Looking at the graph thus far we can tell it is a hyperbola that looks like
this ")(", so its equation is of this form:
The center of the hyperbola will be where its two asymptotes
intersect. So we solve them as a system:
and get x=3, y=6 which means the center is the point (3,6).
The value of "a" is the distance from the vertex to the center, the
transverse axis, which is 3 units, so now we know that the center (h,k) = (3,6),
and that a=3, so we know everything but "b", the conjugate axis. So the equation
so far is
The other vertex is a point on the other side of the center (3,6),
3 units on the other side of the center.
Now we can draw in the defining rectangle for the hyperbola since we
know that the asymptotes are the extended diagonals of the defining
rectangle:
Then we can sketch in the hyperbola:
The value of "b" is the semi-conjugate axis which is half the
height of the height of the defining rectangle, which is 4, so
the equation is
or
Edwin
