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

Explanation:

Step1: Write the standard form of hyperbola

Divide the given equation \(4x^{2}-y^{2}=256\) by \(256\) to get \(\frac{x^{2}}{64}-\frac{y^{2}}{256}=1\). The standard form of a hyperbola is \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\) (for hyperbola opening left - right) or \(\frac{(y - k)^{2}}{a^{2}}-\frac{(x - h)^{2}}{b^{2}}=1\) (for hyperbola opening up - down), where \((h,k)\) is the center. Here \(h = 0,k = 0,a^{2}=64,b^{2}=256\).

Step2: Determine the transverse axis

For the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), the transverse axis is along the \(x\) - axis.

Step3: Find the vertices

The vertices of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) are \((h\pm a,k)\). Since \(h = 0,k = 0,a = 8\), the vertices are \((\pm8,0)\).

Step4: Calculate \(c\) (for foci)

Use the relationship \(c^{2}=a^{2}+b^{2}\). Given \(a^{2}=64,b^{2}=256\), then \(c^{2}=64 + 256=320\), so \(c=\sqrt{320}=8\sqrt{5}\). The foci are \((h\pm c,k)\), which are \((\pm8\sqrt{5},0)\).

Step5: Find the asymptotes

The equations of the asymptotes for the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) are \(y=\pm\frac{b}{a}(x - h)+k\). Substituting \(h = 0,k = 0,a = 8,b = 16\), we get \(y=\pm2x\).

Answer:

The transverse axis is along the \(x\) - axis. The vertices are \((8,0)\) and \((- 8,0)\). The foci are \((8\sqrt{5},0)\) and \((-8\sqrt{5},0)\). The asymptotes are \(y = 2x\) and \(y=-2x\).