables: d2 = the distance from (βc, 0) to (x, y) d1 = the distance from (c, 0) to (x, y) By definition of a hyperbola, d2 β d1 is constant for any point (x, y) on the hyperbola. We know that the difference of these distances is 2a for the vertex (a, 0). It follows that d2 β d1 = 2a for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses. β ββ ( |
x β ( β c))2 + (y β 0)2 β β (x β c)2 + (y β 0)2 = 2a Distance formula β (x + c)2 + y 2 β β β d2 β d1 = β (x β c)2 + y 2 = 2a β β (x + c)2 + y 2 = 2a + β β (x β c)2 + y 2 β (x + c)2 + y 2 = (2a + β x 2 + 2cx + c2 + y 2 = 4a2 + 4a β β (x β c)2 + y 2 )2 β (x β c)2 + y 2 + (x β c)2 + y 2 x 2 + 2cx + c2 + y 2 = 4a2 + 4a β β (x β c)2 + y 2 + x 2 β 2cx + c2 + y 2 2cx = 4a2 + 4a β β (x β c)2 + y 2 β 2cx 4cx β 4a2 = 4a β β (x β c)2 + y 2 β (x β c)2 + y 2 cx β a2 = a β (cx β a2) 2 = a2 ξ’ β (x β c)2 + y 2 ξͺ c2 x 2 β 2a2 cx + a4 = a2(x 2 β 2cx + c2 + y 2) β 2 c2 x 2 β 2a2 cx + a4 = a2 x 2 β 2a2 cx + a2 c2 + a2 y 2 a4 + c2 x 2 = a2 x 2 + a2 c2 + a2 y 2 c2 x 2 β a2 x 2 β a2 y 2 = a2 c2 β a4 x 2 (c2 β a2) β a2 y 2 = a2 (c2 β a2) x 2 b2 β a2 y 2 = a2 b2 a2b2 ____ a2b2 x 2b2 a2y 2 ____ ____ a2b2 β a2b2 = y 2 x 2 _ _ b2 = 1 a2 β Simplify expressions. Move radical to opposite side. Square both sides. Expand |
the squares. Expand remaining square. Combine like terms. Isolate the radical. Divide by 4. Square both sides. Expand the squares. Distribute a2. Combine like terms. Rearrange terms. Factor common terms Set b2 = c2 β a2. Divide both sides by a2b2. This equation defines a hyperbola centered at the origin with vertices (Β±a, 0) and co-vertices (0 Β± b). standard forms of the equation of a hyperbola with center (0, 0) The standard form of the equation of a hyperbola with center (0, 0) and major axis on the x-axis is y 2 x 2 _ _ b2 = 1 a2 β where β’ the length of the transverse axis is 2a β’ the coordinates of the vertices are (Β±a, 0) β’ the length of the conjugate axis is 2b β’ the coordinates of the co-vertices are (0, Β±b) β’ the distance between the foci is 2c, where c2 = a2 + b2 β’ the coordinates of the foci are (Β±c, 0) b __ a x β’ the equations of the asymptotes are y = Β± 882 CHAPTER 10 analytic geometry See Figure 5a. The standard form of the equation of a hyperbola with center (0, 0) and transverse axis on the y-axis is y 2 x 2 _ _ b2 = 1 a2 β where β’ the length of the transverse axis is 2a β’ the coordinates of the vertices are (0, Β± a) β’ the length of the conjugate axis is 2b β’ the coordinates of the co-vertices are (Β±b, 0) β’ the distance between the foci is 2c, where c2 = a2 + b2 β’ the coordinates of the foci are (0, Β± c) a _ β’ the equations of the asymptotes are y = Β± x b See Figure 5b. Note that the vertices, co-vertices, and foci are related by the equation c2 = a2 + b2. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci0, b) y = β a x b (βc, 0) (βa, 0) (c, 0) (a, 0) (0, 0) x (βb, 0 |
) (0, βb) (a) y = a x b x (b, 0) y (0, c ) (0, a) (0, 0) (0, βa) (0, βc) (b) Figure 5 (a) Horizontal hyperbola with center (0, 0) (b) Vertical hyperbola with center (0, 0) How Toβ¦ Given the equation of a hyperbola in standard form, locate its vertices and foci. 1. Determine whether the transverse axis lies on the x- or y-axis. Notice that a2 is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices. a. If the equation has the form y 2 x 2 _ _ b2 = 1, then the transverse axis lies on the x-axis. The vertices are located at a2 β ( Β± a, 0), and the foci are located at (Β± c, 0). b. If the equation has the form y 2 x 2 _ _ b2 = 1, then the transverse axis lies on the y-axis. The vertices are located at a2 β (0, Β± a), and the foci are located at (0, Β± c). 2. Solve for a using the equation a = β 3. Solve for c using the equation c = β β a2. a2 + b2. β SECTION 10.2 the hyperBola 883 Example 1 Locating a Hyperbolaβs Vertices and Foci y 2 _ 49 β = 1. x 2 _ 32 Identify the vertices and foci of the hyperbola with equation Solution The equation has the form y 2 x 2 _ _ b2 = 1, so the transverse axis lies on the y-axis. The hyperbola is centered a2 β at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set x = 0, and solve for y. β β x 2 _ 32 02 _ 32 1 = 1 = y 2 _ 49 y 2 _ 49 y 2 _ 49 y 2 = 49 1 = The foci are located at (0, Β± c). Solving for c |
, y = Β± β β 49 = Β± 7 c = β β a2 + b2 = β β 49 + 32 = β β 81 = 9 Therefore, the vertices are located at (0, Β± 7), and the foci are located at (0, 9). Try It #1 Identify the vertices and foci of the hyperbola with equation x 2 _ 9 β y 2 _ 25 = 1. Writing equations of Hyperbolas in Standard Form Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin. Hyperbolas Centered at the Origin Reviewing the standard forms given for hyperbolas centered at (0, 0), we see that the vertices, co-vertices, and foci are related by the equation c2 = a2 + b2. Note that this equation can also be rewritten as b2 = c2 β a2. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices. How Toβ¦ Given the vertices and foci of a hyperbola centered at (0, 0), write its equation in standard form. 1. Determine whether the transverse axis lies on the x- or y-axis. transverse axis is the x-axis. Use the standard form a. If the given coordinates of the vertices and foci have the form (Β± a, 0) and (Β± c, 0), respectively, then the y 2 x 2 _ _ b2 = 1. a2 β b. If the given coordinates of the vertices and foci have the form (0, Β± a) and (0, Β± c), respectively, then the y 2 x 2 _ _ b2 = 1. a2 β transverse axis is the y-axis. Use the standard form 2. Find b2 using the equation b2 = c2 β a2. 3. Substitute the values for a2 and b2 into |
the standard form of the equation determined in Step 1. 884 CHAPTER 10 analytic geometry Example 2 What is the standard form equation of the hyperbola that has vertices (Β±6, 0) and foci (Β±2 β Finding the Equation of a Hyperbola Centered at (0, 0) Given its Foci and Vertices 10, 0)? β Solution The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form The vertices are (Β±6, 0), so a = 6 and a2 = 36. The foci are (Β±2 β Solving for b2, we have 10, 0), so c = 2 β β β 10 and c2 = 40. b2 = c2 β a2 b2 = 40 β 36 b2 = 4 Substitute for c2 and a2. Subtract. y 2 x 2 _ _ b2 = 1. a2 β Finally, we substitute a2 = 36 and b2 = 4 into the standard form of the equation, hyperbola is β = 1, as shown in Figure 6. x 2 _ 36 b2 = 1. The equation of the a2 β β10 β8 β6 β4 β2 y 6 4 2 0 β2 β4 β6 Figure 6 2 4 6 8 10 x Try It #2 What is the standard form equation of the hyperbola that has vertices (0, Β± 2) and foci (0, Β± 2 β β 5 )? Hyperbolas Not Centered at the Origin Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated h units horizontally and k units vertically, the center of the hyperbola will be (h, k). This translation results in the standard form of the equation we saw previously, with x replaced by (x β h) and y replaced by (y β k). standard forms of the equation of a hyperbola with center (h, k) The standard form of the equation of a hyperbola with center (h, k) and transverse axis parallel to the x-axis is where (x β h)2 _ a2 β (y β k)2 _ b2 = 1 β’ the length of the transverse axis is 2a β’ the coordinates of the vertices are (h Β± a, k) β’ the length of |
the conjugate axis is 2b β’ the coordinates of the co-vertices are (h, k Β± b) β’ the distance between the foci is 2c, where c 2 = a 2 + b 2 β’ the coordinates of the foci are (h Β± c, k) SECTION 10.2 the hyperBola 885 The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle b __ is 2a and its width is 2b. The slopes of the diagonals are Β±, and each diagonal passes through the center (h, k). a b __ Using the point-slope formula, it is simple to show that the equations of the asymptotes are y = Β± (x β h) + k. a See Figure 7a. The standard form of the equation of a hyperbola with center (h, k) and transverse axis parallel to the y-axis is where (y β k)2 _ a2 β (x β h)2 _ b2 = 1 β’ the length of the transverse axis is 2a β’ the coordinates of the vertices are (h, k Β± a) β’ the length of the conjugate axis is 2b β’ the coordinates of the co-vertices are (h Β± b, k) β’ the distance between the foci is 2c, where c 2 = a 2 + b 2 β’ the coordinates of the foci are (h, k Β± c) a __ (x β h) + k. See Figure 7b. Using the reasoning above, the equations of the asymptotes are x β h) + k a (h, k + b) y (h β c, k) (h β a, k) (h β b, k) (h, k) (h + c, k) (h + a, k) x (h, k β b) y = b (x β h) + k a (a) (h, k + c) (h, k + a) (h, k) (h, k β a) (h, k β c) (b) Figure 7 (a) Horizontal hyperbola with center (h, k) (b) Vertical hyperbola with center (h, k) y = a (x β h) + k b (h + b, k) y = β a ( |
x β h) + k b x Like hyperbolas centered at the origin, hyperbolas centered at a point (h, k) have vertices, co-vertices, and foci that are related by the equation c2 = a2 + b2. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given. How Toβ¦ Given the vertices and foci of a hyperbola centered at (h, k), write its equation in standard form. 1. Determine whether the transverse axis is parallel to the x- or y-axis. a. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form (x β h)2 _ a2 β (y β k)2 _ b2 = 1. b. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form (y β k)2 _ a2 β (x β h)2 _ b2 = 1. 2. Identify the center of the hyperbola, (h, k), using the midpoint formula and the given coordinates for the vertices. 886 CHAPTER 10 analytic geometry 3. Find a2 by solving for the length of the transverse axis, 2a, which is the distance between the given vertices. 4. Find c2 using h and k found in Step 2 along with the given coordinates for the foci. 5. Solve for b2 using the equation b2 = c2 β a2. 6. Substitute the values for h, k, a2, and b2 into the standard form of the equation determined in Step 1. Example 3 Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices What is the standard form equation of the hyperbola that has vertices at (0, β2) and (6, β2) and foci at (β2, β2) and (8, β2)? Solution The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis. Thus, the equation of the hyperbola will have the form (x |
β h)2 _ a2 β (y β k)2 _ b2 = 1 First, we identify the center, (h, k). The center is halfway between the vertices (0, β2) and (6, β2). Applying the midpoint formula, we have Next, we find a2. The length of the transverse axis, 2a, is bounded by the vertices. So, we can find a2 by finding the distance between the x-coordinates of the vertices. (h, k) = ξ’ 0 + 6 β2 + (β2) _________ _____, 2 2 ξͺ = (3, β2) 2a = | 0 β 6 | 2a = 6 a = 3 a2 = 9 Now we need to find c2. The coordinates of the foci are (h Β± c, k). So (h β c, k) = (β2, β2) and (h + c, k) = (8, β2). We can use the x-coordinate from either of these points to solve for c. Using the point (8, β2), and substituting h = 3 c2 = 25 Next, solve for b2 using the equation b2 = c2 β a2 : b2 = c2 β a2 = 25 β 9 = 16 Finally, substitute the values found for h, k, a2, and b2 into the standard form of the equation. (x β 3)2 _ 9 β (y + 2)2 _ 16 = 1 Try It #3 What is the standard form equation of the hyperbola that has vertices (1, β2) and (1, 8) and foci (1, β10) and (1, 16)? Graphing Hyperbolas Centered at the Origin When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the y 2 x 2 _ _ b2 = 1 for a2 β transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form y 2 x 2 _ _ b2 = 1 for vertical hyperbolas. a2 β horizontal hyperbolas and the standard form SECTION 10.2 the |
hyperBola 887 How Toβ¦ Given a standard form equation for a hyperbola centered at (0, 0), sketch the graph. 1. Determine which of the standard forms applies to the given equation. 2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the a. If the equation is in the form vertices, co-vertices, and foci; and the equations for the asymptotes. y 2 x 2 _ _ b2 = 1, then a2 β β’ the transverse axis is on the x-axis β’ the coordinates of the vertices are (Β±a, 0) β’ the coordinates of the co-vertices are (0, Β± b) β’ the coordinates of the foci are (Β±c, 0) b _ β’ the equations of the asymptotes are b2 = 1, then a2 β β’ the transverse axis is on the y-axis β’ the coordinates of the vertices are (0, Β± a) β’ the coordinates of the co-vertices are (Β±b, 0) β’ the coordinates of the foci are (0, Β± c) a __ β’ the equations of the asymptotes are y = Β± x b b. If the equation is in the form 3. Solve for the coordinates of the foci using the equation c = Β± β β a2 + b2. 4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola. Example 4 Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form Graph the hyperbola given by the equation asymptotes. y 2 _ 64 x 2 _ 36 β = 1. Identify and label the vertices, co-vertices, foci, and Solution The standard form that applies to the given equation is y-axis y 2 x 2 _ _ b2 = 1. Thus, the transverse axis is on the a2 β The coordinates of the vertices are (0, Β± a) = (0, Β± β β 64 ) = (0, Β± 8) The coordinates of the co-vertices are (Β±b, 0) = (Β± β β 36, 0) = (Β±6, 0) The coordinates of the foci are (0, Β± c |
), where c = Β± β β a2 + b2. Solving for c, we have c = Β± β β a2 + b2 = Β± β β 64 + 36 = Β± β β 100 = Β± 10 Therefore, the coordinates of the foci are (0, Β± 10) a 8 4 __ __ __ The equations of the asymptotes are Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in Figure 8. 888 CHAPTER 10 analytic geometry y (0, 10) (0, 8) y = β 4 x 3 (β6, 8) y = 4 x 3 y (0, 8) (β6, 0) (0, 0) (6, 0) x (β6, 0) (0, 0) (6, 8) (6, 0) x (0, β8) (0, β 10) (0, β8) (β6, β8) (6, β8) Figure 8 Try It #4 Graph the hyperbola given by the equation asymptotes. x 2 _ 144 β y 2 _ 81 = 1. Identify and label the vertices, co-vertices, foci, and Graphing Hyperbolas not Centered at the Origin Graphing hyperbolas centered at a point (h, k) other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms (y β k)2 _ a2 (x β h)2 _ b2 β = 1 for vertical hyperbolas. From these standard form equations we can easily calculate and plot (x β h)2 _ a2 β (y β k)2 _ b2 = 1 for horizontal hyperbolas, and key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes |
. How Toβ¦ Given a general form for a hyperbola centered at (h, k), sketch the graph. 1. Determine which of the standard forms applies to the given equation. Convert the general form to that standard form. 2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes. a. If the equation is in the form β’ the transverse axis is parallel to the x-axis (x β h)2 _ a2 β (y β k)2 _ b2 = 1, then β’ the center is (h, k) β’ the coordinates of the vertices are (h Β± a, k) β’ the coordinates of the co-vertices are (h, k Β± b) β’ the coordinates of the foci are (h Β± c, k) b __ (x β h) + k β’ the equations of the asymptotes are y = Β± a SECTION 10.2 the hyperBola 889 b. If the equation is in the form (y β k)2 _______ a2 β (x β h)2 _______ b2 = 1, then β’ the transverse axis is parallel to the y-axis β’ the center is (h, k) β’ the coordinates of the vertices are (h, k Β± a) β’ the coordinates of the co-vertices are (h Β± b, k) β’ the coordinates of the foci are (h, k Β± c) β’ the equations of the asymptotes are y = Β± 3. Solve for the coordinates of the foci using the equation c = Β± β β a2 + b2. a _ (x β h) + k b 4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola. Example 5 Graphing a Hyperbola Centered at (h, k) Given an Equation in General Form Graph the hyperbola given by the equation 9x 2 β 4y 2 β 36x β 40y β 388 = 0. Identify and label the center, vertices, co-vertices, foci, and asymptotes. Solution Start by expressing the equation in standard form. Group terms that contain the same variable, and move the |
constant to the opposite side of the equation. Factor the leading coefficient of each expression. (9x 2 β 36x) β (4y 2 + 40y) = 388 9(x 2 β 4x) β 4(y 2 + 10y) = 388 Complete the square twice. Remember to balance the equation by adding the same constants to each side. 9(x 2 β 4x + 4) β4(y 2 + 10y + 25) = 388 + 36 β 100 Rewrite as perfect squares. 9(x β 2)2 β 4(y + 5)2 = 324 Divide both sides by the constant term to place the equation in standard form. (x β 2)2 _ 36 β (y + 5)2 _ 81 = 1 The standard form that applies to the given equation is = 1, where a2 = 36 and b2 = 81, or a = 6 (x β h)2 _ a2 β (y β k)2 _ b2 and b = 9. Thus, the transverse axis is parallel to the x-axis. It follows that: β’ the center of the ellipse is (h, k) = (2, β5) β’ the coordinates of the vertices are (h Β± a, k) = (2 Β± 6, β5), or (β4, β5) and (8, β5) β’ the coordinates of the co-vertices are (h, k Β± b) = (2, β 5 Β± 9), or (2, β 14) and (2, 4) a2 + b 2. Solving for c, we have β’ the coordinates of the foci are (h Β± c, k), where c = Β± β β c = Β± β β 36 + 81 = Β± β β 117 = Β± 3 β β 13 Therefore, the coordinates of the foci are (2 β 3 β β 13, β5) and (2 + 3 β β 13, β5). b 3 __ __ The equations of the asymptotes are y = Β± (x β 2) β 5. (x β h) + k = Β± 2 a Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola, as shown in Figure 9. 890 CHAPTER 10 analytic geometry y (2, 4) x (β4 |
, β5) (8, β5) (2, β14) (2, β5) Figure 9 Try It #5 Graph the hyperbola given by the standard form of an equation vertices, co-vertices, foci, and asymptotes. (y + 4)2 _ β 100 (x β 3)2 _ 64 = 1. Identify and label the center, Solving Applied Problems Involving Hyperbolas As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. See Figure 10. For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide! Figure 10 Cooling towers at the Drax power station in north Yorkshire, United Kingdom (credit: les Haines, Flickr) The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In Example 6 we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides. SECTION 10.2 the hyperBola 891 Example 6 Solving Applied Problems Involving Hyperbolas The design layout of a cooling tower is shown in Figure 11. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart. 72 m 60 m 79.6 m 179.6 m Figure 11 Project design for a natural draft cooling tower Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbolaβ indicated by the intersection of dashed perpendicular lines in the figureβis the origin of the coordinate plane. Round final values to four decimal places. Solution We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal y 2 x 2 _ _ b2 = 1, where the branches of the hyperbola form the sides of the cooling tower. a2 β hyperbola centered at the origin |
: We must find the values of a2 and b2 to complete the model. First, we find a2. Recall that the length of the transverse axis of a hyperbola is 2a. This length is represented by the distance where the sides are closest, which is given as 65.3 meters. So, 2a = 60. Therefore, a = 30 and a2 = 900. To solve for b2, we need to substitute for x and y in our equation using a known point. To do this, we can use the dimensions of the tower to find some point (x, y) that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y-axis bisects the tower, our x-value can be represented by the radius of the top, or 36 meters. The y-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore, y 2 x 2 _ _ b2 = 1 a2 β Standard form of horizontal hyperbola. b2 = y 2 _ x 2 _ a2 β 1 = (79.6)2 _ (36)2 ____ β 1 900 Isolate b2 Substitute for a2, x, and y The sides of the tower can be modeled by the hyperbolic equation β 14400.3636 Round to four decimal places x 2 _ 900 β y 2 _ 14400.3636 = 1, or x 2 _ 302 β y 2 _ 120.00152 = 1 892 CHAPTER 10 analytic geometry Try It #6 A design for a cooling tower project is shown in Figure 12. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbolaβindicated by the intersection of dashed perpendicular lines in the figureβis the origin of the coordinate plane. Round final values to four decimal places. 60 m 40 m 67.082 m 167.082 m Figure 12 Access these online resources for additional instruction and practice with hyperbolas. β’ Conic Sections: The Hyperbola Part 1 of 2 (http://openstaxcollege.org/l/hyperbola1) β’ Conic Sections: The Hyperbola Part 2 of 2 (http://openstaxcollege.org/l/hyperbola2) β’ Graph a Hyperbola with Center at Origin (http://open |
staxcollege.org/l/hyperbolaorigin) β’ Graph a Hyperbola with Center not at Origin (http://openstaxcollege.org/l/hbnotorigin) SECTION 10.2 section exercises 893 10.2 SeCTIOn exeRCISeS VeRBAl 1. Define a hyperbola in terms of its foci. 2. What can we conclude about a hyperbola if its asymptotes intersect at the origin? 3. What must be true of the foci of a hyperbola? 4. If the transverse axis of a hyperbola is vertical, what do we know about the graph? 5. Where must the center of hyperbola be relative to its foci? AlGeBRAIC For the following exercises, determine whether the following equations represent hyperbolas. If so, write in standard form. 6. 3y 2 + 2x = 6 8. 5y 2 + 4x 2 = 6x 10. β9x 2 + 18x + y 2 + 4y β 14 = 0 7. β x 2 ___ 36 y 2 __ = 1 9 9. 25x 2 β 16y 2 = 400 For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. 11. 13 ___ ___ 36 25 y 2 x 2 __ ___ β 4 81 (x β 1)2 _______ 9 (x β 2)2 _______ 49 15. 17. β β = 1 = 1 (y β 2)2 _______ 16 (y + 7)2 _______ 49 19. β9x 2 β 54x + 9y 2 β 54y + 81 = 0 21. β4x 2 + 24x + 16y 2 β 128y + 156 = 0 23. x 2 + 2x β 100y 2 β 1000y + 2401 = 0 25. 4x 2 + 24x β 25y 2 + 200y β 464 = 0 12. x 2 ___ 100 β y 2 __ = 1 9 14. 9y 2 β 4x 2 = 1 16. (y β 6)2 _______ 36 β (x + 1)2 _______ 16 = 1 18. 4x 2 β 8x β 9y 2 β 72y + 112 = 0 20. 4x 2 β 24x β 36y 2 β |
360y + 864 = 0 22. β4x 2 + 40x + 25y 2 β 100y + 100 = 0 24. β9x 2 + 72x + 16y 2 + 16y + 4 = 0 For the following exercises, find the equations of the asymptotes for each hyperbola. (y + 4)2 _______ 22 (x β 3)2 _______ 52 β 26. 27. = 1 x 2 y 2 __ __ 32 = 1 32 β (y β 3)2 _______ 32 28. β (x + 5)2 _______ 62 30. 16y 2 + 96y β 4x 2 + 16x + 112 = 0 = 1 29. 9x 2 β 18x β 16y 2 + 32y β 151 = 0 GRAPHICAl For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci. = 1 = 1 31. 33. 35. 37. β β x 2 y 2 ___ ___ 49 16 x 2 y 2 ___ __ 9 25 (y + 5)2 _______ 9 (y β 3)2 _______ 9 β β (x β 4)2 _______ 25 (x β 3)2 _______ 9 = 1 = 1 32. β = 1 x 2 ___ 64 y 2 __ 4 34. 81x 2 β 9y 2 = 1 36. (x β 2)2 _______ 8 β (y + 3)2 _______ 27 = 1 38. β4x 2 β 8x + 16y 2 β 32y β 52 = 0 894 CHAPTER 10 analytic geometry 39. x 2 β 8x β 25y 2 β 100y β 109 = 0 40. βx 2 + 8x + 4y 2 β 40y + 88 = 0 41. 64x 2 + 128x β 9y 2 β 72y β 656 = 0 42. 16x 2 + 64x β 4y 2 β 8y β 4 = 0 43. β100x 2 + 1000x + y 2 β 10y β 2575 = 0 44. 4x 2 + 16x β 4y 2 + 16y + 16 = 0 For the following exercises, given information about the graph of the hyperbola, find its equation. 45. Vertices at (3, 0) and (β3, 0) and one focus at (5, 0). 46. Vertices at (0, 6) and (0, β6 |
) and one focus at (0, β8). 47. Vertices at (1, 1) and (11, 1) and one focus at (12, 1). 48. Center: (0, 0); vertex: (0, β13); one focus: (0, β β 313 ). 49. Center: (4, 2); vertex: (9, 2); one focus: (4 + β β 26, 2). 50. Center: (3, 5); vertex: (3, 11); one focus: (3, 5 + 2 β β 10 ). For the following exercises, given the graph of the hyperbola, find its equation. 52. y Foci Vertices Center (1, 1) Foci x 54. y Foci Center (3, 1) Foci Vertices 51. 53. 55. y 10 8 6 4 2 0 β2 β2 β4 β6 β8 β10 β10 β8 β6 β4 42 6 8 10 x y Foci (β1, 3) Center (β1, 0) (β1, β3) Foci y (β3, β3) Center Vertices Vertices Foci (β8, β3) x x Foci (2, β3) SECTION 10.2 section exercises 895 exTenSIOnS For the following exercises, express the equation for the hyperbola as two functions, with y as a function of x. Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. 56 57 58. (x β 2)2 _ 16 β (y + 3)2 _ 25 = 1 59. β4x 2 β 16x + y 2 β 2y β 19 = 0 60. 4x 2 β 24x β y 2 β 4y + 16 = 0 ReAl-WORlD APPlICATIOnS For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. 61. The hedge will follow the asymptotes y = x and y = βx, and its closest distance to the center fountain is 5 yards. 62. The hedge will follow the asymptotes y = 2x and y = β2x, and its closest distance to the center fountain is 6 yards. 1 __ 63. |
The hedge will follow the asymptotes y = x and 2 1 __ y = β x, and its closest distance to the center 2 2 __ 64. The hedge will follow the asymptotes y = x and 3 2 __ y = β x, and its closest distance to the center 3 fountain is 10 yards. fountain is 12 yards. 3 __ 65. The hedge will follow the asymptotes y = x and 4 3 __ y = β x, and its closest distance to the center 4 fountain is 20 yards. For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information. 66. The object enters along a path approximated by the line y = x β 2 and passes within 1 au (astronomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = βx + 2. 68. The object enters along a path approximated by the line y = 0.5x + 2 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = β0.5x β 2. 67. The object enters along a path approximated by the line y = 2x β 2 and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = β2x + 2. 69. The object enters along a path approximated by the 1 __ x β 1 and passes within 1 au of the sun line y = 3 at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a 1 __ path approximated by the line y = β x + 1. 3 70. The object enters along a path approximated by the line y = 3x β 9 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by |
the line y = β3x + 9. 896 CHAPTER 10 analytic geometry leARnInG OBjeCTIVeS In this section, you will: β’ Graph parabolas with vertices at the origin. β’ Write equations of parabolas in standard form. β’ Graph parabolas with vertices not at the origin. β’ Solve applied problems involving parabolas. 10. 3 THe PARABOlA Figure 1 The Olympic torch concludes its journey around the world when it is used to light the Olympic cauldron during the opening ceremony. (credit: Ken Hackman, U.S. Air Force) Did you know that the Olympic torch is lit several months before the start of the games? The ceremonial method for lighting the flame is the same as in ancient times. The ceremony takes place at the Temple of Hera in Olympia, Greece, and is rooted in Greek mythology, paying tribute to Prometheus, who stole fire from Zeus to give to all humans. One of eleven acting priestesses places the torch at the focus of a parabolic mirror (see Figure 1), which focuses light rays from the sun to ignite the flame. Parabolic mirrors (or reflectors) are able to capture energy and focus it to a single point. The advantages of this property are evidenced by the vast list of parabolic objects we use every day: satellite dishes, suspension bridges, telescopes, microphones, spotlights, and car headlights, to name a few. Parabolic reflectors are also used in alternative energy devices, such as solar cookers and water heaters, because they are inexpensive to manufacture and need little maintenance. In this section we will explore the parabola and its uses, including low-cost, energy-efficient solar designs. Graphing Parabolas with Vertices at the Origin In The Ellipse, we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola. See Figure 2. Figure 2 Parabola SECTION 10.3 the paraBola 897 Like the ellipse and hyperbola, the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the direct |
rix. In Quadratic Functions, we learned about a parabolaβs vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See Figure 3. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus. The line segment that passes through the focus and is parallel to the directrix is called the latus rectum. The endpoints of the latus rectum lie on the curve. By definition, the distance d from the focus to any point P on the parabola is equal to the distance from P to the directrix. y Latus rectum Axis of symmetry Focus Vertex Directrix x Figure 3 Key features of the parabola To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former. y y = βp d (x, y) d x (x, βp) (0, p) (0, 0) Figure 4 Let (x, y) be a point on the parabola with vertex (0, 0), focus (0, p), and directrix y = βp as shown in Figure 4. The distance d from point (x, y) to point (x, βp) on the directrix is the difference of the y-values: d = y + p. The distance from the focus (0, p) to the point (x, y) is also equal to d and can be expressed using the distance formula. d = β = β β (x β 0)2 + (y β p)2 x 2 + (y β p)2 β Set the two expressions for d equal to each other and solve for y to derive the equation of the parabola. We do this because the distance from (x, y) to (0, p) equals the distance from (x, y) to (x, βp). β x 2 + (y β p)2 = y + p β We then square both sides of the equation, expand the squared terms, and simplify by combining like terms. x 2 + (y β p)2 = (y + p)2 x 2 + y 2 β 2py + p2 = y 2 + 2py + p |
2 x 2 β 2py = 2py x 2 = 4py The equations of parabolas with vertex (0, 0) are y 2 = 4px when the x-axis is the axis of symmetry and x 2 = 4py when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features. 898 CHAPTER 10 analytic geometry standard forms of parabolas with vertex (0, 0) Table 1 and Figure 5 summarize the standard features of parabolas with a vertex at the origin. Focus (p, 0) (0, p) Table 1 x Axis of Symmetry Equation x-axis y-axis y 2 = 4px x 2 = 4py y y 2 = 4px p > 0 (p, |2p|) (p, 0) (p, β|2p|) (a) y x2 = 4py p > 0 (0, 0) x = βp Directrix Endpoints of Latus Rectum x = βp y = βp (p, Β± 2p) (Β± 2p, p) y 2 = 4px p < 0 y (p, |2p|) (p, 0) (p, β|2p|) x (0, 0) (b) y (0, 0) x = βp y = βp x (β|2p|, p) (0, p) (|2p|, p) x y = βp (0, 0) (c) (β|2p|, p) (0, p) (|2p|, p) x2 = 4py p < 0 (d) Figure 5 (a) When p > 0 and the axis of symmetry is the x-axis, the parabola opens right. (b) When p < 0 and the axis of symmetry is the x-axis, the parabola opens left. (c) When p < 0 and the axis of symmetry is the y-axis, the parabola opens up. (d) When p < 0 and the axis of symmetry is the y-axis, the parabola opens down. The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. See Figure 5. When given a standard equation for a parabola centered at the origin, we can easily identify the key |
features to graph the parabola. A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in Figure 6. y y 2 = 24x (6, 12) (0, 0) (6, 0) x (6, β12) x = β6 Figure 6 SECTION 10.3 the paraBola 899 How Toβ¦ Given a standard form equation for a parabola centered at (0, 0), sketch the graph. 1. Determine which of the standard forms applies to the given equation: y 2 = 4px or x 2 = 4py. 2. Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum. a. If the equation is in the form y 2 = 4px, then β’ the axis of symmetry is the x-axis, y = 0 β’ set 4p equal to the coefficient of x in the given equation to solve for p. If p > 0, the parabola opens right. If p < 0, the parabola opens left. β’ use p to find the coordinates of the focus, (p, 0) β’ use p to find the equation of the directrix, x = β p β’ use p to find the endpoints of the latus rectum, (p, Β± 2p). Alternately, substitute x = p into the original equation. b. If the equation is in the form x 2 = 4py, then β’ the axis of symmetry is the y-axis, x = 0 β’ set 4p equal to the coefficient of y in the given equation to solve for p. If p > 0, the parabola opens up. If p < 0, the parabola opens down. β’ use p to find the coordinates of the focus, (0, p) β’ use p to find equation of the directrix, y = β p β’ use p to find the endpoints of the latus rectum, (Β±2p, p) 3. Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Example 1 Graphing a Parabola with Vertex (0, 0) and the x-axis |
as the Axis of Symmetry Graph y 2 = 24x. Identify and label the focus, directrix, and endpoints of the latus rectum. Solution The standard form that applies to the given equation is y 2 = 4px. Thus, the axis of symmetry is the x-axis. It follows that: β’ 24 = 4p, so p = 6. Since p > 0, the parabola opens right β’ the coordinates of the focus are (p, 0) = (6, 0) β’ the equation of the directrix is x = βp = β 6 β’ the endpoints of the latus rectum have the same x-coordinate at the focus. To find the endpoints, substitute x = 6 into the original equation: (6, Β± 12) Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 7. x = β6 (0, 0) y 20 16 12 8 4 β10 β8 β6 β4 β2 (6, 12) (6, 0) 4 6 8 10 x 2 0 β4 β8 β12 β16 β20 Figure 7 (6, β12) 900 CHAPTER 10 analytic geometry Try It #1 Graph y 2 = β16x. Identify and label the focus, directrix, and endpoints of the latus rectum. Example 2 Graphing a Parabola with Vertex (0, 0) and the y-axis as the Axis of Symmetry Graph x 2 = β6y. Identify and label the focus, directrix, and endpoints of the latus rectum. Solution The standard form that applies to the given equation is x 2 = 4py. Thus, the axis of symmetry is the y-axis. It follows that: 3 __ β’ β6 = 4p, so p = β Since p < 0, the parabola opens down. 2 3 β’ the coordinates of the focus are (0, p) = ξ’ 0, β ξͺ __ 2 3 __ β’ the equation of the directrix is y = β p = 2 3 3 ξͺ into the original equation, ξ’ Β±3, β __ __ β’ the endpoints of the latus rectum can be found by substituting y = 2 2 Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to |
form the parabola. β3, β 3 2 y (0, 0) 0, β 3 2 Figure 8 y = 3 2 x 3, β 3 2 x2 = β6y Try It #2 Graph x 2 = 8y. Identify and label the focus, directrix, and endpoints of the latus rectum. Writing equations of Parabolas in Standard Form In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features. How Toβ¦ Given its focus and directrix, write the equation for a parabola in standard form. 1. Determine whether the axis of symmetry is the x- or y-axis. a. If the given coordinates of the focus have the form (p, 0), then the axis of symmetry is the x-axis. Use the standard form y 2 = 4px. b. If the given coordinates of the focus have the form (0, p), then the axis of symmetry is the y-axis. Use the standard form x 2 = 4py. 2. Multiply 4p. 3. Substitute the value from Step 2 into the equation determined in Step 1. Example 3 Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix 1 1 What is the equation for the parabola with focus ξ’ β, 0 ξͺ and directrix x = __ __? 2 2 Solution The focus has the form (p, 0), so the equation will have the form y 2 = 4px. 1 _ ) = β2. β’ Multiplying 4p, we have 4p = 4(β 2 β’ Substituting for 4p, we have y 2 = 4px = β2x. Therefore, the equation for the parabola is y 2 = β2x. SECTION 10.3 the paraBola 901 Try It #3 7 7 ξͺ and directrix y = β What is the equation for the parabola with focus ξ’ 0, __ __? 2 2 Graphing Parabolas with Vertices not at the Origin Like other graphs weβve worked with, the graph of a parabola can be translated. If a parabola is translated h units horizontally and k units vertically, the vertex will be (h, k). This |
translation results in the standard form of the equation we saw previously with x replaced by (x β h) and y replaced by (y β k). To graph parabolas with a vertex (h, k) other than the origin, we use the standard form (y β k)2 = 4p(x β h) for parabolas that have an axis of symmetry parallel to the x-axis, and (x β h)2 = 4p(y β k) for parabolas that have an axis of symmetry parallel to the y-axis. These standard forms are given below, along with their general graphs and key features. standard forms of parabolas with vertex (h, k) Table 2 and Figure 9 summarize the standard features of parabolas with a vertex at a point (h, k). Axis of Symmetry Equation Focus Directrix Endpoints of Latus Rectum y = k x = h (y β k)2 = 4p(x β h) (h + p, k) (x β h)2 = 4p(y β k) (h, k + p) x = h βp y = k βp (h + p, k Β± 2p) (h Β± 2p, k + p) Table 2 y (y β k)2 = 4p(x β h) p > 0 y (y β k)2 = 4p(x β h) p < 0 (h + p, k + 2p) y = k (h + p, k) (h + p, k β 2p) (h, k) x = h β p (a) x (h + p, k + |2p|) (h + p, k) (h + p, k β |2p|) y = k x (h, k) x = h β p (b) y (x β h)2 = 4p(y β h) p > 0 y (x β h)2 = 4p(y β k) p < 0 x = h (h, k + ph, k) (h β 2p, k + p) (h + 2p, k + p) y = k β p (h, k) (h β |2p|, k + p) (h + |2p|, k + p) (h, k + p) x x (c) (d) |
Figure 9 (a) When p > 0, the parabola opens right. (b) When p < 0, the parabola opens left. (c) When p > 0, the parabola opens up. (d) When p < 0, the parabola opens down. 902 CHAPTER 10 analytic geometry How Toβ¦ Given a standard form equation for a parabola centered at (h, k), sketch the graph. 1. Determine which of the standard forms applies to the given equation: (y β k)2 = 4p(x β h) or (x β h)2 = 4p(y β k). 2. Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum. a. If the equation is in the form (y β k)2 = 4p(x β h), then: β’ use the given equation to identify h and k for the vertex, (h, k) β’ use the value of k to determine the axis of symmetry, y = k β’ set 4p equal to the coefficient of (x β h) in the given equation to solve for p. If p > 0, the parabola opens right. If p < 0, the parabola opens left. β’ use h, k, and p to find the coordinates of the focus, (h + p, k) β’ use h and p to find the equation of the directrix, x = h β p β’ use h, k, and p to find the endpoints of the latus rectum, (h + p, k Β± 2p) b. If the equation is in the form (x β h)2 = 4p(y β k), then: β’ use the given equation to identify h and k for the vertex, (h, k) β’ use the value of h to determine the axis of symmetry, x = h β’ set 4p equal to the coefficient of (y β k) in the given equation to solve for p. If p > 0, the parabola opens up. If p < 0, the parabola opens down. β’ use h, k, and p to find the coordinates of the focus, (h, k + p) β’ use k and p to find the equation of the directrix, y = k β p β’ use h, k, and p to |
find the endpoints of the latus rectum, (h Β± 2p, k + p) 3. Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Example 4 Graphing a Parabola with Vertex ( h, k) and Axis of Symmetry Parallel to the x-axis Graph (y β 1)2 = β16(x + 3). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solution The standard form that applies to the given equation is (y β k)2 = 4p(x β h). Thus, the axis of symmetry is parallel to the x-axis. It follows that: β’ the vertex is (h, k) = (β3, 1) β’ the axis of symmetry is y = k = 1 β’ β16 = 4p, so p = β4. Since p < 0, the parabola opens left. β’ the coordinates of the focus are (h + p, k) = (β3 + (β4), 1) = (β7, 1) β’ the equation of the directrix is x = h β p = β3 β (β4) = 1 β’ the endpoints of the latus rectum are (h + p, k Β± 2p) = (β3 + (β4), 1 Β± 2(β4)), or (β7, β7) and (β7, 9) Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 10. y (y β 1)2 = β16(x + 3) (β7, 9) (β7, 1) (β3, 1) (β7, β7 ) Figure 10 y = 1 x x = 1 SECTION 10.3 the paraBola 903 Try It #4 Graph (y + 1)2 = 4(x β 8). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Example 5 Graphing a Parabola from an Equation Given in General Form Graph x 2 β 8x β 28y β 208 = 0. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solution |
Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is (x β h)2 = 4p (y β k). Thus, the axis of symmetry is parallel to the y-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable x in order to complete the square. x 2 β 8x β 28y β 208 = 0 x 2 β 8x = 28y + 208 x 2 β 8x + 16 = 28y + 208 + 16 (x β 4)2 = 28y + 224 (x β 4)2 = 28(y + 8) (x β 4)2 = 4 β
7 β
(y + 8) It follows that: β’ the vertex is (h, k) = (4, β8) β’ the axis of symmetry is x = h = 4 β’ since p = 7, p > 0 and so the parabola opens up β’ the coordinates of the focus are (h, k + p) = (4, β8 + 7) = (4, β1) β’ the equation of the directrix is y = k β p = β8 β 7 = β15 β’ the endpoints of the latus rectum are (h Β± 2p, k + p) = (4 Β± 2(7), β8 + 7), or (β10, β1) and (18, β1) Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 11. (β10, β1) y (x β 4)2 = 28(y + 8) x (18, β1) y = β15 (4, β1) (4, β8) x = 4 Figure 11 Try It #5 Graph (x + 2)2 = β20 (y β 3). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solving Applied Problems Involving Parabolas As we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors, such as the one used to light the Olympic torch, have a very unique reflecting property. When rays |
of light parallel to the parabolaβs axis of symmetry are directed toward any surface of the mirror, the light is reflected directly to the focus. See Figure 12. This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror. 904 CHAPTER 10 analytic geometry Parallel rays of sunlight Focus Parabolic reflector Figure 12 Parabolic mirrors have the ability to focus the sunβs energy to a single point, raising the temperature hundreds of degrees in a matter of seconds. Thus, parabolic mirrors are featured in many low-cost, energy efficient solar products, such as solar cookers, solar heaters, and even travel-sized fire starters. Example 6 Solving Applied Problems Involving Parabolas A cross-section of a design for a travel-sized solar fire starter is shown in Figure 13. The sunβs rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds. a. Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane. b. Use the equation found in part ( a) to find the depth of the fire starter. Igniter 1.7 in Depth 4.5 in Figure 13 Cross-section of a travel-sized solar fire starter Solution a. The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form x 2 = 4py, where p > 0. The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have p = 1.7. x 2 = 4py Standard form of upward-facing parabola with vertex (0, 0) x 2 = 4(1.7)y Substitute 1.7 for p. x 2 = 6.8y Multiply. b. The dish extends = 2.25 inches on either side of the origin. We can substitute 2.25 for x in the equation 4.5 ___ 2 from part ( a) to find the depth of the dish. x 2 = 6.8y (2.25)2 = 6.8y Equation found in part ( a). Substitute 2.25 for x. y β 0.74 Solve for y. The dish is |
about 0.74 inches deep. SECTION 10.3 the paraBola 905 Try It #6 Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1,600 mm. The sunβs rays reflect off the parabolic mirror toward the βcooker,β which is placed 320 mm from the base. a. Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the x-axis as its axis of symmetry). b. Use the equation found in part (a) to find the depth of the cooker. Access these online resources for additional instruction and practice with parabolas. β’ Conic Sections: The Parabola Part 1 of 2 (http://openstaxcollege.org/l/parabola1) β’ Conic Sections: The Parabola Part 2 of 2 (http://openstaxcollege.org/l/parabola2) β’ Parabola with Vertical Axis (http://openstaxcollege.org/l/parabolavertcal) β’ Parabola with Horizontal Axis (http://openstaxcollege.org/l/parabolahoriz) 906 CHAPTER 10 analytic geometry 10.3 SeCTIOn exeRCISeS VeRBAl 1. Define a parabola in terms of its focus and directrix. 3. If the equation of a parabola is written in standard form and p is negative and the directrix is a horizontal line, then what can we conclude about its graph? 5. As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix? AlGeBRAIC 2. If the equation of a parabola is written in standard form and p is positive and the directrix is a vertical line, then what can we conclude about its graph? 4. What is the effect on the graph of a parabola if its equation in standard form has increasing values of p? For the following exercises, determine whether the given equation is a parabola. If so, rewrite the equation in standard form. 6. y 2 = 4 β x 2 9. (y β 3)2 = 8(x β 2) 7. y = |
4x 2 10. y 2 + 12x β 6y β 51 = 0 8. 3x 2 β 6y 2 = 12 For the following exercises, rewrite the given equation in standard form, and then determine the vertex (V), focus (F), and directrix (d) of the parabola. 11. x = 8y 2 1 __ 12. y = x 2 4 13. y = β4x 2 1 __ 14. x = y 2 8 15. x = 36y 2 16. x = 1 ___ 36 y 2 17. (x β 1)2 = 4(y β 1) 20. (x + 1)2 = 2(y + 4) 23. y 2 + 12x β 6y + 21 = 0 26. y 2 β 24x + 4y β 68 = 0 29. 3y 2 β 4x β 6y + 23 = 0 GRAPHICAl 4 __ (x + 4) 18. (y β 2)2 = 5 21. (x + 4)2 = 24(y + 1) 24. x 2 β 4x β 24y + 28 = 0 27. x 2 β 4x + 2y β 6 = 0 30. x 2 + 4x + 8y β 4 = 0 19. (y β 4)2 = 2(x + 3) 22. (y + 4)2 = 16(x + 4) 25. 5x 2 β 50x β 4y + 113 = 0 28. y 2 β 6y + 12x β 3 = 0 For the following exercises, graph the parabola, labeling the focus and the directrix. 1 __ 31. x = y 2 8 32. y = 36x 2 33. y = 1 ___ 36 x 2 34. y = β9x 2 37. β6(y + 5)2 = 4(x β 4) 4 __ (x + 2) 35. (y β 2)2 = β 3 38. y 2 β 6y β 8x + 1 = 0 36. β5(x + 5)2 = 4(y + 5) 39. x 2 + 8x + 4y + 20 = 0 40. 3x 2 + 30x β 4y + 95 = 0 41. y 2 β 8x + 10y + 9 = 0 42. x 2 + 4x + 2y + 2 = 0 43. y 2 + 2y β 12x + 61 |
= 0 44. β2x 2 + 8x β 4y β 24 = 0 For the following exercises, find the equation of the parabola given information about its graph. 45. Vertex is (0, 0); directrix is y = 4, focus is (0, β4). 46. Vertex is (0, 0); directrix is x = 4, focus is (β4, 0). 47. Vertex is (2, 2); directrix is x = 2 β β β (2 + β 2, 2). β 2, focus is 1 7, 3 ξͺ., focus is ξ’ β __ __ 48. Vertex is (β2, 3); directrix is x = β 2 2 49. Vertex is ( β 3 ). is (0 ); directrix is x = 2 β β 2, focus 50. Vertex is (1, 2); directrix is y = 1 11 ξͺ., focus is ξ’ 1, __ ___ 3 3 SECTION 10.3 section exercises 907 For the following exercises, determine the equation for the parabola from its graph. 52. y Focus (β1, 2) Vertex (3, 2) Axis of symmetry 54. y Vertex (β3, 5) Focus β3, 319 64 Axis of symmetry x x 51. 53. Axis of symmetry Focus 1 4 0, y y Vertex (0, 0) Vertex (β2, 2) Axis of symmetry Focus β 31, 2 16 x x 55. y Vertex Focus Axis of symmetry x 908 CHAPTER 10 analytic geometry exTenSIOnS For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation. 56. V(0, 0), Endpoints (2, 1), (β2, 1) 57. V(0, 0), Endpoints (β2, 4), (β2, β4) 58. V(1, 2), Endpoints (β5, 5), (7, 5) 59. V(β3, β1), Endpoints (0, 5), (0, β7) 7 7 ξͺ ξͺ, ξ’ 3, β 60. V(4, β3), Endpoints ξ’ 5, β __ __ 2 2 ReAl-WORlD APPlICATIOnS 61. The mirror in an automobile headlight has |
a parabolic cross-section with the light bulb at the focus. On a schematic, the equation of the parabola is given as x 2 = 4y. At what coordinates should you place the light bulb? 62. If we want to construct the mirror from the previous exercise such that the focus is located at (0, 0.25), what should the equation of the parabola be? 63. A satellite dish is shaped like a paraboloid of 64. Consider the satellite dish from the previous revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed? exercise. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver? 65. A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth. 66. If the searchlight from the previous exercise has the light source located 6 inches from the base along the axis of symmetry and the opening is 4 feet, find the depth. 67. An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center. 68. If the arch from the previous exercise has a span of 160 feet and a maximum height of 40 feet, find the equation of the parabola, and determine the distance from the center at which the height is 20 feet. 69. An object is projected so as to follow a parabolic path given by y = βx 2 + 96x, where x is the horizontal distance traveled in feet and y is the height. Determine the maximum height the object reaches. 70. For the object from the previous exercise, assume the path followed is given by y = β0.5x 2 + 80x. Determine how far along the horizontal the object traveled to reach maximum height. SECTION 10.4 rotation oF axis 909 leARnInG OBjeCTIVeS In this section, you will: β’ Identify nondegenerate conic sections given their general form equations. β’ Use rotation of axes formulas. β’ Write |
equations of rotated conics in standard form. β’ Identify conics without rotating axes. 10. 4 ROTATIOn OF AxIS As we have seen, conic sections are formed when a plane intersects two right circular cones aligned tip to tip and extending infinitely far in opposite directions, which we also call a cone. The way in which we slice the cone will determine the type of conic section formed at the intersection. A circle is formed by slicing a cone with a plane perpendicular to the axis of symmetry of the cone. An ellipse is formed by slicing a single cone with a slanted plane not perpendicular to the axis of symmetry. A parabola is formed by slicing the plane through the top or bottom of the double-cone, whereas a hyperbola is formed when the plane slices both the top and bottom of the cone. See Figure 1. Diagonal Slice Horizontal Slice Deep Vertical Slice Vertical Slice Ellipse Circle Hyperbola Parabola Figure 1 The nondegenerate conic sections Ellipses, circles, hyperbolas, and parabolas are sometimes called the nondegenerate conic sections, in contrast to the degenerate conic sections, which are shown in Figure 2. A degenerate conic results when a plane intersects the double cone and passes through the apex. Depending on the angle of the plane, three types of degenerate conic sections are possible: a point, a line, or two intersecting lines. Intersecting Lines Single Line Single Point Figure 2 Degenerate conic sections 910 CHAPTER 10 analytic geometry Identifying nondegenerate Conics in General Form In previous sections of this chapter, we have focused on the standard form equations for nondegenerate conic sections. In this section, we will shift our focus to the general form equation, which can be used for any conic. The general form is set equal to zero, and the terms and coefficients are given in a particular order, as shown below. Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 where A, B, and C are not all zero. We can use the values of the coefficients to identify which type conic is represented by a given equation. You may notice that the general form equation has an xy term that we have not seen in any of the standard form equations. As we will discuss later, the xy term rotates the conic whenever B is not equal to |
zero. Conic Sections Example ellipse circle hyperbola parabola one line intersecting lines parallel lines a point no graph 4x 2 + 9y 2 = 1 4x 2 + 4y 2 = 1 4x 2 β 9y 2 = 1 4x 2 = 9y or 4y 2 = 9x 4x + 9y = 1 (x β 4) (y + 4)= 0 (x β 4)(x β 9) = 0 4x 2 + 4y 2 = 0 4x 2 + 4y 2 = β 1 Table 1 general form of conic sections A conic section has the general form where A, B, and C are not all zero. Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 Table 2 summarizes the different conic sections where B = 0, and A and C are nonzero real numbers. This indicates that the conic has not been rotated. Conic Sections Example ellipse circle hyperbola parabola Ax 2 + Cy 2 + Dx + Ey + F = 0, A β C and AC > 0 Ax 2 + Cy 2 + Dx + Ey + F = 0, A = C Ax 2 β Cy 2 + Dx + Ey + F = 0 or β Ax 2 + Cy 2 + Dx + Ey + F = 0, where A and C are positive Ax 2 + Dx + Ey + F = 0 or Cy 2 + Dx + Ey + F = 0 Table 2 How Toβ¦ Given the equation of a conic, identify the type of conic. 1. Rewrite the equation in the general form, Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. 2. Identify the values of A and C from the general form. a. If A and C are nonzero, have the same sign, and are not equal to each other, then the graph may be an ellipse. b. If A and C are equal and nonzero and have the same sign, then the graph may be a circle. SECTION 10.4 rotation oF axis 911 c. If A and C are nonzero and have opposite signs, then the graph may be a hyperbola. d. If either A or C is zero, then the graph may be a parabola. If B = 0, the conic section will have a vertical and/or horizontal axes. If B does not equal 0, |
as shown below, the conic section is rotated. Notice the phrase βmay beβ in the definitions. That is because the equation may not represent a conic section at all, depending on the values of A, B, C, D, E, and F. For example, the degenerate case of a circle or an ellipse is a point: Ax 2 +By 2=0, when A and B have the same sign. The degenerate case of a hyperbola is two intersecting straight lines: Ax 2 +By 2=0, when A and B have opposite signs. On the other hand, the equation Ax 2 +By 2+1=0, when A and B are positive does not represent a graph at all, since there are no real ordered pairs which satisfy it. Example 1 Identifying a Conic from Its General Form Identify the graph of each of the following nondegenerate conic sections. a. 4x 2 β 9y 2 + 36x + 36y β 125 = 0 b. 9y 2 + 16x + 36y β 10 = 0 c. 3x 2 + 3y 2 β 2x β 6y β 4 = 0 d. β25x 2 β 4y 2 + 100x + 16y + 20 = 0 Solution a. Rewriting the general form, we have Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 4x 2 + 0xy + (β9)y 2 + 36x + 36y + (β125) = 0 A = 4 and C = β9, so we observe that A and C have opposite signs. The graph of this equation is a hyperbola. b. Rewriting the general form, we have Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 0x 2 + 0xy + 9y 2 + 16x + 36y + (β10) = 0 A = 0 and C = 9. We can determine that the equation is a parabola, since A is zero. c. Rewriting the general form, we have Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 3x 2 + 0xy + 3y 2 + (β2)x + (β6)y + (β4) = 0 A = 3 and C = 3. Because A = C, the graph of this equation is a circle. d. Rewriting the general form |
, we have Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 (β25)x 2 + 0xy + (β4)y 2 + 100x + 16y + 20 = 0 A = β25 and C = β4. Because AC > 0 and A β C, the graph of this equation is an ellipse. Try It #1 Identify the graph of each of the following nondegenerate conic sections. a. 16y 2 β x 2 + x β 4y β 9 = 0 b. 16x 2 + 4y 2 + 16x + 49y β 81 = 0 912 CHAPTER 10 analytic geometry Finding a New Representation of the Given Equation after Rotating through a Given Angle Until now, we have looked at equations of conic sections without an xy term, which aligns the graphs with the x- and y- axes. When we add an xy term, we are rotating the conic about the origin. If the x- and y-axes are rotated through an angle, say ΞΈ, then every point on the plane may be thought of as having two representations: (x, y) on the Cartesian plane with the original x-axis and y-axis, and (xβ², yβ²) on the new plane defined by the new, rotated axes, called the xβ²-axis and yβ²-axis. See Figure 3. y yβ xβ ΞΈ x Figure 3 The graph of the rotated ellipse x 2 + y 2 β xy β 15 = 0 We will find the relationships between x and y on the Cartesian plane with xβ² and yβ² on the new rotated plane. See Figure 4. y ΞΈ y' cos ΞΈ βsin ΞΈ x' sin ΞΈ ΞΈ cos ΞΈ x Figure 4 The Cartesian plane with x- and y-axes and the resulting xβ²β and yβ²βaxes formed by a rotation by an angle ΞΈ. The original coordinate x- and y-axes have unit vectors i and j. The rotated coordinate axes have unit vectors iβ² and jβ². The angle ΞΈ is known as the angle of rotation. See Figure 5. We may write the new unit vectors in terms of the original ones. iβ² = cos ΞΈi + sin ΞΈ j jβ² = βsin ΞΈi + cos ΞΈ j y j y' j' cos οΏ½ |
οΏ½ βsin ΞΈ x' sin ΞΈ x i' i ΞΈ cos ΞΈ Figure 5 Relationship between the old and new coordinate planes. SECTION 10.4 rotation oF axis 913 Consider a vector u in the new coordinate plane. It may be represented in terms of its coordinate axes. u = xβ² iβ² + yβ² jβ² u = xβ²(i cos ΞΈ + j sin ΞΈ) + yβ²( β i sin ΞΈ + j cos ΞΈ) u = ix'cos ΞΈ + jx'sin ΞΈ β iy' sin ΞΈ + jy' cos ΞΈ Substitute. Distribute. u = ix'cos ΞΈ β iy' sin ΞΈ + jx'sin ΞΈ + jy' cos ΞΈ Apply commutative property. u = (x'cos ΞΈ β y' sin ΞΈ)i + (x' sin ΞΈ + y' cos ΞΈ) j Factor by grouping. Because u = xβ² iβ² + yβ² jβ², we have representations of x and y in terms of the new coordinate system. x = xβ² cos ΞΈ β yβ² sin ΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ equations of rotation If a point (x, y) on the Cartesian plane is represented on a new coordinate plane where the axes of rotation are formed by rotating an angle ΞΈ from the positive x-axis, then the coordinates of the point with respect to the new axes are (xβ², yβ²). We can use the following equations of rotation to define the relationship between (x, y) and (xβ², yβ²): x = xβ² cos ΞΈ β yβ² sin ΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ How Toβ¦ Given the equation of a conic, find a new representation after rotating through an angle. 1. Find x and y where x = xβ² cos ΞΈ β yβ² sin ΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ. 2. Substitute the expression for x and y into in the given equation, then simplify. 3. Write the equations with xβ² and yβ² in standard form. Example 2 Finding a New Representation of an Equation after Rotating through a Given Angle Find a new representation of the equation 2x 2 β xy + 2y 2 β 30 = 0 after rotating through an |
angle of ΞΈ = 45Β°. Solution Find x and y, where x = xβ² cos ΞΈ β yβ² sin ΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ. Because ΞΈ = 45Β°, and x = xβ² cos(45Β°) β yβ² sin(45Β°) ξͺ β yβ² ξ’ 1 _ β 2 β xβ² β yβ² sin(45Β°) + yβ² cos(45Β°) ξͺ + yβ² ξ’ 1 _ β 2 β xβ² + yβ² _ 2 β β y = Substitute x = xβ² cosΞΈ β yβ² sinΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ into 2x 2 β xy + 2y 2 β 30 = 0. 2 ξ’ 2 xβ² β yβ² _ 2 β ξͺ β β ξ’ xβ² β yβ² _ 2 β ξͺ ξ’ xβ² + yβ² + yβ² _ β 30 = 0 2 β ξͺ β 914 Simplify. CHAPTER 10 analytic geometry / / 2 (xβ² β yβ²)(xβ² β yβ²) __ β / 2 (xβ² + yβ²)(xβ² + yβ²) __ / 2 β 30 = 0 + / 2 (xβ² β yβ²)(xβ² + yβ²) __ 2 (xβ² 2 β yβ² 2) _ 2 xβ² 2 ξ β2xβ² yβ² + yβ² 2 β + xβ² 2 ξ + 2xβ² yβ² + yβ² 2 β 30 = 0 2xβ² 2 + 2yβ² 2 β (xβ² 2 β yβ² 2) _ 2 = 30 FOIL method Combine like terms. Combine like terms. 2 ξ’ 2xβ² 2 + 2yβ² 2 β (xβ² 2 β yβ² 2) _ 2 ξͺ = 2(30) Multiply both sides by 2. 4xβ² 2 + 4yβ² 2 β (xβ² 2 β yβ² 2) = 60 Simplify. 4xβ² 2 + 4yβ² 2 β xβ² 2 + yβ² 2 = 60 Distribute. 3xβ² 2 _ 60 + 5yβ² 2 _ 60 = 60 _ 60 Set equal to 1. Write the equations with xβ² and yβ² in the standard form. This equation is an ellipse |
. Figure 6 shows the graph. xβ² 2 _ 20 + yβ² 2 _ 12 = 1 xβ ΞΈ = 45Β° 2 3 1 4 x yβ β4 β3 β2 β1 4 3 2 1 0 β1 β2 β3 β4 Figure 6 Writing equations of Rotated Conics in Standard Form Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the xβ² and yβ² coordinate system without the xβ²yβ² term, by rotating the axes by a measure of ΞΈ that satisfies We have learned already that any conic may be represented by the second degree equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. cot(2ΞΈ) = A β C ______ B where A, B, and C are not all zero. However, if B β 0, then we have an xy term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle ΞΈ where cot(2ΞΈ) = A β C ______. B β’ If cot(2ΞΈ) > 0, then 2ΞΈ is in the first quadrant, and ΞΈ is between (0Β°, 45Β°). β’ If cot(2ΞΈ) < 0, then 2ΞΈ is in the second quadrant, and ΞΈ is between (45Β°, 90Β°). β’ If A = C, then ΞΈ = 45Β°. How Toβ¦ Given an equation for a conic in the xβ² yβ² system, rewrite the equation without the xβ² yβ² term in terms of xβ² and yβ², where the xβ² and yβ² axes are rotations of the standard axes by ΞΈ degrees. SECTION 10.4 rotation oF axis 915 1. Find cot(2ΞΈ). 2. Find sin ΞΈ and cos ΞΈ. 3. Substitute sin ΞΈ and cos ΞΈ into x = xβ² cos ΞΈ β yβ² sin ΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ. 4. Substitute the expression for x and y into |
in the given equation, and then simplify. 5. Write the equations with xβ² and yβ² in the standard form with respect to the rotated axes. Example 3 Rewriting an Equation with respect to the xΒ΄ and yΒ΄ axes without the xΒ΄yΒ΄ Term Rewrite the equation 8x 2 β 12xy + 17y 2 = 20 in the xβ² yβ² system without an xβ² yβ² term. Solution First, we find cot(2ΞΈ). See Figure 7. 8x 2 β 12xy + 17y 2 = 20 β A = 8, B = β 12 and C = 17 cot(2ΞΈ) = cot(2ΞΈ) = = A β C ______ B β9 ____ β12 3 __ = 4 8 β 17 ______ β12 y h 2ΞΈ 3 Figure 7 4 x 3 __ = cot(2ΞΈ) = 4 adjacent _ opposite 32 + 42 = h2 9 + 16 = h2 25 = h2 h = 5 ββββ βββββ So the hypotenuse is Next, we find sin ΞΈ and cos ΞΈ. ___________ 1 β cos(2ΞΈ) __________ 2 = β 3 __ 1 β 5 _ 2 = β 3 5 __ __ β 5 5 _ 2 = β _________ 5 β 3 1 = β __ _____ β
5 2 ___ 2 = β __ 10 __ 1 __ 5 ββββ βββββ = β 3 __ 1 + 5 _ 2 = β 3 5 __ __ + 5 5 _ 2 = β _________ 5 + 3 1 = β __ _____ β
5 2 ___ 8 = β _ 10 __ 4 __ 5 sin ΞΈ = β sin ΞΈ = cos ΞΈ = β cos β ___________ 1 + cos(2ΞΈ) __________ 2 Substitute the values of sin ΞΈ and cos ΞΈ into x = xβ² cos ΞΈ β yβ² sin ΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ. x = xβ² cos ΞΈ β yβ² sin ΞΈ and 1 2 _ _ ξͺ ξͺ β yβ² ξ’ x = xβ² ξ’ β β 5 5 β β 2xβ² β yβ² sin ΞΈ + yβ² cos ΞΈ 916 CHAPTER 10 analytic geometry Substitute the expressions for x and |
y into in the given equation, and then simplify. 2 1 _ _ ξͺ ξͺ + yβ² ξ’ y = xβ² ξ’ β β 5 5 β β xβ² + 2yβ² _ 5 y = β β 2 2xβ² β yβ² ξͺ 8 ξ’ _ 5 β β β 12 ξ’ 2xβ² β yβ² ξͺ ξ’ _ 5 β xβ² + 2yβ² _ 5 β β β ξͺ + 17 ξ’ 2 xβ² + 2yβ² ξͺ _ = 20 5 β β 8 ξ’ (2xβ² β yβ²)(2xβ² β yβ²) __ 5 ξͺ = 20 8 (4xβ² 2 β 4xβ² yβ² + yβ² 2) β 12(2xβ² 2 + 3xβ² yβ² β 2yβ² 2) + 17(xβ² 2 + 4xβ² yβ² + 4yβ² 2) = 100 (2xβ² β yβ²)(xβ² + 2yβ²) __ 5 (xβ² + 2yβ²)(xβ² + 2yβ²) __ 5 ξͺ + 17 ξ’ ξͺ β 12 ξ’ 32xβ² 2 β 32xβ² yβ² + 8yβ² 2 β 24xβ² 2 β 36xβ² yβ² + 24yβ² 2 + 17xβ² 2 + 68xβ² yβ² + 68yβ² 2 = 100 25xβ² 2 + 100yβ² 2 = 100 100 ___ 100 Write the equations with xβ² and yβ² in the standard form with respect to the new coordinate system. 100 ___ 100 yβ² 2 = xβ² 2 + 25 ____ 100 Figure 8 shows the graph of the ellipse. xβ² 2 _ 4 + y3 β2 β1 1 2 3 x β1 β2 Figure 8 Try It #2 Rewrite the 13x 2 β 6 β β 3 xy + 7y 2 = 16 in the xβ² yβ² system without the xβ² yβ² term. Example 4 Graphing an Equation That Has No xΒ΄yΒ΄ Terms Graph the following equation relative to the xβ² yβ² system: x 2 + 12xy β 4y 2 = 30 Solution First, we find cot(2ΞΈ). x 2 + 12xy β 4y 2 = 20 β A = 1, B = 12, and C |
= β4 cot(2ΞΈ) = A β C ______ B 1 β (β4) ________ 12 cot(2ΞΈ) = cot(2ΞΈ) = 5 ___ 12 SECTION 10.4 rotation oF axis 917 Because cot(2ΞΈ) =, we can draw a reference triangle as in Figure 9. 5 __ 12 y 12 cot(2ΞΈ) = 5 12 x 2ΞΈ 5 Figure 9 cot(2ΞΈ) = = 5 ___ 12 adjacent _ opposite Thus, the hypotenuse is 52 + 122 = h2 25 + 144 = h2 169 = h2 h = 13 Next, we find sin ΞΈ and cos ΞΈ. We will use half-angle identities. ββββ βββββ sin ΞΈ = β ___________ 1 β cos(2ΞΈ) __________ 2 cos ΞΈ = β ___________ 1 + cos(2ΞΈ) __________ 2 = β = β 5 __ 1 β 13 _ 2 ββββ 5 __ 1 + 13 _ 2 = β = β 5 _ 13 13 _ β 13 _ 2 βββββ = β ______ 8 1 __ _____ = β
2 13 2 _ β β 13 5 _ 13 13 _ + 13 _ 2 = β ______ 18 1 __ ___ = β
2 13 3 _ β β 13 Now we find x and y. x = xβ² cos ΞΈ β yβ² sin ΞΈ and ξͺ β yβ² ξ’ ξͺ 2 _ β 13 β x = xβ² ξ’ 3 _ β 13 β 3xβ² β 2yβ² _ 13 β β x = y = xβ² sin ΞΈ + yβ² cos ΞΈ ξͺ + yβ² ξ’ ξͺ 3 _ β 13 β y = xβ² ξ’ 2 _ β 13 β 2xβ² + 3yβ² _ 13 β β y = Now we substitute x = and y = into x 2 + 12xy β 4y 2 = 30. β 3xβ² β 2yβ² _ 13 β 2 3xβ² β 2yβ² _ 13 ξͺ β β ξ’ β 2xβ² + 3yβ² _ 13 β 3xβ² β 2yβ² _ 13 ξͺ ξ’ β β + 12 ξ’ 2xβ² + |
3yβ² _ 13 ξͺ β 4 ξ’ β β 2 2xβ² + 3yβ² _ = 30 13 ξͺ β β 918 CHAPTER 10 analytic geometry 1 ξ’ ___ 13 ξͺ [(3xβ² β 2yβ² )2 + 12(3xβ² β 2yβ² )(2xβ² + 3yβ² ) β 4 (2xβ² + 3yβ² )2] = 30 1 ξͺ [9xβ² 2 β 12xβ² yβ² + 4yβ² 2 + 12 (6xβ² 2 + 5xβ² yβ² β 6yβ² 2) β 4 (4xβ² 2 + 12xβ² yβ² + 9yβ² 2)] = 30 ξ’ ___ 13 1 ξͺ [9xβ² 2 β 12xβ² yβ² + 4yβ² 2 + 72xβ² 2 + 60xβ² yβ² β 72yβ² 2 β 16xβ² 2 β 48xβ² yβ² β 36yβ² 2] = 30 ξ’ ___ 13 ξͺ [65xβ² 2 β 104yβ² 2] = 30 1 ξ’ ___ 13 Factor. Multiply. Distribute. Combine like terms. Figure 10 shows the graph of the hyperbola xβ² 2 _ 6 β 4yβ² 2 _ 15 = 1. 65xβ² 2 β 104yβ² 2 = 390 Multiply. xβ² 2 _ β 6 4yβ² 2 _ 15 = 1 Divide by 390. xβ 1 2 3 4 5 x yβ β5 β4 β3 β2 β1 y 5 4 3 2 1 0 β1 β2 β3 β4 β5 Identifying Conics without Rotating Axes Figure 10 Now we have come full circle. How do we identify the type of conic described by an equation? What happens when the axes are rotated? Recall, the general form of a conic is If we apply the rotation formulas to this equation we get the form Aβ² xβ² 2 + Bβ²xβ²yβ² + Cβ²yβ² 2 + Dβ²xβ² + Eβ²yβ² + Fβ² = 0 Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 It may be shown that B2 β 4AC = Bβ² 2 β 4Aβ² Cβ². The expression does not vary after rotation, so we call the expression invariant. The discriminant, B2 β 4AC |
, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section. using the discriminant to identify a conic If the equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 is transformed by rotating axes into the equation Aβ²xβ² 2 + Bβ²xβ²yβ² + Cβ²yβ² 2 + Dβ²xβ² + Eβ²yβ² + Fβ² = 0, then B2 β 4AC = Bβ² 2 β 4Aβ²Cβ². The equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these. If the discriminant, B2 β 4AC, is β’ < 0, the conic section is an ellipse β’ = 0, the conic section is a parabola β’ > 0, the conic section is a hyperbola SECTION 10.4 rotation oF axis 919 Example 5 Identifying the Conic without Rotating Axes Identify the conic for each of the following without rotating axes. a. 5x 2 + 2 β β b. 5x 2 + 2 β 3 xy + 2y 2 β 5 = 0 3 xy + 12y 2 β 5 = 0 β Solution a. Letβs begin by determining A, B, and C xy + 2 C { y 2 β 5 = 0 Now, we find the discriminant. β B2 β 4AC = ξ’ 2 β 2 3 ξͺ β 4(5)(2) = 4(3) β 40 = 12 β 40 = β 28 < 0 3 xy + 2y 2 β 5 = 0 represents an ellipse. β Therefore, 5x 2 + 2 β b. Again, letβs begin by determining A, B, and. β 3 x 2 + 2 β B Now, we find the discriminant. 5 A { xy + 12 C { y 2 β 5 = 0 B2 β 4AC = ξ’ 2 β β 2 3 ξͺ β 4(5)(12) = 4(3) β 240 = 12 β 240 = β 228 < 0 3 xy + 12y 2 β 5 = 0 represents an ellipse. β Therefore, 5x 2 + |
2 β Try It #3 Identify the conic for each of the following without rotating axes. a. x 2 β 9xy + 3y 2 β 12 = 0 b. 10x 2 β 9xy + 4y 2 β 4 = 0 Access this online resource for additional instruction and practice with conic sections and rotation of axes. β’ Introduction to Conic Sections (http://openstaxcollege.org/l/introconic) 920 CHAPTER 10 analytic geometry 10.4 SeCTIOn exeRCISeS VeRBAl 1. What effect does the xy term have on the graph of a conic section? 3. If the equation of a conic section is written in the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0, and B2 β 4AC > 0, what can we conclude? 2. If the equation of a conic section is written in the form Ax 2 + By 2 + Cx + Dy + E = 0 and AB = 0, what can we conclude? 4. Given the equation ax 2 + 4x + 3y 2 β 12 = 0, what can we conclude if a > 0? 5. For the equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0, the value of ΞΈ that satisfies cot(2ΞΈ) = gives us A β C ______ B what information? AlGeBRAIC For the following exercises, determine which conic section is represented based on the given equation. 6. 9x 2 + 4y 2 + 72x + 36y β 500 = 0 8. 2x 2 β 2y 2 + 4x β 6y β 2 = 0 10. 4y 2 β 5x + 9y + 1 = 0 7. x 2 β 10x + 4y β 10 = 0 9. 4x 2 β y 2 + 8x β 1 = 0 11. 2x 2 + 3y 2 β 8x β 12y + 2 = 0 12. 4x 2 + 9xy + 4y 2 β 36y β 125 = 0 13. 3x 2 + 6xy + 3y 2 β 36y β 125 = 0 14. β3x 2 + 3 β β 3 xy β 4y 2 + 9 = 0 16. βx 2 + 4 β β 2 xy + 2y 2 β 2y + 1 = 0 15. 2x 2 + 4 |
β β 3 xy + 6y 2 β 6x β 3 = 0 17. 8x 2 + 4 β β 2 xy + 4y 2 β 10x + 1 = 0 For the following exercises, find a new representation of the given equation after rotating through the given angle. 18. 3x 2 + xy + 3y 2 β 5 = 0, ΞΈ = 45Β° 20. 2x 2 + 8xy β 1 = 0, ΞΈ = 30Β° 22. 4x 2 + β 2 xy + 4y 2 + y + 2 = 0, ΞΈ = 45Β° β 19. 4x 2 β xy + 4y 2 β 2 = 0, ΞΈ = 45Β° 21. β2x 2 + 8xy + 1 = 0, ΞΈ = 45Β° For the following exercises, determine the angle ΞΈ that will eliminate the xy term and write the corresponding equation without the xy term. 23. x 2 + 3 β β 3 xy + 4y 2 + y β 2 = 0 β 25. 9x 2 β 3 β 27. 16x 2 + 24xy + 9y 2 + 6x β 6y + 2 = 0 3 xy + 6y 2 + 4y β 3 = 0 29. x 2 + 4xy + y 2 β 2x + 1 = 0 GRAPHICAl 24. 4x 2 + 2 β β 3 xy + 6y 2 + y β 2 = 0 β 3 xy β 2y 2 β x = 0 26. β3x 2 β β 28. x 2 + 4xy + 4y 2 + 3x β 2 = 0 30. 4x 2 β 2 β β 3 xy + 6y 2 β 1 = 0 For the following exercises, rotate through the given angle based on the given equation. Give the new equation and graph the original and rotated equation. 31. y = β x 2, ΞΈ = β 45Β° 32. x = y 2, ΞΈ = 45Β° 33, ΞΈ = 45Β° 1 y 2 _ 16 34. + x 2 _ = 1, ΞΈ = 45Β° 9 37. x = (y β 1)2, ΞΈ = 30Β° 35. y 2 β x 2 = 1, ΞΈ = 45Β° y 2 _ = 1, ΞΈ = 30Β° 4 x 2 _ 9 + 38. 36. y = x 2 _, ΞΈ |
= 30Β° 2 SECTION 10.4 section exercises 921 For the following exercises, graph the equation relative to the xβ² yβ² system in which the equation has no xβ² yβ² term. 39. xy = 9 41. x 2 β 10xy + y 2 β 24 = 0 43. 6x 2 + 2 β β 3 xy + 4y 2 β 21 = 0 45. 21x 2 + 2 β 3 xy + 19y 2 β 18 = 0 β 47. 16x 2 + 24xy + 9y 2 β 60x + 80y = 0 49. 4x 2 β 4xy + y 2 β 8 β β 5 x β 16 β β 5 y = 0 40. x 2 + 10xy + y 2 β 6 = 0 42. 4x 2 β 3 β β 3 xy + y 2 β 22 = 0 44. 11x 2 + 10 β β 3 xy + y 2 β 64 = 0 46. 16x 2 + 24xy + 9y 2 β 130x + 90y = 0 48. 13x 2 β 6 β β 3 xy + 7y 2 β 16 = 0 For the following exercises, determine the angle of rotation in order to eliminate the xy term. Then graph the new set of axes. 50. 6x 2 β 5 β β 3 xy + y 2 + 10x β 12y = 0 52. 6x 2 β 8 β β 3 xy + 14y 2 + 10x β 3y = 0 51. 6x 2 β 5xy + 6y 2 + 20x β y = 0 53. 4x 2 + 6 β β 3 xy + 10y 2 + 20x β 40y = 0 54. 8x 2 + 3xy + 4y 2 + 2x β 4 = 0 55. 16x 2 + 24xy + 9y 2 + 20x β 44y = 0 For the following exercises, determine the value of k based on the given equation. 56. Given 4x 2 + kxy + 16y 2 + 8x + 24y β 48 = 0, 57. Given 2x 2 + kxy + 12y 2 + 10x β 16y + 28 = 0, find k for the graph to be a parabola. find k for the graph to be an ellipse. 58. Given 3x 2 + kxy + 4y 2 β 6x + 20y + |
128 = 0, 59. Given kx 2 + 8xy + 8y 2 β 12x + 16y + 18 = 0, find k for the graph to be a hyperbola. find k for the graph to be a parabola. 60. Given 6x 2 + 12xy + ky 2 + 16x + 10y + 4 = 0, find k for the graph to be an ellipse. 922 CHAPTER 10 analytic geometry leARnInG OBjeCTIVeS In this section, you will: β’ Identify a conic in polar form. β’ Graph the polar equations of conics. β’ Define conics in terms of a focus and a directrix. 10. 5 COnIC SeCTIOnS In POlAR COORDInATeS Figure 1 Planets orbiting the sun follow elliptical paths. (credit: nASA Blueshift, Flickr) Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planetsβ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits. In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial bodyβs gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system. Identifying a Conic in Polar Form Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola x |
= 2 + y 2 shown in Figure 2. P(r, θ) D r θ Polar axis Directrix F, Focus @ pole x = 2 + y2 Figure 2 SECTION 10.5 conic sections in polar coordinates 923 In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus P(r, θ) at the pole, and a line, the directrix, which is perpendicular to the polar axis. If F is a fixed point, the focus, and D is a fixed line, the directrix, then we can let e be a fixed positive number, called the eccentricity, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points P such that e = is a conic. In other words, we can define PF ___ PD a conic as the set of all points P with the property that the ratio of the distance from P to F to the distance from P to D is equal to the constant e. For a conic with eccentricity e, ⒠if 0 †e < 1, the conic is an ellipse ⒠if e = 1, the conic is a parabola ⒠if e > 1, the conic is an hyperbola With this definition, we may now define a conic in terms of the directrix, x = ± p, the eccentricity e, and the angle θ. Thus, each conic may be written as a polar equation, an equation written in terms of r and θ. the polar equation for a conic For a conic with a focus at the origin, if the directrix is x = ± p, where p is a positive real number, and the eccentricity is a positive real number e, the conic has a polar equation r = ep _ 1 ± e cos θ For a conic with a focus at the origin, if the directrix is y = ± p, where p is a positive real number, and the eccentricity is a positive real number e, the conic has a polar equation r = ep _ 1 ± e sin θ How To⦠Given the polar equation for a conic, identify |
the type of conic, the directrix, and the eccentricity. 1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form. 2. Identify the eccentricity e as the coefficient of the trigonometric function in the denominator. 3. Compare e with 1 to determine the shape of the conic. 4. Determine the directrix as x = p if cosine is in the denominator and y = p if sine is in the denominator. Set ep equal to the numerator in standard form to solve for x or y. Example 1 Identifying a Conic Given the Polar Form For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity. a. r = 6 _________ 3 + 2 sin ΞΈ b. r = 12 _________ 4 + 5 cos ΞΈ c. r = 7 _________ 2 β 2 sin ΞΈ Solution For each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and 1 __, where c is that constant. denominator by the reciprocal of the constant of the original equation, c 1 __. a. Multiply the numerator and denominator by 3 r = 1 ξͺ ξ’ __ 3 6 _ _________ = β
3 + 2sin ΞΈ 1 ξͺ ξ’ __ 3 1 ξͺ 6 ξ’ __ 3 __ = 1 1 ξͺ sin ΞΈ ξͺ + 2 ξ’ 3 ξ’ __ __ 3 3 2 __ 2 _ 1 + sin ΞΈ 3 924 CHAPTER 10 analytic geometry 2 __ Because sin ΞΈ is in the denominator, the directrix is y = p. Comparing to standard form, note that e =. 3 Therefore, from the numerator, 2 = ep 2 __ = ξ’ ξ’ __ __ __ p 3 2 2 3 = p 2 __ and the directrix is y = 3. Since e < 1, the conic is an ellipse. The eccentricity is e = 3 1 __. b. Multiply the numerator and denominator by 4 r = r = 1 ξͺ ξ’ _ 12 4 _ _ οΏ½ |
οΏ½οΏ½ 4 + 5 cos ΞΈ 1 ξͺ ξ’ _ 4 1 _ ξͺ 12 ξ’ 4 __ 1 1 _ _ ξͺ cos __ Because cos ΞΈ is in the denominator, the directrix is x = p. Comparing to standard form + cos ΞΈ 4 Therefore, from the numerator, 3 = ep 5 __ = ξ’ ξ’ __ __ __ p 4 5 5 12 ___ = p 5 5 __ and the directrix is x = Since e > 1, the conic is a hyperbola. The eccentricity is e = 4 12 ___ 5 = 2.4. 1 __. c. Multiply the numerator and denominator by sin __ r = 1 1 ξͺ sin β sin ΞΈ r = Because sine is in the denominator, the directrix is y = βp. Comparing to standard form, e = 1. Therefore, from the numerator, 7 __ = ep 2 7 __ = (1)p 2 7 __ = p 2 7 __ = β3.5. Because e = 1, the conic is a parabola. The eccentricity is e = 1 and the directrix is y = β 2 SECTION 10.5 conic sections in polar coordinates 925 Try It #1 Identify the conic with focus at the origin, the directrix, and the eccentricity for r = 2 ________. 3 β cos ΞΈ Graphing the Polar equations of Conics When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine e and, therefore, the shape of the curve. The next step is to substitute values for ΞΈ and solve for r to plot Ο __, Ο, and a few key points. Setting ΞΈ equal to 0, 2 3Ο ___ provides the vertices so we can create a rough sketch of the graph. 2 Example 2 Graphing a Parabola in Polar Form Graph r = 5 _________ 3 + 3 cos ΞΈ. Solution First, we rewrite |
the conic in standard form by multiplying the numerator and denominator by the 1 __. reciprocal of 3, which is 3 1 ξͺ 5 ξ’ _ 3 __ 1 1 ξͺ cos cos ΞΈ 5 __ 3 _ 1 + cos ΞΈ Because e = 1, we will graph a parabola with a focus at the origin. The function has a cos ΞΈ, and there is an addition sign in the denominator, so the directrix is x = p. 5 __ = ep 3 5 __ = (1)p 3 5 __ = p 3 5 __ The directrix is x =. 3 Plotting a few key points as in Table 1 will enable us to see the vertices. See Figure 3. ΞΈ A 0 r = 5 _________ 3 + 3 cos ΞΈ 5 __ β 0.83 6 B Ο __ 2 5 __ β 1.67 3 Table 1 C Ο undefined D 3Ο ___ 2 5 __ β 1.67 Directrix Figure 3 926 CHAPTER 10 analytic geometry Analysis We can check our result with a graphing utility. See Figure 4. Example 3 Graphing a Hyperbola in Polar Form 2 3 4 5 Figure 4 Graph r = 8 _________ 2 β 3 sin ΞΈ. Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the 1 __. reciprocal of 2, which is sin ΞΈ 1 ξͺ 8 ξ’ _ 2 __ 1 1 ξͺ sin β cos ΞΈ 2 3 __ Because e =, e > 1, so we will graph a hyperbola with a focus at the origin. The function has a sin ΞΈ term and there 2 is a subtraction sign in the denominator, so the directrix is y = βp. 4 = ep 3 4 = ξ’ ξͺ p __ 2 2 ξͺ = p 4 ξ’ __ 3 8 __ = p 3 8 __ The directrix is y = β. 3 Plotting a few key points as in Table 2 will enable us to see the vertices. See Figure 5. ΞΈ r = 8 _________ 2 β 3 sin ΞΈ A 0 4 B Ο __ 2 β8 Table 2 C Ο 4 D 3Ο ___ 2 8 __ = 1. 10 C D B Figure 5 SECTION 10.5 conic sections in polar coordinates 927 Example 4 Graphing an Ellipse |
in Polar Form Graph r = 10 __________ 5 β 4 cos ΞΈ. Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the 1 __. reciprocal of 5, which is 5 r = 10 _________ 5 β 4 cos ΞΈ = 1 _ ξͺ 10 ξ’ 5 __ 1 1 _ _ ξͺ cos __ 1 β sin ΞΈ 5 4 __, e < 1, so we will graph an ellipse with a focus at the origin. The function has a cos ΞΈ, and there is a Because e = 5 subtraction sign in the denominator, so the directrix is x = βp. 2 = ep 4 ξͺ p 2 = ξ’ __ 5 5 ξͺ = p 2 ξ’ __ 4 5 __ = p 2 5 __ The directrix is x = β. 2 Plotting a few key points as in Table 3 will enable us to see the vertices. See Figure 6. ΞΈ r = 10 _________ 5 β 4 cos ΞΈ A 0 10 B Ο __ 2 2 Table 3 C Ο 10 ___ 9 β 1.1 D 3Ο ___ 2 2 B C D 2 4 68 1 A 0 12 r x = β 5 2 Directrix Figure 6 928 CHAPTER 10 analytic geometry Analysis We can check our result with a graphing utility. See Figure 7. Figure 7 r = of [β3, 12, 1] by [ β4, 4, 1], ΞΈ min = 0 and ΞΈ max = 2Ο. graphed on a viewing window 10 _ 5 β 4 cos ΞΈ Try It #2 Graph r = 2 _________ 4 β cos ΞΈ. Defining Conics in Terms of a Focus and a Directrix So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation. How Toβ¦ Given the focus, eccentricity, and directrix of a conic, determine the polar equation. 1. Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of y, we use the general polar form in terms of sine. If the directrix is given in terms of x, we use the general polar form in terms of cosine. 2. Determine the sign in the denominator. |
If p < 0, use subtraction. If p > 0, use addition. 3. Write the coefficient of the trigonometric function as the given eccentricity. 4. Write the absolute value of p in the numerator, and simplify the equation. Example 5 Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix Find the polar form of the conic given a focus at the origin, e = 3 and directrix y = β 2. Solution The directrix is y = βp, so we know the trigonometric function in the denominator is sine. Because y = β2, β2 < 0, so we know there is a subtraction sign in the denominator. We use the standard form of and e = 3 and | β2 | = 2 = p. Therefore, r = ep ________ 1 β esin ΞΈ r = (3)(2) _________ 1 β 3 sin ΞΈ r = 6 _________ 1 β 3 sin ΞΈ Example 6 Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix 3 _, and directrix x = 4. Find the polar form of a conic given a focus at the origin, e = 5 Solution Because the directrix is x = p, we know the function in the denominator is cosine. Because x = 4, 4 > 0, so we know there is an addition sign in the denominator. We use the standard form of 3 __ and |4| = 4 = p. and e = 5 r = ep _________ 1 + e cos ΞΈ SECTION 10.5 conic sections in polar coordinates 929 Therefore, r = 3 ξͺ (4) ξ’ _ 5 ___________ 3 _ 1 + cos ΞΈ 5 r = 12 _ 5 ___________ 3 _ 1 + cos ΞΈ 5 r = 12 _ 5 _______________ 3 5 ξͺ + 1 ξ’ _ _ cos ΞΈ 5 5 r = 12 _ 5 ____________ 3 5 _ _ + cos ΞΈ 5 5 r = 12 _ 5 β
5 _________ 5 + 3 cos ΞΈ r = 12 _________ 5 + 3 cos ΞΈ Try It #3 Find the polar form of the conic given a focus at the origin, e = 1, and directrix x = β1. Example 7 Converting a Con |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.