4.3. Conics
Objectives Recognize the four basic conics: circle, ellipse, parabola, and hyperbola. Recognize, graph, and write equations of parabolas (vertex at origin). Recognize, graph, and write equations of ellipses (center at origin). Recognize, graph, and write equations of hyperbolas (center at origin).
Introduction Conic sections were discovered during the classical Greek period, 600 to 300 B.C. This early Greek study was largely concerned with the geometric properties of conics. It was not until the early 17th century that the broad applicability of conics became apparent and played a prominent role in the early development of calculus.
Introduction A conic section (or simply conic) is the intersection of a plane and a double napped cone. Notice in Figure 4.16 that in the formation of the four basic conics, the intersecting plane does not pass through the vertex of the cone. Circle Ellipse Parabola Hyperbola Basic Conics Figure 4.16
Introduction When the plane does pass through the vertex, the resulting figure is a degenerate conic, as shown in Figure 4.17. Point Line Two Intersecting Lines Degenerate Conics Figure 4.17
Introduction 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 + Bxy + Cy 2 + Dx + Ey + F = 0. However, you will study a third approach, in which each of the conics is defined as a locus (collection) of points satisfying a certain geometric property.
Introduction For example, the definition of a circle as the collection of all points (x, y) that are equidistant from a fixed point (h, k) led easily to the standard form of the equation of a circle (x h) 2 + (y k) 2 = r 2. Equation of a circle We know that the center of a circle is at (h, k) and that the radius of the circle is r.
Circle
standard form of the equation of a circle
Circle with center at origin x 2 + y 2 = r 2. Example: Find the equation of a circle whose center is at the origin and which passes (2,0).
Answer x 2 + y 2 = 2 2 because the radius is 2.
Example 1. Find the distance between the point (3,4) and (-1,2). 2. Find the equation of a circle which passes (3,4) and whose center is at (-1,2).
Answer 1. Distance d between x 1, y 1 and x 2, y 2 = x 1 x 2 2 + y 1 y 2 2 Thus, d = 3 1 2 + 4 2 2 = 3 + 1 2 + 2 2 = 4 2 + 4 = 16 + 4 = 20 2. (x + 1) 2 + (y 2) 2 = 20.
Ellipses
Ellipses The line through the foci intersects the ellipse at two points (vertices). The chord joining the vertices is the major axis, and its midpoint is the center of the ellipse. The chord perpendicular to the major axis at the center is the minor axis. (See Figure 4.21.) Figure 4.21
Ellipses You can visualize the definition of an ellipse by imagining two thumbtacks placed at the foci, as shown in Figure 4.22. If the ends of a fixed length of string are fastened to the thumbtacks and the string is drawn taut with a pencil, the path traced by the pencil will be an ellipse. Figure 4.22 The standard form of the equation of an ellipse takes one of two forms, depending on whether the major axis is horizontal or vertical.
Ellipses (a) Major axis is horizontal; minor axis is vertical. (b) Major axis is vertical; minor axis is horizontal.
Ellipses In Figure (a), note that because the sum of the distances from a point on the ellipse to the two foci is constant, it follows that (Sum of distances from (0, b) to foci) = (sum of distances from (a, 0) to foci)
Example 3 Finding the Standard Equation of an Ell Example Find the standard form of the equation of the ellipse shown in Figure 4.23. Figure 4.23
Solution From Figure 4.23, the foci occur at ( 2, 0) and (2, 0). So, the center of the ellipse is (0, 0), the major axis is horizontal, and the ellipse has an equation of the form Standard form, horizontal major axis Also from Figure 4.23, the length of the major axis is 2a = 6. This implies that a = 3.
Solution
Sketch the ellipse x 2 + 4y 2 = 16. Example
Solution By dividing by 16 on both sides, the given equation x 2 + 4y 2 = 16 is equivalent to x 2 16 + 4y2 16 = 16 16 x 2 + y2 = 1 16 4 Thus, a = 4, b = 2 andc = 2 3 Because c 2 = a 2 b 2 = 4 2 2 2 = 16 4 = 12. The foci are (2 3, 0) and ( 2 3, 0) because the major axis is x-axis.
Solution x-intercepts: (4,0) and ( 4,0) y-intercepts: (0,2) and 0, 2 Since the major axis is x-axis, the vertices are x-intercepts.
Sketch the ellipse 4x 2 + y 2 = 36. Example
Solution By dividing by 36 on both sides, the equation 4x 2 + y 2 = 36 is equivalent to 4x2 + y2 = 36. Thus, 36 36 36 x 2 9 + y2 36 = 1 We know that a=6 and b=3. c 2 = a 2 b 2 = 6 2 3 2 = 36 9 = 27. Thus, c = 3 3. Since the major axis is y-axis, the foci are (0, 3 3) and 0, 3 3.
Solution x-intercepts: (3,0) and (-3,0) y-intercepts:(0,6) and (0,-6). Since the major axis is the y-axis, the vertices are the y-intercepts.
Hyperbolas
Hyperbolas The definition of a hyperbola is similar to that of an ellipse. For an ellipse, the sum of the distances between the foci and a point on the ellipse is constant, whereas for a hyperbola, the absolute value of the difference of the distances between the foci and a point on the hyperbola is constant.
Hyperbolas d 2 d 1 is a constant. Figure 4.25(a)
Hyperbolas The graph of a hyperbola has two disconnected parts (branches). The line through the two foci intersects the hyperbola at two points (vertices). The line segment connecting the vertices is the transverse axis, and the midpoint of the transverse axis is the center of the hyperbola. See Figure 4.25(b). Figure 4.25(b)
Hyperbolas (a) Transverse axis is horizontal (b) Transverse axis is vertical
Hyperbolas Be careful when finding the foci of ellipses and hyperbolas. Notice that the relationships among a, b and c differ slightly. Finding the foci of an ellipse: c 2 = a 2 b 2 Finding the foci of a hyperbola: c 2 = a 2 + b 2
Example Find the standard form of the equation of the hyperbola with foci at ( 3, 0) and (3, 0) and vertices at ( 2, 0) and (2, 0), as shown in Figure 4.26. Figure 4.26
Solution From the graph, you can determine that c = 3, because the foci are three units from the center. Moreover, a = 2 because the vertices are two units from the center. So, it follows that b 2 = c 2 a 2 = 3 2 2 2 = 9 4 = 5.
Solution cont d Because the transverse axis is horizontal, the standard form of the equation is Standard form, horizontal transverse axis Finally, substitute a 2 = 2 2 and to obtain Write in standard form. Simplify.
Example Sketch the hyperbola 4x 2 y 2 = 16. Find the vertices, foci and asymptotes.
Solution x intercepts: Thus, x = ±2. Vertices: (2,0) and (-2,0) 4x 2 0 2 = 16 4x 2 = 16 x 2 = 4
Solution Asymptotes: To convert the given form to x2 a 2 y2 b 2 = 1, 4x 2 y 2 = 16 Thus, a = 2 and b = 4. 4x 2 16 y2 16 = 16 16 x 2 4 y2 16 = 1 Therefore, y = 4 x and y = 4 x. That is, y = 2x and y = 2x 2 2
Solution Foci: c 2 = a 2 + b 2 = 4 + 16 = 20 c = ± 20 = ±2 5
Example Find the standard form of the equation of the hyperbola that has vertices at (0, 3) and 0,3, asymptotes y = 2x and y = 2x.
Solution Since vertices are (0,-3) and (0,3), we know a = 3 in the standard form of the equation y2 x2 = 1. a 2 b 2 Thus, we need to find b in y 2 3 2 x2 b 2 = 1. Since one of the asymptotes is y = 2x and a = 3, we have b = 3 from 2 y = a x = 3 x = 2x. b b
Hyperbolas An important aid in sketching the graph of a hyperbola is the determination of its asymptotes, as shown in below. (a) Transverse axis is horizontal; conjugate axis is vertical. (b) Transverse axis is vertical; conjugate axis is horizontal. Figure 4.33
Hyperbolas Each hyperbola has two asymptotes that intersect at the center of the hyperbola. Furthermore, the asymptotes pass through the corners of a rectangle of dimensions 2a by 2b. The line segment of length 2b joining (0, b) and (0, b) (b, 0)] is the conjugate axis of the hyperbola. [or ( b, 0) and
Parabolas
Parabolas You have learned that the graph of the quadratic function f(x) = ax 2 + bx + 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.
Parabolas Note in the figure above that a parabola is symmetric with respect to its axis. Using the definition of a parabola, you can derive the following standard form of the equation of a parabola whose directrix is parallel to the x-axis or to the y- axis.
Parabolas Notice that a parabola can have a vertical or a horizontal axis. Examples of each are shown below. (a) Parabola with vertical axis (b) Parabola with horizontal axis
Example Find the focus and directrix of the parabola y = 2x 2. Then, sketch the parabola.
Solution x 2 = 1 2 y = 4py = 4 1 8 y Thus, p = 1. The focus is 8 0, 1 and directrix is y = 1. 8 8
Example 2 Finding the Standard Equation of a Parabola Example Find the standard form of the equation of the parabola with vertex at the origin and focus at (2, 0). Figure 4.19
Example 2 Finding the Standard Equation of a Parabola Solution The axis of the parabola is horizontal, passing through (0, 0) and (2, 0), as shown in Figure 4.19. So, the standard form is y 2 = 4px. Because the focus is p = 2 units from the vertex, the equation is y 2 = 4(2)x y 2 = 8x. Figure 4.19
Real Life Application of Parabola Parabolas occur in a wide variety of applications. For instance, a parabolic reflector can be formed by revolving a parabola about its axis. The resulting surface has the property that all incoming rays parallel to the axis are reflected through the focus of the parabola. This is the principle behind the construction of the parabolic mirrors used in reflecting telescopes.
Parabolas Conversely, the light rays emanating from the focus of a parabolic reflector used in a flashlight are all parallel to one another, as shown in Figure 4.20. Figure 4.20