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Introduction to Conics: Parabolas 735 333202_1002.qxd 12/8/05 9:00 AM Page 735 Section 10.2 Introduction to Conics: Parabolas 735 10.2 Introduction to Conics: Parabolas What you should learn •Recognize a conic as the Conics intersection of a plane and Conic sections were discovered during the classical Greek period, 600 to 300 B.C. a double-napped cone. The early Greeks were concerned largely with the geometric properties of conics. •Write equations of parabolas It was not until the 17th century that the broad applicability of conics became in standard form and graph apparent and played a prominent role in the early development of calculus. parabolas. A conic section (or simply conic) is the intersection of a plane and a double- •Use the reflective property of napped cone. Notice in Figure 10.8 that in the formation of the four basic parabolas to solve real-life conics, the intersecting plane does not pass through the vertex of the cone. When problems. the plane does pass through the vertex, the resulting figure is a degenerate conic, Why you should learn it as shown in Figure 10.9. Parabolas can be used to model and solve many types of real-life problems. For instance, in Exercise 62 on page 742, a parabola is used to model the cables of the Golden Gate Bridge. Circle Ellipse Parabola Hyperbola FIGURE 10.8 Basic Conics Cosmo Condina/Getty Images Point Line Two Intersecting FIGURE 10.9 Degenerate Conics Lines There are several ways to approach the study of conics. You could begin by defining conics in terms of the intersections of planes and cones, as the Greeks did, or you could define them algebraically, in terms of the general second- degree equation Ax 2 ϩ Bxy ϩ Cy 2 ϩ Dx ϩ Ey ϩ F ϭ 0. This study of conics is from a locus-of- However, you will study a third approach, in which each of the conics is defined points approach, which leads to the development of the standard equation as a locus (collection) of points satisfying a geometric property. For example, in for each conic.Your students should Section 1.2, you learned that a circle is defined as the collection of all points know the standard equations of all ͑x, y͒ that are equidistant from a fixed point ͑h, k͒. This leads to the standard form conics well. Make sure they understand of the equation of a circle the relationship of h and k to the horizontal and vertical shifts. ͑x Ϫ h͒ 2 ϩ ͑y Ϫ k͒ 2 ϭ r 2. Equation of circle 333202_1002.qxd 12/8/05 9:00 AM Page 736 736 Chapter 10 Topics in Analytic Geometry Parabolas In Section 2.1, you learned that the graph of the quadratic function f ͑x͒ ϭ ax2 ϩ bx ϩ c is a parabola that opens upward or downward. The following definition of a parabola is more general in the sense that it is independent of the orientation of the parabola. Definition of Parabola A parabola is the set of all points ͑x, y͒ in a plane that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line. y The midpoint between the focus and the directrix is called the vertex, and the line passing through the focus and the vertex is called the axis of the parabola. Note in Figure 10.10 that a parabola is symmetric with respect to its axis. Using d2 Focus the definition of a parabola, you can derive the following standard form of the d1 equation of a parabola whose directrix is parallel to the x-axis or to the y-axis. d2 Vertex d1 Standard Equation of a Parabola Directrix The standard form of the equation of a parabola with vertex at ͑h, k͒ is as x follows. ͑x Ϫ h͒2 ϭ 4p͑y Ϫ k͒, p 0 Vertical axis, directrix: y ϭ k Ϫ p FIGURE 10.10 Parabola ͑y Ϫ k͒2 ϭ 4p͑x Ϫ h͒, p 0 Horizontal axis, directrix: x ϭ h Ϫ p The focus lies on the axis p units (directed distance) from the vertex. If the vertex is at the origin ͑0, 0͒, the equation takes one of the following forms. x 2 ϭ 4py Vertical axis y 2 ϭ 4px Horizontal axis See Figure 10.11. For a proof of the standard form of the equation of a parabola, see Proofs in Mathematics on page 807. Axis: Axis: x = h Directrix: Directrix: x = h − p xh= Directrix: y = k − p xhp= − p < 0 > Focus: Vertex: (h, k) p 0 (,h k + p ) p < 0 Focus: (h + p, k) Axis: Axis: > p 0 Focus: y=k= y = k hp k Vertex: Directrix: (+,) Focus: Vertex: (,h k ) ykp= − (h, k + p) Vertex: (h , k ) (h, k) 4pͧx ؊ hͨ ؍ 4pͧx ؊ hͨ (d) ͧy ؊ kͨ2 ؍ 4pͧy ؊ kͨ (c) ͧy ؊ kͨ2 ؍ 4pͧy ؊ kͨ (b) ͧx ؊ hͨ2 ؍ a) ͧx ؊ hͨ2) Vertical axis: p > 0 Vertical axis: p < 0 Horizontal axis: p > 0 Horizontal axis: p < 0 FIGURE 10.11 333202_1002.qxd 12/8/05 9:00 AM Page 737 Section 10.2 Introduction to Conics: Parabolas 737 Technology Example 1 Vertex at the Origin Use a graphing utility to confirm Find the standard equation of the parabola with vertex at the origin and focus the equation found in Example 1. ͑2, 0͒. In order to graph the equation, you may have to use two separate Solution equations: The axis of the parabola is horizontal, passing through ͑0, 0͒ and ͑2, 0͒, as shown in Figure 10.12. ϭ Ί y1 8x Upper part and y ϭϪΊ y2 8x. Lower part 2 yx2 = 8 1 Focus Vertex (2, 0) x 1234 (0, 0) −1 −2 You may want to review the technique of completing the FIGURE 10.12 square found in Appendix A.5, which will be used to rewrite So, the standard form is y 2 ϭ 4px, where h ϭ 0, k ϭ 0, and p ϭ 2. So, the each of the conics in standard equation is y 2 ϭ 8x. form. Now try Exercise 33. Example 2 Finding the Focus of a Parabola ϭ Ϫ1 2 Ϫ ϩ 1 Find the focus of the parabola given by y 2 x x 2. Solution To find the focus, convert to standard form by completing the square. ϭ Ϫ1 2 Ϫ ϩ 1 y y 2 x x 2 Write original equation. Ϫ2y ϭ x 2 ϩ 2x Ϫ 1 Multiply each side by –2. 2 Vertex (− 1, 1) 1 Ϫ 2y ϭ x 2 ϩ 2x Add 1 to each side. − 1 Focus() 1, 2 1 1 ϩ 1 Ϫ 2y ϭ x 2 ϩ 2x ϩ 1 Complete the square. 2 Ϫ 2y ϭ x2 ϩ 2x ϩ 1 Combine like terms. x − − − 3 2 11 Ϫ2͑y Ϫ 1͒ ϭ ͑x ϩ 1͒2 Standard form −1 Comparing this equation with yxx=+−−112 22 ͑x Ϫ h͒2 ϭ 4p͑ y Ϫ k͒ −2 ϭϪ ϭ ϭ Ϫ 1 you can conclude that h 1, k 1, and p 2. Because p is negative, FIGURE 10.13 the parabola opens downward, as shown in Figure 10.13. So, the focus of the ͑ ϩ ͒ ϭ ͑Ϫ 1͒ parabola is h, k p 1, 2 . Now try Exercise 21. 333202_1002.qxd 12/8/05 9:00 AM Page 738 738 Chapter 10 Topics in Analytic Geometry y Example 3 Finding the Standard Equation of a Parabola 8 − 2 − (x 2) = 12(y 1) Find the standard form of the equation of the parabola with vertex ͑2, 1͒ and 6 Focus focus ͑2, 4͒. (2, 4) 4 Solution Vertex ͑ ͒ ͑ ͒ (2, 1) Because the axis of the parabola is vertical, passing through 2, 1 and 2, 4 , consider the equation x −4 −22468 ͑x Ϫ h͒2 ϭ 4p͑ y Ϫ k͒ −2 where h ϭ 2, k ϭ 1, and p ϭ 4 Ϫ 1 ϭ 3. So, the standard form is −4 ͑x Ϫ 2͒2 ϭ 12͑ y Ϫ 1͒. FIGURE 10.14 You can obtain the more common quadratic form as follows. ͑x Ϫ 2͒2 ϭ 12͑ y Ϫ 1͒ Write original equation. x2 Ϫ 4x ϩ 4 ϭ 12y Ϫ 12 Multiply. x2 Ϫ 4x ϩ 16 ϭ 12y Add 12 to each side. 1 ͑x 2 Ϫ 4x ϩ 16͒ ϭ y Divide each side by 12. Light source 12 at focus The graph of this parabola is shown in Figure 10.14. Now try Exercise 45. Focus Axis Application A line segment that passes through the focus of a parabola and has endpoints on the parabola is called a focal chord. The specific focal chord perpendicular to the axis of the parabola is called the latus rectum. Parabolic reflector: Parabolas occur in a wide variety of applications. For instance, a parabolic Light is reflected reflector can be formed by revolving a parabola around its axis. The resulting in parallel rays. surface has the property that all incoming rays parallel to the axis are reflected FIGURE 10.15 through the focus of the parabola. This is the principle behind the construction of the parabolic mirrors used in reflecting telescopes. Conversely, the light rays Axis emanating from the focus of a parabolic reflector used in a flashlight are all parallel to one another, as shown in Figure 10.15. P A line is tangent to a parabola at a point on the parabola if the line intersects, but does not cross, the parabola at the point. Tangent lines to parabolas have spe- α Focus cial properties related to the use of parabolas in constructing reflective surfaces.
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