OHHS Pre-Calculus Mr. J. Focht

8.3 Hyperbolas Geometry of a Hyperbola Translations of Hyperbolas Eccentricity 8.3

A Hyperbola is a Conic Section 8.3

Hyperbola Definition Set of all points whose difference of the distances to 2 fixed points is constant. Focus (x,y) d1d1 d2d2 d 1 – d 2 is the same for whatever (x,y) you choose on the blue curves. 8.3

Hyperbola Terms Focus Center Transverse Axis Conjugate Axis Vertex Asymptote Focal Axis 8.3

Hyperbola Terms Focus Center Asymptote a = distance from center to vertex a b b = distance from vertex to asymptote c = distance from center to focus c c 2 = a 2 + b 2 8.3

Hyperbola Equation Focus Center Asymptote a b c (h,k) 8.3

Hyperbola Equation Focus Asymptote a b c (h,k) 8.3

Asymptote Equations Focus Center Asymptote a b c (h,k) 8.3

Asymptote Equations Focus Asymptote a b c (h,k) 8.3

Example Find the vertices and the foci of the hyperbola 4x 2 - 9y 2 = 36. Divide by sides by

Example: Find the Center, Vertices, and Foci a 2 = 9 a = 3 Vertices (-3, 0), (3, 0) (h, k) = (0, 0) b 2 = 4 c 2 = a 2 + b 2 = = 13 Foci 8.3

Now You Try Find the center, vertices, and foci 8.3

Find the equation of the hyperbola 4 (1,-5) (1, 1) The center is halfway between the foci. (1, -2) 2 2 a c = distance from center to a focus c = 3 c 2 = a 2 + b 2 9 = 4 + b 2 b 2 =

Now You Try P. 663, #23: Find the equation that satisfies these conditions: Foci (±3, 0), transverse axis length 4 8.3

Example Find the coordinates of the center, foci, and vertices, and the equations of the asymptotes of the graph of 4x 2 – y x + 4y + 28 = 0 4x x – y 2 + 4y = -28 4(x 2 + 6x ) - (y 2 -4y ) = -28 4(x+3) 2 – (y-2) 2 =

Example 4(x+3) 2 – (y-2) 2 = 4 a = 1 b = 2 c 2 = a 2 + b 2 c 2 = c 2 = 5 Now let’s find the vertices, foci, and asymptotes on the graph. 8.3

Example a=1 b=2 (-3, 2) (-2, 2) (-4, 2) 8.3

Now You Try P. 664, #49: Find the center, vertices, and foci of 9x 2 – 4y 2 – 36x + 8y – 4 =0 8.3

Eccentricity Hyperbolas have eccentricities too. Since c > a, E > 1 8.3

Example Write the equation of the hyperbola with center at (-2, -4), a focus of (2,-4) and eccentricity c = distance from center to focus = 4 Since c= 4, a = 3 c 2 = a 2 + b = b 2 b 2 =

Now You Try p. 663, #37: Find an equation in standard form for the hyperbola that satisfies the conditions: Center(-3,6), a=5, e=2, vertical transverse axis 8.3

Home Work P , #2, 6, 24, 28, 32, 38, 44, 50, #