First get it in standard form, which is either
if the hyperbola opens right and left,
and the slopes of the asymptotes are 
or
if the hyperbola opens upward and downward.
and the slopes of the asymptotes are 
Get the 65 off the left side:
Factor out the coefficient of 
out of the
first two terms on the left.
Factor out the coefficient of 
out of the
last two terms on the left.
Complete the square on 
by multiplying
the coefficient of x, which is -10, by 
getting -5,
and then squaring -5, getting 25. And we add that inside the
first parentheses. However since there is a 4 in front of the
first parentheses, adding 25 inside the parentheses amounts
to adding 4*25 or 100 to the left side, so we must add 100
to the right side:
Complete the square on 
by multiplying
the coefficient of y, which is 2, by getting 1,
and then squaring 1, getting 1. And we add that inside the
second parentheses. However since there is a -7 in front of the
second parentheses, adding 1 inside the parentheses amounts
to adding 
or -7 to the left side, so we must add -7
to the right side:
Factor the parentheses as squares of binomials, and combine
the numbers on the right:
Get a 1 on the right by dividing through by 4
Since the variable x comes first in the standard form, the
hyperbola opens right and left.
So we compare that to:
The center is at (5,-1). So we plot the center:
so 
, the semi-transverse axis is
units long, and that's about 2.6. So we draw the
complete transverse axis right and left about 2.6 unit from the
center. The tranverse axis is the horizontal green line below:
"Trans" means "across" and "VERse" means "VERtices", so the
transverse axis goes across from vertex to vertex, and so the
ends of the transverse axis are the vertices and they are found
by adding and subtracting from the x-coordinate
of the center, so
The VERTICES are (5+,-1) and (5-,-1)
so 
, the semi-conjugate axis is 2 units
long, so we draw the complete conjugate axis up and down
2 units from the center, that is, the conjugate axis is the
vertical green line below:
Now we draw in the defining rectangle
Now we can draw the asymptotes which are the extended diagonals
of the defining rectangle:
and we can sketch in the hyperbola:
The slopes of the asymptotes are 
They pass through the center (5,-1) so we use the point-slope
formula:
That's the equation of one of the asymptotes. The other one:
The foci are the two points "c" units right and left of the
vertex, where
c² = a² + b²
c² = (sqrt(7))² + 2²
c² = 7 + 4
c² = 11
c = ±
So the foci are (5+,-1) and (5-,-1),
Here they are plotted. They are inside the branches of they
hyperbola:
I don't have time to do the other one. It's a hyperbola
that opens upward and downward. But I think you can do it.
Just remember that the transverse axis is vertical and the
conjugate axis is horizontal.
Edwin
