Find the center, vertices,
foci, eccentricity, and asymptotes of the hyperbola with the given equation,
and sketch:

Since the y part
of the equation is added, then the center, foci, and vertices will be
above and below the center (on a line paralleling the y-axis),
rather than side by side.

Looking at the denominators, I see that
a^{2} =
25 and b^{2}
= 144, so a
= 5 and b
= 12. The equation c^{2}
– a^{2} = b^{2}
tells me that c^{2}
= 144 + 25 = 169, so c
= 13, and the eccentricity is e
= 13/5. Since x^{2}
= (x – 0)^{2} and
y^{2} =
(y – 0)^{2}, then
the center is at (h,
k) = (0, 0). The vertices
and foci are above and below the center, so the foci are at (0,
–13) and (0,
13), and the vertices are at (0,
5) and (0,
–5).

Because the y
part of the equation is dominant (being added, not subtracted),
then the slope of the asymptotes has the a
on top, so the slopes will be m
= ± 5/12. To graph, I start
with the center, and draw the asymptotes through it, using dashed
lines:

Then I draw in the vertices,
and rough in the graph, rotating the paper as necessary and "eye-balling"
for smoothness:

Then I draw in the final graph
as a neat, smooth, heavier line:

And the rest of my answer is:

center (0,
0), vertices
(0,
–5) and (0,
5), foci
(0,
–13) and
(0,
13), eccentricity
,
and asymptotes

Give the center, vertices,
foci, and asymptotes for the hyperbola with equation:

Since the x
part is added, then a^{2}
= 16 and b^{2}
= 9, so a
= 4 and b
= 3. Also, this hyperbola's foci
and vertices are to the left and right of the center, on a horizontal
line paralleling the x-axis.

Since the a^{2}
went with the x part
of the equation, then a
is in the denominator of the slopes of the asymptotes, giving me m
= ± 3/4. Keeping in mind that the
asymptotes go through the center of the hyperbola, the asymptes are
then given by the straight-line equationsy – 2 = ± (3/4)(x
+ 3).

center (–3,
2), vertices
(–7,
2) and (1,
2), foci
(–8,
2) and (2,
2), and
asymptotes

Find the center, vertices,
and asymptotes of the hyperbola with equation 4x^{2}
– 5y^{2} + 40x
– 30y – 45 = 0.

To find the information I need, I'll
first have to convert this equation to "conics" form by completing
the square.

Then the center is at (h,
k) = (–5, –3). Since the x
part of the equation is added, then the center, foci, and vertices lie
on a horizontal line paralleling the x-axis;
a^{2} =
25 and b^{2}
= 20, so a =
5 and b
= 2sqrt[5]. The equation a^{2}
+ b^{2} = c^{2}
gives me c^{2}
= 25 + 20 = 45, so c
= sqrt[45] = 3sqrt[5].
The slopes of the two asymptotes will be m
= ± (2/5)sqrt[5]. Then my
complete answer is:

center (–5,
–3), vertices
(–10,
–3) and (0,
–3),

foci
and ,

and asymptotes

If I had needed to graph this hyperbola,
I'd have used a decimal approximation of ±
0.89442719... for the slope, but would
have rounded the value to something reasonable like m
= ± 0.9.

Stapel, Elizabeth.
"Conics: Hyperbolas: Finding Information From the Equation."
Purplemath. Available from http://www.purplemath.com/modules/hyperbola2.htm.
Accessed