Question

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

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

(simplify your answer. type ordered pairs type exact answers, using radicals as needed. use a comma to separate answers as needed.)

the foci are at \\( ( - 5,0 ), ( 5,0 ) \\)
(simplify your answer. type ordered pairs. type exact answers, using radicals as needed. use a comma to separate answers as needed )

the asymptote with positive slope is \\( y = \frac { 3 } { 4 } x \\), and the asymptote with negative slope is \\( y = - \frac { 3 } { 4 } x \\)
(simplify your answers. type equations. use integers or fractions for any numbers in the equations.)

which of the following is the correct graph?

oa.

ob.

oc.

od.

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 \(h = 0,k = 0,a^{2}=16,a = 4,b^{2}=9,b = 3\).

Step2: Find the center

Since \(h = 0,k = 0\), the center \((h,k)=(0,0)\).

Step3: Find the transverse axis

Because the \(x^{2}\) term is positive, the transverse axis is along the \(x\) - axis.

Step4: Find the vertices

The vertices of a hyperbola \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) are \((h\pm a,k)\). Substituting \(h = 0,k = 0,a = 4\), the vertices are \((- 4,0),(4,0)\).

Step5: Find \(c\) (for foci)

We use the relationship \(c^{2}=a^{2}+b^{2}\). Given \(a^{2}=16,b^{2}=9\), then \(c^{2}=16 + 9=25\), so \(c = 5\). The foci of a hyperbola \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) are \((h\pm c,k)\). Substituting \(h = 0,k = 0,c = 5\), the foci are \((-5,0),(5,0)\).

Step6: Find the asymptotes

The equations of the asymptotes of a hyperbola \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) are \(y - k=\pm\frac{b}{a}(x - h)\). Substituting \(h = 0,k = 0,a = 4,b = 3\), the asymptotes are \(y=\frac{3}{4}x\) and \(y =-\frac{3}{4}x\).

For the graph:
Since the hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\) has center \((0,0)\), transverse axis along the \(x\) - axis, vertices \((-4,0),(4,0)\) and asymptotes \(y=\pm\frac{3}{4}x\).

• Option A: The hyperbola opens left - right (correct orientation as \(x^{2}\) term is positive), has center at the origin, and the general shape with asymptotes \(y=\pm\frac{3}{4}x\).
• Option B: Opens up - down (incorrect, since \(x^{2}\) term is positive).
• Option C: Opens up - down (incorrect, since \(x^{2}\) term is positive).
• Option D: The scale or the relationship between \(a\) and \(b\) (related to the steepness of asymptotes) may be incorrect.

Answer:

• Center: \((0,0)\)
• Transverse axis: \(x\) - axis
• Vertices: \((-4,0),(4,0)\)
• Foci: \((-5,0),(5,0)\)
• Asymptotes: \(y=\frac{3}{4}x,y =-\frac{3}{4}x\)
• Graph: A.