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 \boxed{} = 1. (simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Determine the standard 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 \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1\), and \(a = 12\) (distance from center to vertex).

Step2: Use the asymptote formula

For a hyperbola \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\), the asymptotes are \(y=\pm\frac{a}{b}x\). Given \(y = 4x\) is an asymptote, so \(\frac{a}{b}=4\). Substitute \(a = 12\) into \(\frac{a}{b}=4\), we get \(12 = 4b\), then \(b = 3\).

Step3: Write the equation of the hyperbola

Substitute \(a = 12\) (\(a^{2}=144\)) and \(b = 3\) (\(b^{2}=9\)) into the standard form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\). The equation is \(\frac{y^{2}}{144}-\frac{x^{2}}{9}=1\).

Answer:

\(\frac{y^{2}}{144}-\frac{x^{2}}{9}\)