1. **State the problem:** We are given the ellipse equation $$\frac{(x-1)^2}{9} + \frac{(y+4)^2}{16} = 1$$ and need to find the center, major and minor axes with their lengths, vertices, and foci. 2. **Identify the center:** The ellipse is in standard form $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$ where $(h,k)$ is the center. Here, $h=1$ and $k=-4$, so the center is $$\boxed{(1,-4)}$$. 3. **Determine $a$ and $b$:** The denominators are $9$ and $16$, so $a^2=9$ and $b^2=16$ or vice versa. Since $16 > 9$, $b^2=16$ is the larger denominator, so: - $a = 3$ - $b = 4$ 4. **Major and minor axes:** The major axis corresponds to the larger denominator, so the major axis length is $$2b = 2 \times 4 = 8$$. The minor axis length is $$2a = 2 \times 3 = 6$$. Since $b^2$ is under the $(y+4)^2$ term, the major axis is vertical. 5. **Vertices:** Vertices lie along the major axis from the center by $b$ units: - Vertices: $$\left(1, -4 \pm 4\right) = (1,0) \text{ and } (1,-8)$$ 6. **Foci:** The focal distance $c$ is given by $$c = \sqrt{b^2 - a^2} = \sqrt{16 - 9} = \sqrt{7}$$. Foci lie along the major axis from the center by $c$ units: - Foci: $$\left(1, -4 \pm \sqrt{7}\right)$$ **Final answers:** - Center: $(1,-4)$ - Major axis length: $8$ (vertical) - Minor axis length: $6$ (horizontal) - Vertices: $(1,0)$ and $(1,-8)$ - Foci: $(1, -4 + \sqrt{7})$ and $(1, -4 - \sqrt{7})$