P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 900 900 Chapter 9 Conic Sections and Analytic Geometry 79. Write 4x2 - 6xy + 2y2 - 3x + 10y - 6 = 0 as a quadratic In Exercises 85–88, determine whether each statement is true or false. equation in y and then use the quadratic formula to express y If the statement is false, make the necessary change(s) to produce a in terms of x. Graph the resulting two equations using a true statement. graphing utility in a -50, 70, 10 by -30, 50, 10 viewing 85. If one branch of a hyperbola is removed from a graph, then 3 4 3 4 rectangle. What effect does the xy-term have on the graph of the branch that remains must define y as a function of x. the resulting hyperbola? What problems would you 86. All points on the asymptotes of a hyperbola also satisfy the encounter if you attempted to write the given equation in hyperbola’s equation. standard form by completing the square? x2 y2 2 87. The graph of - = 1 does not intersect the line y =- x. x2 y2 x ƒ x ƒ y ƒ y ƒ 9 4 3 80. Graph - = 1 and - = 1 in the same viewing 88. Two different hyperbolas can never share the same asymptotes. 16 9 16 9 x2 y2 rectangle. Explain why the graphs are not the same. - = 89. What happens to the shape of the graph of 2 2 1 as c a b : q, where c2 = a2 + b2? a Critical Thinking Exercises 90. Find the standard form of the equation of the hyperbola with vertices 5, -6 and (5, 6), passing through (0, 9). Make Sense? In Exercises 81–84, determine whether each 1 2 statement makes sense or does not make sense, and explain your 91. Find the equation of a hyperbola whose asymptotes are reasoning. perpendicular. 81. I changed the addition in an ellipse’s equation to subtraction and this changed its elongation from horizontal to vertical. Preview Exercises Exercises 92–94 will help you prepare for the material covered in 82. I noticed that the definition of a hyperbola closely resembles the next section. that of an ellipse in that it depends on the distances between a set of points in a plane to two fixed points, the foci. In Exercises 92–93, graph each parabola with the given equation. 92.y = x2 + 4x - 5 93. y =-3 x - 1 2 + 2 83. I graphed a hyperbola centered at the origin that had 1 2 y-intercepts, but no x-intercepts. 94. Isolate the terms involving y on the left side of the equation: 2 + + - = 84. I graphed a hyperbola centered at the origin that was y 2y 12x 23 0. symmetric with respect to the x-axis and also symmetric Then write the equation in an equivalent form by completing with respect to the y-axis. the square on the left side. Section 9.3 The Parabola Objectives At first glance, this image looks like columns of smoke rising from a ᕡ Graph parabolas with vertices fire into a starry sky.Those are, indeed, stars in the background, but at the origin. you are not looking at ordinary smoke columns. These stand almost 6 trillion miles high and are 7000 light-years from Earth—more ᕢ Write equations of parabolas than 400 million times as far away as the sun. in standard form. his NASA photograph is one of a series of ᕣ Graph parabolas with vertices stunning images captured from the ends of not at the origin. T the universe by the Hubble Space ᕤ Solve applied problems Telescope. The image shows infant involving parabolas. star systems the size of our solar system emerging from the gas and dust that shrouded their creation. Using a parabolic mirror that is 94.5 inches in diameter, the Hubble has provided answers to many of the profound mysteries of the cosmos: How big and how old is the universe? How did the galaxies come to exist? Do other Earth-like planets orbit other sun-like stars? In this section, we study parabolas and their applications, including parabolic shapes that gather distant rays of light and focus them into spectacular images. Definition of a Parabola In Chapter 2, we studied parabolas, viewing them as graphs of quadratic functions in the form y = a x - h 2 + k or y = ax2 + bx + c. 1 2 P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 901 Section 9.3 The Parabola 901 Study Tip Here is a summary of what you should already know about graphing parabolas. ax 2 ؉ bx ؉ c ؍ a(x ؊ h)2 ؉ k and y ؍ Graphing y 1. If a 7 0, the graph opens upward. If a 6 0, the graph opens downward. 2. The vertex of y = a x - h 2 + k is h, k . 1 2 1 2 y y x = h y = a(x − h)2 + k (h, k) a < 0 x x y = a(x − h)2 + k a > 0 (h, k) x = h 2 b 3. The x-coordinate of the vertex of y = ax + bx + c is x =- . 2a Parabola Parabolas can be given a geometric definition that enables us to include graphs that open to the left or to the right, as well as those that open obliquely.The Directrix definitions of ellipses and hyperbolas involved two fixed points, the foci. By Axis of contrast, the definition of a parabola is based on one point and a line. Focus symmetry Definition of a Parabola Vertex A parabola is the set of all points in a plane that are equidistant from a fixed line, the directrix, and a fixed point, the focus, that is not on the line (see Figure 9.29). Figure 9.29 In Figure 9.29, find the line passing through the focus and perpendicular to the directrix. This is the axis of symmetry of the parabola. The point of intersection of the parabola with its axis of symmetry is called the vertex. Notice that the vertex is midway between the focus and the directrix. Standard Form of the Equation of a Parabola The rectangular coordinate system enables us y d1 to translate a parabola’s geometric definition M(−p, y) into an algebraic equation. Figure 9.30 is our P(x, y) starting point for obtaining an equation. We place the focus on the x-axis at the point d2 p, 0 . The directrix has an equation given by 1 2 x x =-p. The vertex, located midway between the focus and the directrix, is at the origin. Focus (p, 0) Directrix: x = −p What does the definition of a parabola tell us about the point x, y in Figure 9.30? 1 2 For any point x, y on the parabola, the 1 2 distance d1 to the directrix is equal to the distance d to the focus.Thus, the point x, y Figure 9.30 2 1 2 is on the parabola if and only if d1 = d2 . 4 x + p 2 + y - y 2 = 4 x - p 2 + y - 0 2 Use the distance formula. 1 2 1 2 1 2 1 2 x + p 2 = x - p 2 + y2 Square both sides of the 1 2 1 2 equation. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 902 902 Chapter 9 Conic Sections and Analytic Geometry x2 + 2px + p2 = x2 - 2px + p2 + y2 Square x + p and x - p. 2px =-2px + y2 Subtract x 2 + p2 from both sides of the equation. y2 = 4px Solve for y 2. This last equation is called the standard form of the equation of a parabola with its vertex at the origin. There are two such equations, one for a focus on the x-axis and one for a focus on the y-axis. Standard Forms of the Equations of a Parabola The standard form of the equation of a parabola with vertex at the origin is y2 = 4px or x2 = 4py. Figure 9.31(a) illustrates that for the equation on the left, the focus is on the x-axis, which is the axis of symmetry. Figure 9.31(b) illustrates that for the equation on the right, the focus is on the y-axis, which is the axis of symmetry. y y x2 = 4py y2 = 4px Focus (0, p) Vertex Vertex Study Tip x x Focus (p, 0) It is helpful to think of p as the Directrix: x = −p directed distance from the vertex to Directrix: y = −p the focus. If p 7 0, the focus lies p units to the right of the vertex or p units above the vertex. If p 6 0, the focus lies ƒ p ƒ units to the left of the Figure 9.31(a) Parabola with the x-axis as the Figure 9.31(b) Parabola with the y-axis axis of symmetry. If p 7 0, the graph opens to the as the axis of symmetry. If p 7 0, the graph vertex or ƒ p ƒ units below the vertex. right. If p 6 0, the graph opens to the left. opens upward. If p 6 0, the graph opens downward. ᕡ Graph parabolas with vertices at Using the Standard Form of the Equation of a Parabola the origin. We can use the standard form of the equation of a parabola to find its focus and directrix. Observing the graph’s symmetry from its equation is helpful in locating the focus. y2=4px x2=4py The equation does not change if The equation does not change if y is replaced with −y.