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, we first need to rewrite the equation in the standard form:
(16x^2) - (4y^2) = 64
Divide both sides of the equation by 64:
(16x^2) / 64 - (4y^2) / 64 = 64 / 64
(x^2) / 4 - (y^2) / 16 = 1
Now, we can see that the equation is in the standard form for a hyperbola with a horizontal transverse axis:
(x^2) / a^2 - (y^2) / b^2 = 1
In this case, a^2 = 4 and b^2 = 16, which means that:
a = 2
b = 4
The center of the hyperbola is given by (0, 0), since there are no translations in the standard form equation.
Verticies:
The vertices of a hyperbola with horizontal transverse axis are given by (±a, 0). So, the vertices in this case are:
(±2, 0) or (-2, 0) and (2, 0)
Foci:
The distance from the center to each focus, c, can be found using the relationship c^2 = a^2 + b^2. In this case:
c^2 = 2^2 + 4^2 = 4 + 16 = 20
So, c = √20. The foci for a hyperbola with a horizontal transverse axis are given by (±c, 0). Therefore, the foci are:
(±√20, 0) or (-√20, 0) and (√20, 0)
Asymptotes:
The asymptotes for a hyperbola in standard form with a horizontal transverse axis are given by the equations y = ±(b/a)x. In this case, the equations are:
y = ±(4/2)x = ±2x
So, the asymptotes are y = 2x and y = -2x.