Conics Instructional Unit
Day 8 - Hyperbolas
by
Mandy Stein
Hyperbola
The locus of all points P(x,y) such that the difference of the distance from P to two fixed points, called foci, are constant
Transverse axis – contains the vertices as endpoints
Conjugate axis – contains the co-vertices as endpoints
Standard Equation of a Hyperbola
Standard equation of an hyperbola centered at (h , k)
Horizontal Transverse Axis - = 1
Vertices: (h – a, k) & (h + a, k) Co-Vertices: (h, k + b) & (h, k – b) Foci: (h – c, k) & (h + c, k) Slope of asymptote = + and - Transverse axis = 2a Conjugate axis = 2b a2 + b2 = c2 |
Vertical Transverse Axis - = 1
Vertices: (h, k – a) & (h, k + a) Co-Vertices: (h – b, k) & (h + b, k) Foci: (h, k – c) & (h, k + c) Slope of asymptote = + and - Transverse axis = 2a Conjugate axis = 2b a2 + b2 = c2 |
To graph a hyperbola:
- = 1
Horizontal transverse axis
Center: (2,1)
Vertices: (0,1) & (4,1)
Co-Vertices: (2,1 + ) & (2,1 - )
Foci: (-1,1) & (5,1)
- = 1
Vertical transverse axis
Center: (-4,3)
Vertices: (-4,7) & (-4,-1)
Co-Vertices: (1,3) & (7,3)
Foci: (-4,-2) & (-4,8)
To graph an equation not in standard form:
-2x2 + y2 +4x + 6y + 3 = 0
First, we put the equation in standard form by completing the square
-2x2 + y2 + 4x + 6y = -3
(y2 + 6y + 9) - 2(x2 - 2x + 1) = -3 + 9 – 2
(y + 3) 2 + -2(x – 1) 2 = 4
- = 1
Vertical transverse axis
Center: (1,-3)
Vertices: (1,-5) & (1,-1)
Co-Vertices: (1 - ,-3) & (1 + ,-3)
Foci: (1,-3 - ) & (1,-3 + )
To write the equation of a hyperbola
Horizontal transverse axis
Center: (0,0)
Vertices: (0,-3) & (0,3)
Co-vertices: (-5,0) & (5,0)
Foci: (0,- ) & (0, )
Asymptotes: y = - & y =