1. **Problem Statement:** Given a hyperbola centered at the origin with equation $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ where $$a^2=9$$ and $$b^2=4$$, find the key points including vertices, co-vertices, and foci. 2. **Formula and Definitions:** - The standard form of a horizontal hyperbola centered at the origin is $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$. - Vertices are at $$(\pm a, 0)$$. - Co-vertices are at $$(0, \pm b)$$. - The focal distance $$c$$ satisfies $$c^2 = a^2 + b^2$$. - Foci are at $$(\pm c, 0)$$. 3. **Calculate values:** - Given $$a^2 = 9$$, so $$a = \sqrt{9} = 3$$. - Given $$b^2 = 4$$, so $$b = \sqrt{4} = 2$$. - Calculate $$c^2 = a^2 + b^2 = 9 + 4 = 13$$. - Therefore, $$c = \sqrt{13}$$. 4. **Find key points:** - Vertices: $$(\pm a, 0) = (\pm 3, 0)$$. - Co-vertices: $$(0, \pm b) = (0, \pm 2)$$. - Foci: $$(\pm c, 0) = (\pm \sqrt{13}, 0)$$. 5. **Explanation:** The hyperbola opens horizontally because the positive term is under $$x^2$$. The vertices mark the closest points on the hyperbola to the center along the transverse axis. The co-vertices lie along the conjugate axis and help define the rectangle used to sketch the asymptotes. The foci are points inside each branch of the hyperbola, located farther from the center than the vertices. **Final answer:** - Vertices: $$(3, 0)$$ and $$(-3, 0)$$ - Co-vertices: $$(0, 2)$$ and $$(0, -2)$$ - Foci: $$(\sqrt{13}, 0)$$ and $$(-\sqrt{13}, 0)$$