1. **State the problem:** We need to find the coordinates of the foci and the lengths of the major and minor axes of the ellipse given by the equation $$\frac{x^2}{49} + \frac{y^2}{36} = 1.$$ 2. **Identify the ellipse parameters:** The standard form of an ellipse centered at the origin is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$$ where $a$ is the semi-major axis and $b$ is the semi-minor axis. 3. **Determine $a$ and $b$:** Since 49 and 36 are denominators, we have $$a^2 = 49 \implies a = 7$$ and $$b^2 = 36 \implies b = 6.$$ 4. **Identify major and minor axis:** Because $a^2 > b^2$, the major axis is along the x-axis with length $2a = 14$, and the minor axis is along the y-axis with length $2b = 12$. 5. **Find the foci:** The foci lie on the major axis at distances $$c = \sqrt{a^2 - b^2} = \sqrt{49 - 36} = \sqrt{13}.$$ 6. **Coordinates of the foci:** The foci are at $$ (\pm c, 0) = (\pm \sqrt{13}, 0).$$ **Final answer:** - Foci are at $(\pm \sqrt{13}, 0)$. - Length of major axis is 14. - Length of minor axis is 12.