13.1 2/20/2018. Conic Sections. Conic Sections: Parabolas and Circles

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1 13 Conic Sections 13.1 Conic Sections: Parabolas and Circles 13.2 Conic Sections: Ellipses 13.3 Conic Sections: Hyperbolas 13.4 Nonlinear Systems of Equations 13.1 Conic Sections: Parabolas and Circles Parabolas Circles This section and the next two examine curves formed by cross sections of cones. These are all graphs of Ax 2 + By 2 + Cxy + Dx + Ey + F = 0. The constants A, B, C, D, E, and F determine which of the following shapes will serve as the graph. Slide

2 Slide 13-4 Parabolas When a cone is cut as shown in the first figure on the previous slide, the conic section formed is a parabola. Parabolas have many applications in electricity, mechanics, and optics. Slide 13-5 Equation of a Parabola A parabola with a vertical axis of symmetry opens upward or downward and has an equation that can be written in the form y = ax 2 + bx + c. A parabola with a horizontal axis of symmetry opens right or left and has an equation that can be written in the form x = ay 2 + by + c. Slide

3 Parabolas with equations of the form f (x) = ax 2 + bx + c were graphed in Chapter 11. Graph: y = x 2 6x Solution Complete the square to find the vertex: y = (x 2 6x) + 10 y = (x 2 6x + 9 9) + 10 y = (x 2 6x + 9) + ( ) y = (x 3) The vertex is at (3, 1). Slide 13-7 Solution continued Calculate and plot some points on each side of the vertex. As expected for a positive coefficient of x 2, the graph opens upward. x y Slide 13-8 A second way to find the vertex of a parabola is to recall that the x-coordinate of the vertex of the parabola given by y = ax 2 + bx + c is b/(2a). For the last example, y = x 2 6x + 10, the x-coordinate of the vertex is ( 6)/[2(1)] = 6/2 = 3. To find the y-coordinate of the vertex, we substitute 3 for x: y = 3 2 6(3) + 10 = 1. Slide

4 To Graph an Equation of the Form y = ax 2 + bx + c 1. Find the vertex (h, k) either by completing the square to find an equivalent equation y = a(x h) 2 + k; or by using b/(2a) to find the x-coordinate and substituting to find the y-coordinate. Slide Choose other values for x on each side of the vertex, and compute the corresponding y-values. 3. The graph opens upward for a > 0 and downward for a < 0. Slide Equations of the form x = ay 2 + by + c represent horizontal parabolas. These parabolas open to the right for a > 0, open to the left for a < 0, and have axes of symmetry parallel to the x-axis. Slide

5 Graph: x = y 2 + 6y 7. Solution Complete the square to find the vertex: x = y 2 + 6y 7 x = (y 2 6y ) 7 x = (y 2 6y + 9) 7 ( 9) x = (y 3) The vertex is at (2, 3) and the parabola opens to the left. Slide Solution continued To plot points, choose values for y and compute the values for x. x y Slide To Graph an Equation of the Form x = ay 2 + by + c 1. Find the vertex (h, k) either by completing the square to find an equivalent equation x = a(y k) 2 + h; or by using b/(2a) to find the y-coordinate and substituting to find the x-coordinate. Slide

6 2. Choose other values for y that are above and below the vertex, and compute the corresponding x-values. 3. The graph opens to the right if a > 0 and to the left if a < 0. Slide Circles One conic section, the circle, is a set of points in a plane that are a fixed distance r, called the radius (plural, radii), from a fixed point (h, k), called the center. If (x, y) is a point on the circle, then by the definition of a circle and the distance formula, it follows that r ( x h) ( y k). Slide Squaring both sides gives the equation of a circle in standard form. Slide

7 Equation of a Circle (Standard Form) The equation of a circle, centered at (h, k), with radius r, is given by 2 ( x h) ( y k) r. Slide Note that for h = 0 and k = 0, the circle is centered at the origin. Otherwise, the circle is translated h units horizontally and k units vertically. Slide Find an equation of the circle having center (1, 7) and radius 4. Solution Using the standard form, we obtain or (x 1) 2 + (y 7) 2 = 4 2, (x 1) 2 + (y 7) 2 = 16. Slide

8 Find the center and the radius and then graph the circle x 2 + y 2 + 2x 6y + 6 = 0. Solution Complete the square twice to write the equation in standard form. x 2 + 2x + y 2 6y = 6 x 2 + 2x y 2 6y + 9 = (x + 1) 2 + (y 3) 2 = 4 (x ( 1)) 2 + (y 3) 2 = Slide Solution continued The center is at ( 1, 3) and the radius is 2. y ( 1, 3) x Slide Conic Sections: Ellipses Ellipses Centered at (0, 0) Ellipses Centered at (h, k) 8

9 When a cone is cut at an angle, as shown below, the conic section formed is an ellipse. To draw an ellipse, tie a loose string between two tacks and draw an oval by moving the pencil while stretching the string tight. Slide Ellipses Centered at (0, 0) An ellipse is defined as the set of all points in a plane for which the sum of the distances from two fixed points F 1 and F 2 is constant. The points F 1 and F 2 are called foci, the plural of focus. The midpoint of the segment F 1 F 2 is the center. Slide Equation of an Ellipse Centered at the Origin The equation of an ellipse centered at the origin and symmetric with respect to both axes is x y 1, ab, 0. (Standard form) a b Slide

10 To graph an ellipse centered at the origin, it helps to first find the intercepts. The y- intercepts are (0, b) and (0, b). The x- intercepts are (a, 0) and ( a, 0). If a > b, the ellipse is said to be horizontal and (a, 0) and ( a, 0) are referred to as vertices (singular vertex). If b > a, the ellipse is said to be vertical and (0, b) and (0, b) are then the vertices. Slide Plotting these four points and drawing an ovalshaped curve, we graph the ellipse. If a more precise graph is desired, we can plot more points. (0, b) (0, b) (-a, 0) (a, 0) (-a, 0) (a, 0) (0, -b) (0, -b) Slide Using a and b to Graph an Ellipse For the ellipse x y 1, a b the x-intercepts are ( a, 0) and (a, 0). The y-intercepts are (0, b) and (0, b). For a 2 > b 2, the ellipse is horizontal. For b 2 > a 2, the ellipse is vertical. Slide

11 Solution Graph the ellipse Here a = 4 and b = 2. Since a > b, the ellipse is horizontal. The x-intercepts are ( 4, 0) and (4, 0), and the y-intercepts are (0, 2) and (0, 2). Plot these points and connect them with an oval-shaped curve. x y (0, 2) ( 4, 0) (4, 0) (0, 2) Slide Ellipses Centered at (h, k) Horizontal and vertical translations, similar to those used in Chapter 8, can be used to graph ellipses that are not centered at the origin. Slide Equation of an Ellipse Centered at (h, k) The standard form of a horizontal or vertical ellipse centered at (h, k) is ( x h) ( y k) 1. a b The vertices are (h + a, k) and (h a, k) if horizontal; (h, k + b) and (h, k b) if vertical. Slide

12 ( x 2) ( y 1) Graph the ellipse Solution Here a = 2 and b = 4. The center is at (2, 1). Plot 2 units to the left and right of the center, as well as 4 units above and below the center. These points are (2, 3), (2, 5), (0, 1), and (4, 1). (2, 3) (0, 1) (4, 1) (2, 1) (2, 5) Slide Ellipses have many applications. Communications satellites move in elliptical orbits with the earth as a focus while the earth itself follows an elliptical path around the sun. Slide A medical instrument, the lithotripter, uses shock waves originating at one focus to crush a kidney stone located at the other focus. Slide

13 13.3 Conic Sections: Hyperbolas Hyperbolas Hyperbolas (Nonstandard Form) Classifying Graphs of Equations Hyperbolas A hyperbola looks like a pair of parabolas, but the shapes are actually different. A hyperbola has two vertices and the line through the vertices is known as the axis. The point halfway between the vertices is called the center. The two curves that comprise a hyperbola are called branches. Slide Slide

14 Equation of a Hyperbola Centered at the Origin The equation of a hyperbola centered at the origin* has its equation as follows: x y 1 a b y x 1 b a (Horizontal axis); (Vertical axis). Slide Note that both equations have a 1 on the right-hand side and a subtraction symbol between the terms. For the discussion that follows, we assume a, b > 0. To graph a hyperbola, it helps to begin by graphing two lines called asymptotes. The asymptotes are not part of the graph, but serve as guidelines for an accurate graph. Slide Asymptotes of a Hyperbola For hyperbolas with equations as shown below, the asymptotes are the lines b b y x and y x. a a Slide

15 For hyperbolas, a and b are used to determine the base and height of a rectangle that can be used as an aid in sketching asymptotes and locating vertices. Slide Solution Graph Here a = 4 and b = 2. The asymptotes are 1 1 y x and y x. The asymptotes pass through (4, 2), (4, 2), ( 4, 2) and ( 4, 2). x y y= (-1/2) x ( 4, 0) (4, 0) x y y= (1/2) x Slide Solution Graph Here a = 1 and b = 4. The asymptotes are y 4 x and y 4 x. The asymptotes pass through (1, 4), ( 1, 4), (1, 4) and ( 1, 4). 2 y 2 x y= -4x 2 y 2 x 1 16 (0, 4) (0, -4) y= 4x Slide

16 Equation of a Hyperbola in Nonstandard Form Hyperbolas having the x- and y-axes as asymptotes have equations as follows: xy = c, where c is a nonzero constant. Slide Graph xy = 2. Solution We first solve for y: y = 2/x. Find some solutions and keep them in a table. Note that x x y y cannot be 0 ½ 4 and for large 1 2 values of x, y 2 1 will be close 4 ½ to 0. The x- -½ -4 and y- axes -1-2 serve as -2-1 asymptotes. -4 -½ x Slide Hyperbolas have many applications. A jet breaking the sound barrier creates a sonic boom with a wave front the shape of a cone. The intersection of the cone with the ground is one branch of a hyperbola. Slide

17 Some comets travel in hyperbolic orbits, and a cross section of many lenses is hyperbolic in shape. Slide Classifying Graphs of Equations Graph Standard Form Ax 2 + By 2 + Cxy + Dx + Ey + F = 0 Parabola y = ax 2 + bx + c (Vertical parabola) Either A = 0 or B = 0, but not both. x = ay 2 + by + c (Horizontal parabola) Circle x 2 + y 2 = r 2 (Center at origin) A = B (x h) 2 + (y k) 2 = r 2 (Center at (h, k)) Slide Classifying Graphs of Equations Graph Standard Form Ax 2 + By 2 + Cxy + Dx + Ey + F = 0 Ellipse x y 1; A B, and A and B have the a b same sign. (Center at origin) ( x h) ( y k) 1; a b (Center at (h, k)) Slide

18 Classifying Graphs of Equations Graph Standard Form Ax 2 + By 2 + Cxy + Dx + Ey + F = 0 Hyperbola A and B have opposite signs x y 1; a b (Horizontal hyperbola) y x 1; b a (Vertical hyperbola) xy = c (Asymptotes are axes) Only C and F are nonzero. Slide Classify the graph of each equation as a circle, an ellipse, a parabola, or a hyperbola. a. 3x 2 = 5 3y 2 b. 2x y = y 2 c. 4x 2 = 2 + y 2 d. 28 3x 2 = 7 + 6y 2 Slide Solution a. 3x 2 = 5 3y 2 Add 3y 2 to both sides: 3x 2 + 3y 2 = 5. Since A = B, this is an equation of a circle. b. 2x y = y 2 Since 2x y = y 2 has only one variable squared, the graph is a parabola. Slide

19 Solution c. 4x 2 = 2 + y 2 Subtract y 2 from both sides: 4x 2 y 2 = 2. Since A and B have opposite signs, the graph is a hyperbola. d. 28 3x 2 = 7 + 6y 2 Subtract 28 from both sides and also subtract 6y 2 from both sides: 3x 2 6y 2 = 21. Multiplying both sides by 1 we get 3x 2 + 6y 2 = 21. Since A is not equal to B and they have the same sign, the graph is an ellipse. Slide Nonlinear Systems of Equations Systems Involving One Nonlinear Equation Systems of Two Nonlinear Equations Problem Solving Systems Involving One Nonlinear Equation Suppose that a system consists of an equation of a circle and an equation of a line. In what ways can the circle and the line intersect? The figures on the next slide represent three ways in which the situation can occur. Slide

20 y y y x x 0 real solutions 1 real solution 2 real solutions Slide Recall that graphing, elimination, and substitution were all used to solve systems of linear equations. To solve systems in which one equation is of first degree and one is of second degree, it is preferable to use the substitution method. Slide Solve the system x 2 + y 2 = 20, (1) (The graph is a circle.) y 2x = 0. (2) (The graph is a line.) Solution First solve the linear equation, (2), for y: y = 2x. (3) We could have solved for x instead. Then substitute 2x for y in equation (1) and solve for x: x 2 + (2x) 2 = 20 x 2 + 4x 2 = 20 5x 2 = 20 Slide

21 Solution continued x 2 = 4 x 2. Using the principle of square roots Now substitute these numbers for x in equation (3) and solve for y: For x = 2, y = 2(2) = 4; For x = 2, y = 2( 2) = 4. Check: It is left to the student to confirm that the pairs (2, 4) and ( 2, 4) check in both equations. The pairs (2, 4) and ( 2, 4) are solutions to the system. Slide Systems of Two Nonlinear Equations We now consider systems of two seconddegree equations. Graphs of such systems can involve any two conic sections. On the next slide are some ways in which a circle and a hyperbola can intersect. Slide Slide

22 To solve systems of two second-degree equations, we either substitute or eliminate. The elimination method is generally better when both equations are of the form Ax 2 + By 2 = C. Then we can eliminate an x 2 -or y 2 -term in a manner similar to the procedure used in Chapter 8. Slide Solve the system 3x 2 + 2y 2 = 66, (1) x 2 y 2 = 7. (2) Solution Multiply equation (2) by 2 and add: 3x 2 + 2y 2 = 66 2x 2 2y 2 = 14 5x 2 = 80 x 2 = 16 x = 4 Slide Solution continued There is no x-term, and x 2 = 16 whether x = 4 or 4, thus we can substitute 4 and 4 for x simultaneously in equation (2): ( 4) 2 y 2 = 7 16 y 2 = 7 y 2 = 9 y 2 = 9 y = 3. Thus if x = 4, then y = 3 or y = 3; and if x = 4, then y = 3 or y = 3. The possible solutions are (4, 3), (4, 3), ( 4, 3), and ( 4, 3). Slide

23 Solution continued Check: 3x 2 + 2y 2 = 66 x 2 y 2 = 7 3( 4) 2 + 2( 3) 2 66 ( 4) 2 ( 3) 2 7 3(16) + 2(9) = 7 66 = 66 True True The solutions are (4, 3), (4, 3), ( 4, 3), and ( 4, 3). Slide Problem Solving We now consider applications that can be modeled by a system of equations in which at least one equation is nonlinear. Slide A picture frame with a picture takes up an area of 216 in 2. The perimeter of the frame is 60 in. Find the dimensions of the frame. Solution 1. Familiarize. We draw and label a sketch, letting l = the length and w = width, both in inches. Area = lw = 216 w Perimeter = 2l + 2w = 60 l Slide

24 2. Translate. We have the following translation: Perimeter: 2l + 2w = 60; Area: lw = Carry out. We solve the system 2l + 2w = 60, lw = 216 Completing the solution is left to the student. to get (18, 12) and (12, 18). Since length is usually considered to be longer than width, we have the solution l = 18 and w = 12. Slide Check. If l = 18 and w = 12, the perimeter is 2(12) + 2(18), or 60. The area is (18)(12), or 216. The numbers check. 5. State. The length is 18 in. and the width is 12 in. Slide