What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x2–4y2=64
16x^2–4y^2=64
To identify the vertices, foci, and asymptotes of the hyperbola with the given equation, we first need to rewrite the equation in standard form:
16x^2 - 4y^2 = 64
Divide all terms by 64:
x^2/4 - y^2/16 = 1
Comparing this with the standard form of a hyperbola:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
where (h, k) is the center of the hyperbola and a and b are the distances from the center to the vertices.
The center of the hyperbola is at (h, k) = (0, 0).
a = 2 and b = 4
Vertices:
The vertices of the hyperbola are on the x-axis and are at (±a, 0).
So the vertices are (2, 0) and (-2, 0).
Foci:
The distance from the center to the foci is given by c = √(a^2 + b^2).
c = √(2^2 + 4^2) = √(4 + 16) = √20 = 2√5
The foci are located along the x-axis and are at (±c, 0).
So the foci are (2√5, 0) and (-2√5, 0).
Asymptotes:
The slopes of the asymptotes are given by ±b/a = ±4/2 = ±2.
The equations of the asymptotes passing through the center (0, 0) are y = ±2x.
Therefore, the vertices are (2, 0) and (-2, 0), the foci are (2√5, 0) and (-2√5, 0), and the asymptotes are y = 2x and y = -2x.