Hyperpolas no asymptotes.notebook

Hyperpolas no asymptotes.notebook
February 25, 2014
HYPERBOLAS
- the set of all points (x, y) in a plane, thedifference of whose distances from
two distinct fixed points (called Foci) is a positive constant.
EQUATIONS AND GRAPHS OF HYPERBOLAS
b
a
C(h, k)
f(h-c, k) V(h-a, k)
V(h+a, k)
where c2 = a2 + b2
f(h+c, k)
This hyperbola opens horizontal
since the x variable is positive!
Major axis is called the Transverse Axis for hyperbolas.
To determine which way the hyperbola lies,
look at which variable is positive!
The denominator does NOT determine
orientation with hyperbolas!!!!
f(h, k+c)
V(h, k+b)
a
where c = a + b
2
2
2
b
C(h, k)
This hyperbola opens vertical since
the y variable is positive!
V(h, k-b)
f(h, k-c)
Find the center, vertices and foci. Then sketch the graph, be sure to include asymptotes.
4y2 - 9x2 = 36
Hyperpolas no asymptotes.notebook
February 25, 2014
Sketch the graph of the hyperbola given by
9x2 - 4y2 + 8y - 40 = 0.
Find the standard form of the equation of the hyperbola
with foci at (0, 0) and (8, 0) and vertices at (3, 0) and (5, 0).
By definition, the center is the midpoint between verticies
and between foci.
This puts the center at (4, 0).
This hyperbola is horizontal. Because the foci, verticies and
center all lie on the transverse axis (aka major axis) we can
see that the orientation is horizontal.
Since the hyperbola is horizontal, the x variable is positive!
Distance from vertex to
the center is 1.
So, we square that and
put it under the x.
Use
c is the distance from the center to the focus. We can see from our picture that it is 4.
Hyperpolas no asymptotes.notebook
February 25, 2014
Find the standard form of the equation of the
hyperbola having vertices (2, -1) and (2, 1) and
passes through the point (5, 4).
By plotting points and looking at the picture, we can see
that the hyperbola must open vertical. This means our y variable is positive!
Since the center must be midway between vertices, the center must be at (2, 0).
The distance from the center to a vertex is 1. So we square 1, and put it under the y variable.