Question

find the center, transverse axis, vertices, foci, and asymptotes. graph the equation
4x² - y² = 256
the center of the hyperbola is at (0,0).
(type an ordered pair.)
the transverse axis is along the x - axis.
the vertices are at (8,0),(-8,0).
(type an ordered pair. use a comma to separate answers as needed.)
the foci are at (8√5,0),(-8√5,0).
(simplify your answer. type an ordered pair. type an exact answer, using radicals as needed. use a com
separate answers as needed.)
the asymptotes are y = ±
(simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Write the standard form of hyperbola

Divide the given equation \(4x^{2}-y^{2}=256\) by \(256\) to get \(\frac{x^{2}}{64}-\frac{y^{2}}{256}=1\). For a hyperbola of the form \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\) (center \((h,k)\)), here \(h = 0,k = 0,a^{2}=64\), so \(a = 8\), \(b^{2}=256\), so \(b = 16\).

Step2: Find the asymptotes formula

The formula for the asymptotes of a hyperbola \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\) is \(y-k=\pm\frac{b}{a}(x - h)\). Substituting \(h = 0,k = 0,a = 8,b = 16\) into the formula.

Step3: Calculate the asymptotes

\(y=\pm\frac{16}{8}x\)

Answer:

\(y=\pm2x\)