What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x2–4y2=64
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To find the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2 - 4y^2 = 64, we first need to rewrite the equation in standard form:
Divide by 64 to get x^2 - y^2/16 = 1.
Now we can see that this is a hyperbola centered at the origin with a horizontal transverse axis. The standard form equation for a hyperbola with a horizontal transverse axis is given by (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where (h,k) is the center of the hyperbola, a is the distance from the center to the vertices, and b is the distance from the center to the co-vertices.
Comparing the standard form equation to our equation, we have:
(h,k) = (0,0)
a^2 = 1
b^2 = 16
Therefore, a = 1 and b = 4. The vertices of the hyperbola are located at (h+a, k) and (h-a, k), which in this case are (1,0) and (-1,0), respectively. The foci of the hyperbola are located at (h+c, k) and (h-c, k), where c is given by c^2 = a^2 + b^2. In this case, c^2 = 1 + 16 = 17, so c = √17. Therefore, the foci are located at (√17,0) and (-√17, 0).
Finally, the equation for the asymptotes of the hyperbola can be found using the formula y = ±(b/a)x. Substituting in our values for a and b, we get the asymptotes y = ±4x.