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 x - axis.

the vertices are at (-4,0),(4,0).
(simplify your answer. type ordered pairs. type exact answers, using radicals as needed. use a c
answers as needed.)

the foci are at
(simplify your answer. type ordered pairs. type exact answers, using radicals as needed. use a co
answers as needed.)

Explanation:

Step1: Recall the standard form of hyperbola

The standard form of a hyperbola is \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\), where \((h,k)\) is the center. For \(\frac{x^2}{16}-\frac{y^2}{9}=1\), \(h = 0,k = 0,a^2=16,b^2 = 9\).

Step2: Calculate \(c\)

Use the formula \(c^2=a^2 + b^2\). Substitute \(a^2 = 16\) and \(b^2=9\), then \(c^2=16 + 9=25\), so \(c = 5\).

Step3: Find the foci

The foci of the hyperbola \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) are \((h\pm c,k)\). Here \(h = 0,k = 0,c = 5\), so the foci are \((- 5,0)\) and \((5,0)\).

Answer:

\((-5,0),(5,0)\)