1. **State the problem:** We are given the hyperbola equation $$\frac{(y - 2)^2}{25} - \frac{x^2}{16} = 1$$ and need to understand its properties and graph it. 2. **Identify the type of conic:** This is a vertical hyperbola because the positive term involves $y$. 3. **Standard form:** The equation matches $$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$$ where the center is at $(h,k)$. 4. **Find center, $a$, and $b$:** - Center: $(0, 2)$ - $a^2 = 25 \Rightarrow a = 5$ - $b^2 = 16 \Rightarrow b = 4$ 5. **Vertices:** Located $a=5$ units above and below the center along the $y$-axis: $$ (0, 2+5) = (0,7), \quad (0, 2-5) = (0,-3) $$ 6. **Foci:** Distance $c$ from center where $$c^2 = a^2 + b^2 = 25 + 16 = 41 \Rightarrow c = \sqrt{41}$$ Foci at: $$ (0, 2 + \sqrt{41}), \quad (0, 2 - \sqrt{41}) $$ 7. **Asymptotes:** Equations are $$ y - 2 = \pm \frac{a}{b} (x - 0) = \pm \frac{5}{4} x $$ 8. **Summary:** The hyperbola opens vertically with center at $(0,2)$, vertices at $(0,7)$ and $(0,-3)$, foci at $(0, 2 \pm \sqrt{41})$, and asymptotes $y = 2 \pm \frac{5}{4}x$. Final answer: The hyperbola is $$\frac{(y - 2)^2}{25} - \frac{x^2}{16} = 1$$ with center $(0,2)$, vertices $(0,7)$ and $(0,-3)$, and asymptotes $y = 2 \pm \frac{5}{4}x$.