710 CHAPTER 10 Analytic Geometry Figure 42 − = y x 4 5 1 2 2 x V1 = (0, –2) V2 = (0, 2) y –5 5 5 2 5( ) , 3 2 5 (– ) , 3 2 5( ) , –3 2 5 (– ) , –3 F1 = (0, –3) F2 = (0, 3) –5 Solution Now Work PROBLEMS 17 AND 23 Look at the equations of the hyperbolas in Examples 3 and 5. For the hyperbola in Example 3, a 16 2 = and b 4, 2 = so a b; > for the hyperbola in Example 5, a 4 2 = and b 5, 2 = so a b. < This indicates that for hyperbolas, there are no requirements involving the relative sizes of a and b. Contrast this situation to the case of an ellipse, in which the relative sizes of a and b dictate which axis is the major axis. Hyperbolas have another feature to distinguish them from ellipses and parabolas: hyperbolas have asymptotes. 2 Find the Asymptotes of a Hyperbola Recall from Section 4.5 that a horizontal or oblique asymptote of a graph is a line with the property that the distance from the line to points on the graph approaches 0 as x →−∞ or as x . →∞ Asymptotes provide information about the end behavior of the graph of a hyperbola. THEOREM Asymptotes of a Hyperbola; Transverse Axis along the x -Axis The hyperbola x a y b 1 2 2 2 2 − = has the two oblique asymptotes y b a x y b a x and = = − (4) Proof Begin by solving for y in the equation of the hyperbola. x a y b y b x a y b x a 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 ( ) − = = − = − Since x 0, ≠ the right side can be factored and rewritten as y b x a a x y bx a a x 1 1 2 2 2 2 2 2 2 2 ( ) = − = ± − Now, as x →−∞ or as x , →∞ the term a x 2 2 approaches 0, so the expression under the radical approaches 1. So as x →−∞ or as x , →∞ the value of y approaches bx a ; ± that is, the graph of the hyperbola approaches the lines y b a x y b a x and = − = These lines are oblique asymptotes of the hyperbola. ■ Since the foci are at 0, 3 ( ) − and 0,3 , ( ) the center of the hyperbola, which is at their midpoint, is the origin. Also, the transverse axis is along the y -axis. This information tells us that c a 3, 2, = = and b c a 9 4 5. 2 2 2 = − = − = The form of the equation of the hyperbola is given by equation (3): y a x b y x 1 4 5 1 2 2 2 2 2 2 − = − = Let y 3 = ± to obtain points on the graph on either side of each focus. See Figure 42.
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