Question

find the center, transverse axis, vertices, foci, and asymptotes. graph the equation

$$\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$$

the center is at $(0,0)$
(type an ordered pair. type exact answers for each coordinate, using radicals as needed.)

the transverse axis is along the

Explanation:

Step1: Identify the standard form of hyperbola

The standard form of a hyperbola is \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) (opens left - right) or \(\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1\) (opens up - down), where \((h,k)\) is the center. For the given equation \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\), we have \(h = 0,k = 0,a^{2}=16,b^{2}=9\).

Step2: Determine the transverse axis

Since the \(x^{2}\) term is positive, the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\) opens left - right. The transverse axis of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is along the \(x\) - axis.

Answer:

The transverse axis is along the \(x\) - axis.