What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2–4y^2=64
Given equation of hyperbola is:
16x^2 - 4y^2 = 64
Dividing by 64 on both sides, we get:
x^2/4 - y^2/16 = 1
This equation can be rearranged into the standard form of a hyperbola:
(x^2/2^2) - (y^2/4^2) = 1
The equation is now in the standard form of a hyperbola.
Comparing the equation to standard form, we see that the vertices of the hyperbola are:
Vertices: (±2, 0)
And, the center of the hyperbola is at the origin (0, 0).
The foci can be found using the formula for hyperbolas: c^2 = a^2 + b^2, where a = 2 and b = 4.
c^2 = 4 + 16
c^2 = 20
c = sqrt(20) ≈ 4.47
Therefore, the foci of the hyperbola are located at approximately (+-4.47, 0).
The asymptotes of the hyperbola are given by the equations:
y = ±(b/a)x
y = ±(4/2)x
y = ±2x
Therefore, the asymptotes are the lines y = 2x and y = -2x.