5 The graph of the equation y = ax 2 + bx + c is a U-shaped curve called a parabola that opens either upward or downward, depending on whether the sign of a is positive or negative. In this section we study parabolas from a geometric rather than an algebraic point of view. We begin with the geometric definition of a parabola and show how this leads to the algebraic formula that we are already familiar with.
6 Geometric Definition of a Parabola This definition is illustrated in Figure 1. The vertex V of the parabola lies halfway between the focus and the directrix, and the axis of symmetry is the line that runs through the focus perpendicular to the directrix. Figure 1
7 Geometric Definition of a Parabola In this section we restrict our attention to parabolas that are situated with the vertex at the origin and that have a vertical or horizontal axis of symmetry. If the focus of such a parabola is the point F(0, p), then the axis of symmetry must be vertical, and the directrix has the equation y = –p. Figure 2 illustrates the case p > 0. Figure 2
8 Geometric Definition of a Parabola If P(x, y) is any point on the parabola, then the distance from P to the focus F (using the Distance Formula) is The distance from P to the directrix is By the definition of a parabola these two distances must be equal:
9 Geometric Definition of a Parabola x 2 + (y – p) 2 = 0| y + p | 2 = (y + p) 2 x 2 + y 2 – 2py + p 2 = y 2 + 2py + p 2 x 2 – 2py = 2py x 2 = 4py If p > 0, then the parabola opens upward; but if p < 0, it opens downward. When x is replaced by –x, the equation remains unchanged, so the graph is symmetric about the y-axis. Square both sides Expand Simplify
11 Equations and Graphs of Parabolas The following box summarizes about the equation and features of a parabola with a vertical axis.
12 Example 1 – Finding the Equation of a Parabola Find an equation for the parabola with vertex V(0, 0) and focus F(0, 2), and sketch its graph. Solution: Since the focus is F(0, 2), we conclude that p = 2 (so the directrix is y = –2). Thus the equation of the parabola is x 2 = 4(2)y x 2 = 8y x 2 = 4py with p = 2
13 Example 1 – Solution Since p = 2 > 0, the parabola opens upwards. See Figure 3. cont’d Figure 3
14 Example 2 – Finding the Focus and Directrix of a Parabola from Its Equation Find the focus and directrix of the parabola y = –x 2, and sketch the graph. Solution: To find the focus and directrix, we put the given equation in the standard form x 2 = –y. Comparing this to the general equation x 2 = 4py, we see that 4p = –1, so p = –. Thus the focus is F(0, – ), and the directrix is y =.
15 Example 2 – Solution The graph of the parabola, together with the focus and the directrix, is shown in Figure 4(a). We can also draw the graph using a graphing calculator as shown in Figure 4(b). cont’d (a) (b) Figure 4
16 Equations and Graphs of Parabolas Reflecting the graph in Figure 2 about the diagonal line y = x has the effect of interchanging the roles of x and y. This results in a parabola with horizontal axis. Figure 2
17 Equations and Graphs of Parabolas By the same method as before, we can prove the following properties.
18 Example 3 – A Parabola with Horizontal Axis A parabola has the equation 6x + y 2 = 0. (a) Find the focus and directrix of the parabola and sketch the graph. (b) Use a graphing calculator to draw the graph. Solution: To find the focus and directrix, we put the given equation in the standard form y 2 = –6x. Comparing this to the general equation y 2 = 4px we see that 4p = –6, so p = –. Thus the focus is F(–, 0), and the directrix is x =.
19 Example 3 – Solution Since p < 0, the parabola opens to the left. The graph of the parabola, together with the focus and the directrix, is shown in Figure 5(a) below. cont’d (a) Figure 5
20 Example 3 – Solution (b) To draw the graph using a graphing calculator, we need to solve for y. 6x + y 2 = 0 y 2 = –6x y = cont’d Subtract 6x Take square roots
21 Example 3 – Solution To obtain the graph of the parabola, we graph both functions y = and y = – as shown in Figure 5(b). cont’d (b) Figure 5
22 Equations and Graphs of Parabolas We can use the coordinates of the focus to estimate the “width” of a parabola when sketching its graph. The line segment that runs through the focus perpendicular to the axis, with endpoints on the parabola, is called the latus rectum, and its length is the focal diameter of the parabola.
23 Equations and Graphs of Parabolas From Figure 6 we can see that the distance from an endpoint Q of the latus rectum to the directrix is |2p|. Thus the distance from Q to the focus must be |2p| as well (by the definition of a parabola), so the focal diameter is |4p|. In the next example we use the focal diameter to determine the “width” of a parabola when graphing it. Figure 6
24 Example 4 – The Focal Diameter of a Parabola Find the focus, directrix, and focal diameter of the parabola y = x 2, and sketch its graph. Solution: We first put the equation in the form x 2 = 4py. y = x 2 x 2 = 2y From this equation we see that 4p = 2, so the focal diameter is 2. Solving for p gives p =, so the focus is (0, ) and the directrix is y = –. Multiply by 2, switch sides
25 Example 4 – Solution Since the focal diameter is 2, the latus rectum extends 1 unit to the left and 1 unit to the right of the focus. The graph is sketched in Figure 7. cont’d Figure 7
26 Equations and Graphs of Parabolas In the next example we graph a family of parabolas, to show how changing the distance between the focus and the vertex affects the “width” of a parabola.
27 Example 5 – A Family of Parabolas (a) Find equations for the parabolas with vertex at the origin and foci F 1 (0, ), F 2 (0, ), F 3 (0, 1) and F 4 (0, 4). (b) Draw the graphs of the parabolas in part (a). What do you conclude?
28 Example 5 – Solution (a) Since the foci are on the positive y-axis, the parabolas open upward and have equations of the form x 2 = 4py. This leads to the following equations. cont’d
29 Example 5 – Solution (b) The graphs are drawn in Figure 8. We see that the closer the focus is to the vertex, the narrower the parabola. y = 0.0625x 2 y = 0.5x 2 y = 0.25x 2 y = 2x 2 A family of parabolas Figure 8 cont’d
31 Applications Parabolas have an important property that makes them useful as reflectors for lamps and telescopes. Light from a source placed at the focus of a surface with parabolic cross section will be reflected in such a way that it travels parallel to the axis of the parabola (see Figure 9). Figure 9 Parabolic reflector
32 Applications Thus, a parabolic mirror reflects the light into a beam of parallel rays. Conversely, light approaching the reflector in rays parallel to its axis of symmetry is concentrated to the focus. This reflection property, which can be proved by using calculus, is used in the construction of reflecting telescopes.
33 Example 6 – Finding the Focal Point of a Searchlight Reflector A searchlight has a parabolic reflector that forms a “bowl,” which is 12 in. wide from rim to rim and 8 in. deep, as shown in Figure 10. If the filament of the light bulb is located at the focus, how far from the vertex of the reflector is it? Figure 10 A Parabolic reflector
34 Example 6 – Solution We introduce a coordinate system and place a parabolic cross section of the reflector so that its vertex is at the origin and its axis is vertical (see Figure 11). Then the equation of this parabola has the form x 2 = 4py. cont’d Figure 11
35 Example 6 – Solution From Figure 11 we see that the point (6, 8) lies on the parabola. We use this to find p. 6 2 = 4p(8) 36 = 32p p = The focus is F(0, ), so the distance between the vertex and the focus is = in. Because the filament is positioned at the focus, it is located in. from the vertex of the reflector. cont’d The point (6, 8) satisfies the equation x 2 = 4py