706 CHAPTER 10 Analytic Geometry [ ] ( ) ( ) ( ) ( ) ( ) ( ) − = − + − + = − + + + = + + − − = − − − = − cx a a x c y cx caxa ax cxc y c x a a x a c a y c a x a y a c a c a x a y a c a 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2 2 2 Figure 36 − = = − x a y b b c a 1, 2 2 2 2 2 2 2 F2 5 (c, 0) F1 5 (2c, 0) x y V2 5 (a, 0) V1 5 (2a, 0) Transverse axis Square both sides. Multiply. Distribute and simplify. Rearrange terms. Factor a2 on the right side. (1) To obtain points on the hyperbola off the x -axis, we must have a c. < To see why, look again at Figure 35. ( ) ( ) ( ) ( ) ( ) ( ) < + − < < < d F P d F P d F F d F P d F P d F F a c a c , , , , , , 2 2 1 2 1 2 1 2 1 2 Since a c, < we also have a c , 2 2 < so c a 0. 2 2 − > Let b c a b , 0. 2 2 2 = − > Then equation (1) can be written as b x a y a b x a y b 1 2 2 2 2 2 2 2 2 2 2 − = − = To find the vertices of the hyperbola, substitute y 0 = in the equation.The vertices satisfy the equation x a 1, 2 2 = the solutions of which are x a. = ± Consequently, the vertices of the hyperbola are V a, 0 1 ( ) = − and V a, 0 . 2 ( ) = Notice that the distance from the center 0, 0 ( ) to either vertex is a. Use triangle F PF . 1 2 P is on the right branch, so ( ) ( ) − = d F P d F P a , , 2 ; 1 2 ( ) = d F F c , 2 . 1 2 Divide both sides by a b . 2 2 THEOREM Equation of a Hyperbola: Center at ( ) 0, 0 ; Transverse Axis along the x -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 x a y b b c a 1 where 2 2 2 2 2 2 2 − = = − (2) The transverse axis is the x- axis. See Figure 36. The hyperbola defined by equation (2) is symmetric with respect to the x -axis, y -axis, and origin.To find the y -intercepts, if any, let x 0 = in equation (2). This results in the equation y b 1, 2 2 = − which has no real solution, so the hyperbola defined by equation (2) has no y -intercepts. In fact, since x a y b 1 0, 2 2 2 2 − = ≥ it follows that x a 1. 2 2 ≥ There are no points on the graph for a x a. − < < Finding and Graphing an Equation of a Hyperbola Find an equation of the hyperbola with center at the origin, one focus at 3, 0 ( ) and one vertex at 2, 0 . ( ) − Graph the equation. Solution EXAMPLE 1 The hyperbola has its center at the origin. Plot the center, focus, and vertex. Since they all lie on the x -axis, the transverse axis coincides with the x -axis. One focus is at c, 0 3, 0 , ( ) ( ) = so c 3. = One vertex is at a, 0 2, 0 , ( ) ( ) − = − so a 2. = From
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