What are the vertices, foci, and asymptotes of the hyperbola with the equation
16x^2 - 4y^2 = 64
To find the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2 - 4y^2 = 64, we need to rewrite the equation in standard form:
Divide by 64 on both sides:
x^2/4 - y^2/16 = 1
Now the equation is in standard form: (x - h)^2/a^2 - (y - k)^2/b^2 = 1
Comparing the given equation to the standard form, we can see that a^2 = 4, b^2 = 16, h = 0, and k = 0.
Thus, a = 2 and b = 4.
Vertices:
The vertices are given by (h ± a, k), 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 = √(4 + 16) = √20 = 2√5
The foci are located at (h ± c, k), so the foci are (√20, 0) and (-√20, 0).
Asymptotes:
The slopes of the asymptotes are given by ±b/a, so the slopes of the asymptotes are ±4/2 = ±2.
The equations of the asymptotes are y = ±(4/2)x, or y = ±2x.