HW 111 Hyperbolas A hyperbola is a set of points in a plane the difference of whose distances from two fixed points, called foci, is a constant. * For any point P that is on the hyperbola, d2 - d, is always the same. F1 In this example, the origin is the center of the hyperbola. It is midway between the foci. *A line through the foci intersects the hyperbola at two points, called the vertices. V The segment connecting the vertices is called the transverse C axis of the hyperbola. * The center of the hyperbola is located at the midpoint of the transverse axis. As x and y get larger the branches of the hyperbola approach a pair of intersecting lines called the asymptotes of the hyperbola. These asymptotes pass through the center of the hyperbola.The figure at the left is an example of a hyperbola whose branches open up and down instead of right and left. * Since the transverse axis is vertical, this type of hyperbola is often referred V to as a vertical hyperbola. TI When the transverse axis is horizontal, the hyperbola is referred to as a horizontal hyperbola. Standard Form Equation of a Hyperbola (x - h)2 _ (y -k)2 = 1 (v - k)2 _ (x - h)2 = 1 a2 b2 a2 62 Horizontal Vertical Hyperbola Hyperbola The center of a hyperbola is at the point (h, k) in either form * For either hyperbola, a2 + b2 = c2 Where c is the distance from the center to a focus point. * The equations of the asymptotes for a HORIZONTAL HYPERBOLA are y = = ( x - h ) + k and y = = ( x - h ) + k The equation of the asymptotes for a vertical hyperbola are:y = " (x - h) + kandy = -" (x - h) + k Eccentricity of hyperbola is e = =Some worked out examples: Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with equation . 9 - 4= 1. Then graph the hyperbola. The center of this hyperbola is at the origin. According to the equation, a2 = 9 and b2 = 4, so a = 3 and b = 2. The coordinates of the vertices are (3, 0) and (-3, 0). c2 = a2 + 62 Equation relating a, b, and c for a hyperbola c2 = 32 + 22 a = 3, b= 2 c2 = 13 Simplify. c = V13 Take the square root of each side. The foci are at (V13, 0) and (-V13, 0). The equations of the asymptotes are y = 1- x ory = + zx. Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with equation 4x2 - 92 - 32x - 18y + 19 = 0. Then graph the hyperbola. Complete the square for each variable to write this equation in standard form. 4x2 - 9y2 - 32x - 18y + 19 = 0 Original equation 4(x2 - 8x + ) - 9(y2 + 2y + )=-19 +4(0)-9(0) Complete the squares. 4(x2 - 8x + 16) - 9(y2 + 2y + 1) = -19 + 4(16) -9(1) 4(x - 4)2 - 9(y + 1)2 = 36 Write the trinomials as perfect squares (x - 4)2 (y + 1)2 4 = 1 Divide each side by 36.Answer the following questions: For the given equation of hyperbola, if it is not in the standard form write the equation in standard form, find the center, a value, b value and c value, decide if it is horizontal or vertical, find the two vertices, find 2 foci, find 2 equations of asymptotes, and eccentricity of hyperbola. 20. (y - 4)2 (x+ 2)2 (y -3)2 (x - 2)2 21. 16 9 = 1 25 16 1 22. x2 - 2yz = 2 23. x2 - 12 =4 28. y2 - 3x2 + 6y + 6x - 18 = 0 29. 4x2 - 25y2 - 8x - 96 = 0