1015 10.3 Hyperbolas Suppose a hyperbola has center at the origin and foci at F′1-c, 02 and F1c, 02. See Figure 29. The midpoint of the segment F′F is the center of the hyperbola and the points V′1-a, 02 and V1a, 02 are the vertices of the hyperbola. The line segment V′V is the transverse axis of the hyperbola. x y F9(–c, 0) F(c, 0) P(x, y) V9(–a, 0) Transverse axis Vertices V(a, 0) Figure 29 For a hyperbola, d1V, F′2 - d1V, F2 = 1c + a2 - 1c - a2 = 2a, so the constant in the definition is 2a, and 0 d1P, F′2 - d1P, F20 = 2a for any point P1x, y2 on the hyperbola. The distance formula and algebraic manipulation similar to that used for finding an equation for an ellipse produce the following result. x2 a2 - y2 c2 - a2 = 1 Replacing c2 - a2 with b2 gives an equation of the hyperbola in Figure 29. x2 a2 − y2 b 2 =1 Letting y = 0 shows that the x-intercepts are 1{a, 02. If x = 0, the equation becomes y2 = -b2, which has no real number solutions, showing that this hyperbola has no y-intercepts. To develop an aid for sketching the graph of a hyperbola, we start with the equation for a hyperbola and solve for y. x2 a2 - y2 b2 = 1 Hyperbola with transverse axis on the x-axis x2 a2 - 1 = y2 b2 Subtract 1. Add y2 b2 . x2 - a2 a2 = y2 b2 Write the left side as a single fraction. y = { b a 2x2 - a2 Take the square root on each side. Multiply by b, and rewrite. If x2 is very large in comparison to a2, the difference x2 - a2 is very close to x2. If this happens, then the points satisfying the final equation above are very close to one of the lines y = t b a x. Thus, as 0 x 0 increases without bound, the points of the hyperbola x 2 a2 - y2 b2 = 1 approach the lines y = { b a x. These lines are asymptotes of the hyperbola and are useful when sketching the graph. Remember both the positive and negative square roots.
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