SECTION 10.4 The Hyperbola 709 An equation of the form of equation (2), x a y b 1, 2 2 2 2 − = is the equation of a hyperbola with center at the origin, foci on the x-axis at c, 0 ( ) − and c, 0 , ( ) where b c a , 2 2 2 = − and transverse axis along the x-axis. An equation of the form of equation (3), y a x b 1, 2 2 2 2 − = is the equation of a hyperbola with center at the origin, foci on the y-axis at c 0, ( ) − and c 0, , ( ) where b c a , 2 2 2 = − and transverse axis along the y-axis. Notice the difference in the forms of equations (2) and (3). When the y -term 2 is subtracted from the x -term, 2 the transverse axis is along the x-axis. When the x -term 2 is subtracted from the y -term, 2 the transverse axis is along the y-axis. Analyzing the Equation of a Hyperbola Analyze the equation y x 4 4. 2 2 − = EXAMPLE 4 Solution To put the equation in proper form, divide both sides by 4: y x 4 1 2 2 − = Since the x -term 2 is subtracted from the y -term, 2 the equation is that of a hyperbola with center at the origin and transverse axis along the y-axis. Also, comparing the equation to equation (3), note that a b 1, 4, 2 2 = = and c a b 5. 2 2 2 = + = The vertices are at a 0, 0, 1, ( ) ( ) ± = ± and the foci are at c 0, 0, 5 . ( ) ( ) ± = ± To locate points on the graph to the left and right of the foci, let y 5 = ± in the equation. Then y x x x x x 4 4 4 5 4 20 4 16 4 2 2 2 2 2 2 ( ) − = ± − = − = = = ± =± y 5 Four other points on the graph are 4, 5 ( ) ± and 4, 5 . ( ) ± − See Figure 41(a) for the graph drawn by hand. Figure 41(b) shows the graph using a TI-84 Plus CE graphing utility. Figure 41 − = y x 4 4 2 2 x y –6 6 6 –6 V1 = (0, –1) V2 = (0, 1) F1 =(0,– 5) F2 = (0, 5) (–4, 5) (4, – 5) (4, 5) (–4, – 5) (a) 5 25 28 8 (b) x2 4 Y2 5 2 1 1 x2 4 Y1 5 1 1 Finding an Equation of a Hyperbola Find an equation of the hyperbola that has one vertex at 0, 2 ( ) and foci at 0, 3 ( ) − and 0, 3 . ( ) Graph the equation. EXAMPLE 5 (continued)
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