SECTION 10.4 The Hyperbola 707 Figure 37 − = x y 4 5 1 2 2 x y F2 5 (3, 0) F1 5 (23, 0) 25 5 5 25 V2 5 (2, 0) ( ) 3, V1 5 (22, 0) 5 – 2 ( ) 3, 2 5 – 2 ( ) 23, 2 5 – 2 ( ) 23, 5 – 2 equation (2), it follows that b c a 9 4 5, 2 2 2 = − = − = so an equation of the hyperbola is x y 4 5 1 2 2 − = To graph a hyperbola, it is helpful to locate and plot other points on the graph. For example, to find the points above and below the foci, let x 3. = ± Then x y y y y y y 4 5 1 3 4 5 1 9 4 5 1 5 5 4 25 4 5 2 2 2 2 2 2 2 2 ( ) − = ± − = − = = = = ± =± x 3 The points above and below the foci are 3, 5 2 ( ) ± and 3, 5 2 . ( ) ± − These points determine the “opening” of the hyperbola. See Figure 37. Using a Graphing Utility to Graph a Hyperbola Using a graphing utility, graph the hyperbola: x y 4 5 1 2 2 − = EXAMPLE 2 Solution To graph the hyperbola x y 4 5 1 2 2 − = using some graphing utilities, we need to graph the two functions Y x 5 4 1 1 2 = − and Y x 5 4 1. 2 2 = − − As with graphing circles, parabolas, and ellipses on a graphing utility, use a square screen setting so that the graph is not distorted. Figure 38(a) shows the graph of the hyperbola on a TI-84 Plus CE. Figure 38(b) Shows the graph obtained using GeoGebra. Again, note that Desmos and GeoGebra can graph equations in an implicit form. Now Work PROBLEM 21 An equation of the form of equation (2) is the equation of a hyperbola with center at the origin, foci on the x-axis at c, 0 ( ) − and c, 0 , ( ) where c a b , 2 2 2 = + and transverse axis along the x-axis. For the next two examples of this section, the direction “Analyze the equation” will mean to find the center, transverse axis, vertices, and foci of the hyperbola and graph it. (b) Figure 38 5 25 28 8 x2 4 Y1 5 2 1 5 x2 4 Y2 5 2 2 1 5 (a)
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