708 CHAPTER 10 Analytic Geometry Figure 39 − = x y 16 4 1 2 2 x V1 = (–4, 0) V2 = (4, 0) y 5 4 –4 (–2 5, 1) (–2 , –1) 5 (2 , 1) (2 , –1) 5 (a) (b) –5 5 F1 = (–2 5, 0) F 2 =(2 5,0) Analyzing the Equation of a Hyperbola Analyze the equation: x y 16 4 1 2 2 − = Solution EXAMPLE 3 The given equation is of the form of equation (2), with a 16 2 = and b 4. 2 = The graph of the equation is a hyperbola with center at 0, 0 ( ) and transverse axis along the x -axis. Also, c a b 16 4 20. 2 2 2 = + = + = The vertices are at a, 0 4, 0 , ( ) ( ) ± = ± and the foci are at c, 0 2 5, 0 . ( ) ( ) ± = ± To locate the points on the graph above and below the foci, let x 2 5. = ± Then x y y y y y y 16 4 1 2 5 16 4 1 20 16 4 1 5 4 4 1 4 1 4 1 2 2 2 2 2 2 2 ( ) − = ± − = − = − = = = ± =± x 2 5 The points above and below the foci are 2 5, 1 ( ) ± and 25, 1. ( ) ± − See Figure 39(a) for the graph drawn by hand. Figure 39(b) shows the graph obtained using Desmos. THEOREM Equation of a Hyperbola: Center at ( ) 0, 0 ; Transverse Axis along the y -Axis An equation of the hyperbola with center at 0, 0 , ( ) foci at c 0, ( ) − and c 0, , ( ) and vertices at a 0, ( ) − and a 0, ( ) is y a x b b c a 1 where 2 2 2 2 2 2 2 − = = − (3) The transverse axis is the y -axis. Figure 40 shows the graph of a typical hyperbola defined by equation (3). Let’s compare equations (2) and (3). Figure 40 − = = − y a x b b c a 1, 2 2 2 2 2 2 2 x V1 5 (0, 2a) V2 5 (0, a) y F2 5 (0, c) F1 5 (0, 2c)
RkJQdWJsaXNoZXIy NjM5ODQ=