Find an equation for
the hyperbola with center (2,
3), vertex
(0,
3), and focus
(5,
3).

The center, focus, and vertex all lie
on the horizontal line y
= 3 (that is, they're side by side
on a line paralleling the x-axis),
so the branches must be side by side, and the x
part of the equation must be added. The a^{2}
will go with the x part
of the equation, and the y
part will be subtracted. The vertex is 2 units
from the center, so a
= 2; the focus is 3 units
from the center, so c =
3. Then a^{2}
+ b^{2} = c^{2}
gives me b^{2}
= 9 – 4 = 5. I don't need
to bother with the value of b
itself, since they only asked me for the equation, which is:

Find an equation for
the hyperbola with center (0,
0), vertex (0,
5), and
asymptotes y
= ± (5/3)x.

Advertisement

The vertex and the center are both on
the vertical line x
= 0 (that is, on the y-axis),
so the hyperbola's branches are above and below each other, not side
by side. Then the y
part of the equation will be added, and will get the a^{2}
as its denominator. Also, the slopes of the two asymptotes will be of
the form m = ± a/b.

The vertex they gave me is 5
units above the center, so a
= 5 and a^{2}
= 25. The slope of the asymptotes
(ignoring the "plus-minus" part) is a/b
= 5/3 = 5/b, so b
= 3 and b^{2}
= 9. And this is all I need in order
to find my equation:

Find an equation of
the hyperbola with x-intercepts
at x
= –5 and x
= 3, and
foci at (–6,
0) and (4,
0).

The center lies on the x-axis,
so the two x-intercepts
must then also be the hyperbola's vertices. Since the intercepts are
4
units to either side of the center, then a
= 4 and a^{2}
= 16. Then a^{2}
+ b^{2} = c^{2}
tells me that b^{2}
= 25 – 16 = 9, and my equation is:

Find an equation for
the hyperbola with vertices at (–2,
15) and (–2,
–1), and having
eccentricity e
= 17/8.

The vertices are above and below each
other, so the center, foci, and vertices lie on a vertical line paralleling
the y-axis.
Then the a^{2}
will go with the y
part of the hyperbola equation, and the x
part will be subtracted.

The center is midway between the two
vertices, so (h,
k) = (–2, 7). The vertices
are 8 units
above and below the center, so a
= 8 and a^{2}
= 64. The eccentricity is e
= c/a = 17/8 = c/8,
so c = 17 and
c^{2} =
289. The equation a^{2}
+ b^{2} = c^{2}
tells me that b^{2}
= 289 – 64 = 225. Then
my equation is:

Stapel, Elizabeth.
"Conics: Hyperbolas: Finding the Equation From Information."
Purplemath. Available from https://www.purplemath.com/modules/hyperbola3.htm.
Accessed