Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y = ± 1/2 x.
Solution:
The standard form of the equation of a vertical hyperbola with center (h,k) is given by
[(y - k)2 / a2] - [(x - h)2 / b2] = 1 --- (1)
Where,
The coordinates of the vertices are (k, h ± a)
The coordinates of the foci are (k, h ± c)
Given, vertices at (0, ±8)
Asymptotes, y = ± (1/2)x.
The points where the hyperbola intersects the axis are called the vertices. The vertices of the hyperbola are (0, a) and (0, -a).
Asymptotes is a pair of straight lines drawn parallel to the hyperbola and assumed to touch the hyperbola at infinity.
The equation of asymptotes is y = k ± [a(x - h) / b]
We know, vertices are at (0, 8) and (0, -8), it's a vertical hyperbola with center at (0,0) i.e. (h, k) = (0, 0) and the transverse axis is 8 units long.
Here, a = 8
From the equation of asymptotes,
k ± [a(x - h) / b] = ± x / 2
0 ± [8(x - 0) / b] = ± x / 2
± 8x / b = ± x / 2
8x / b = x / 2
8 / b = 1 / 2
b = 8(2)
b = 16
The equation of hyperbola can be written as
[(y - 0)2 / (8)2] - [(x - 0)2 / (16)2] = 1
[y2 / 64] - [x2 / 256] = 1
Therefore, the equation of hyperbola is [y2 / 64] - [x2 / 256] = 1.
Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y = ± 1/2 x.
Summary:
An equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y = ± 1/2 x is [y2 / 64] - [x2 / 256] = 1.