Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x2–4y2=64
?
To find the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2 - 4y^2 = 64, we first need to write the equation in standard form.
16x^2 - 4y^2 = 64
Divide by 64 to get:
x^2/4 - y^2/16 = 1
Now we can see that this is a hyperbola with a horizontal transverse axis, since x^2 is positive and y^2 is negative. To find the vertices, foci, and asymptotes, we need to identify the following values:
a^2 = 4 => a = 2 (distance from center to vertices)
b^2 = 16 => b = 4
c^2 = a^2 + b^2 = 20 => c ≈ 4.47 (distance from center to foci)
The center of the hyperbola is at (0,0). The vertices are located at (±2,0). The foci are located at (±√20, 0), which is approximately (±4.47, 0).
The equation of the asymptotes for a hyperbola with a horizontal transverse axis is y = ±(b/a)x, so in this case the asymptotes are y = ±(4/2)x or y = ±2x.
Therefore, the vertices are (±2,0), the foci are (±4.47,0), and the asymptotes are y = ±2x.