Question

find an equation for the hyperbola described. graph the equation. vertices at (0, - 12) and (0,12); asymptote the line y = 4x. an equation of the hyperbola is \\( \frac { y ^ { 2 } } { 144 } - \frac { x ^ { 2 } } { 9 } = 1 \\). (simplify your answer. use integers or fractions for any numbers in the expression.) choose the correct graph below. a. b. c. d.

Explanation:

Step1: Determine the form of the hyperbola

Since the vertices are at \((0,-12)\) and \((0,12)\), the hyperbola has a vertical transverse axis. The standard form is \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1\), where \(a = 12\) (distance from center to vertex), so \(a^{2}=144\).

Step2: Find \(b\) using the asymptote

The asymptote of a vertical - axis hyperbola is \(y=\pm\frac{a}{b}x\). Given \(y = 4x\) (one of the asymptotes), \(\frac{a}{b}=4\). Substituting \(a = 12\), we get \(12/b = 4\), so \(b = 3\) and \(b^{2}=9\).

Step3: Analyze the graph

For a hyperbola \(\frac{y^{2}}{144}-\frac{x^{2}}{9}=1\) (vertical - axis):

• The vertices are at \((0,\pm12)\).
• The asymptotes are \(y=\pm4x\).
• A vertical - axis hyperbola opens up and down.

Answer:

The equation of the hyperbola is \(\frac{y^{2}}{144}-\frac{x^{2}}{9}=1\). For the graph, we need to look for a hyperbola that opens up and down (vertical - axis) with vertices at \((0,\pm12)\) and asymptotes \(y = \pm4x\). Among the options (assuming the standard graph - matching logic for hyperbola forms):

• Option B (assuming in the given options, the graph with vertical opening, vertices at \((0,\pm12)\) and appropriate asymptotes is labeled as B. Without seeing the exact visual details of the options in a more precise way, but based on the standard form analysis, for a vertical - axis hyperbola \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\) with \(a = 12\), \(b = 3\))