1. 11.3 The Hyperbola

2. Hyperbola: the set of all points P in a plane such that the absolute value of the difference of the distances from two fixed points is a constant. (foci) Standard Equation (centered at origin) x-int: (±a, 0) y-int: none transverse axis: 2a conjugate axis: 2b foci: (±c, 0) c 2 = a 2 + b 2 asymptotes: x-int: none y-int: (0, ±a) transverse axis: 2a conjugate axis: 2b foci: (0, ±c) c 2 = a 2 + b 2 asymptotes:

3. eqtns of asym.: y in front  vertices (0, ±3) endpoints of conjugate axis: (±4, 0) Foci: c 2 = 9 + 16 = 25 c = 5 (on y-axis) (0, ±5) Eccentricity is still ratio: e > 1 iff it is a hyperbola Ex 1) Determine the vertices, endpoints of conjugate axis, the foci, the asymptotes and eccentricity of. Graph it. *Hint: Graph as you go!

4. What if center is not the origin? Standard form of a hyperbola with center (h, k) & with axes parallel to the coordinate axes is OR Ex 2) Graph. Determine center, vertices, foci, & asymptotes. Center (4, –3) Eqtns of asym – use point slope with center & x in front – vertices: (4 ± 5, –3)  (9, –3) (–1, –3) endpoints of conjugate axis: (4, –3 ± 4)  (4, 1) (4, –7)

5. Ex 3) Determine the center, vertices, foci and asymptotes of 3x 2 – y 2 – 12x – 6y = 0. Then graph it. 3(x 2 – 4x + 4 ) – (y 2 + 6y + 9 ) = 0 + 12 + –9 3(x – 2) 2 – (y + 3) 2 = 3 Center (2, –3) Eqtns of asym: x in front – vertices: (2 ± 1, –3)  (3, –3) (1, –3) endpts of conj axis: Foci: c 2 = 1 + 3 = 4 c = 2 (2 ± 2, –3)  (4, –3) (0, –3)

6. Ex 4) Determine an equation in standard form for a hyperbola with vertices at (0, 2) and (0, –8) with points (12, –16) and being two points on the hyperbola. center halfway between vertices ½ (2 – 8) = ½(–6) = –3 C(0, –3)a = 5 on y-axis (12, –16): plug in one of the points to solve for b 2

7. Homework #1103 Pg 555 #1, 5, 9, 14, 17, 21, 29, 33, 36, 37