1. **State the problem:** We are given the vertices and foci of a conic section and need to identify and analyze it. 2. **Identify the conic:** Vertices: $(8, 11)$ and $(8, -15)$ Foci: $(8, -2 + \sqrt{105})$ and $(8, -2 - \sqrt{105})$ Since the vertices and foci share the same $x$-coordinate, the conic is vertical. 3. **Find the center:** The center is the midpoint of the vertices: $$\left(8, \frac{11 + (-15)}{2}\right) = (8, -2)$$ 4. **Calculate $a$ (distance from center to vertex):** $$a = 11 - (-2) = 13$$ 5. **Calculate $c$ (distance from center to focus):** $$c = \sqrt{105}$$ 6. **Find $b$ using the relationship for hyperbolas:** $$c^2 = a^2 + b^2 \implies b^2 = c^2 - a^2 = 105 - 169 = -64$$ Since $b^2$ is negative, this is not a hyperbola but an ellipse. 7. **Check ellipse relationship:** For ellipses, $c^2 = a^2 - b^2$, so $$b^2 = a^2 - c^2 = 169 - 105 = 64$$ 8. **Equation of ellipse:** Center at $(h,k) = (8,-2)$, vertical major axis: $$\frac{(x-8)^2}{64} + \frac{(y+2)^2}{169} = 1$$ **Final answer:** The conic is an ellipse centered at $(8,-2)$ with equation $$\frac{(x-8)^2}{64} + \frac{(y+2)^2}{169} = 1$$