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What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2–4y^2=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 put the equation in standard form:
Divide the equation by 64 to get: x^2/4 - y^2/16 = 1
The standard form of a hyperbola can be written as (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 a vertex along the x-axis, and b is the distance from the center to a vertex along the y-axis.
Comparing the standard form with our equation, we can see that h=0, k=0, a=2, and b=4.
Thus, the center of the hyperbola is at (0,0) and the vertices are located at (2,0) and (-2,0).
To find the foci of the hyperbola, we use the formula c = sqrt(a^2 + b^2), where c is the distance from the center to a focus.
Substitute a=2 and b=4 into the formula: c = sqrt(2^2 + 4^2) = sqrt(4 + 16) = sqrt(20) = 2√5.
Therefore, the foci are located at (2√5,0) and (-2√5,0).
To find the asymptotes of the hyperbola, we use the formula for the slopes of the asymptotes: ±b/a.
Substitute a=2 and b=4 into the formula: ±4/2 = ±2.
The equation of the asymptotes with slope ±2 passing through the center (0,0) is y = ±2x.
Therefore, the vertices of the hyperbola 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.