Write the equation of a hyperbola with the follow- ing characteristics: Vertices: \( (-2,9) \) and \( (-2,-7) \) Asymptotes: \( y-1= \pm \frac{4}{3}(x+2) \)

To find the equation of the hyperbola with the given characteristics, let's break down the information step by step. ### Given: - **Vertices:** \((-2, 9)\) and \((-2, -7)\) - **Asymptotes:** \(y - 1 = \pm \frac{4}{3}(x + 2)\) ### Step 1: Determine the Center The center \((h, k)\) of the hyperbola is the midpoint of the vertices. \[ h = \frac{-2 + (-2)}{2} = -2 \] \[ k = \frac{9 + (-7)}{2} = 1 \] So, the center is \((-2, 1)\). ### Step 2: Find the Distance Between the Center and Vertices (a) The distance \(a\) from the center to each vertex along the transverse axis (vertical in this case) is: \[ a = 9 - 1 = 8 \] ### Step 3: Determine the Slope of the Asymptotes and Find b The asymptotes for a vertical hyperbola are given by: \[ y = k \pm \frac{a}{b}(x - h) \] Comparing this with the given asymptotes \(y - 1 = \pm \frac{4}{3}(x + 2)\), we can identify: \[ \frac{a}{b} = \frac{4}{3} \implies \frac{8}{b} = \frac{4}{3} \implies b = 6 \] ### Step 4: Write the Standard Equation For a vertical hyperbola, the standard form is: \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \] Substituting the values we found: \[ \frac{(y - 1)^2}{8^2} - \frac{(x + 2)^2}{6^2} = 1 \] \[ \frac{(y - 1)^2}{64} - \frac{(x + 2)^2}{36} = 1 \] ### Final Equation: \[ \boxed{\dfrac{(y - 1)^2}{64}\ -\ \dfrac{(x + 2)^2}{36}\ =\ 1} \]