1. **State the problem:** Find the equation of an ellipse with vertices at $(-5,0)$ and $(5,0)$ and passing through the point $(-3,4)$. 2. **Recall the standard form of an ellipse centered at the origin with horizontal major axis:** $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $a$ is the semi-major axis length and $b$ is the semi-minor axis length. 3. **Identify $a$ from the vertices:** The vertices are at $\pm 5$ on the x-axis, so $a = 5$. 4. **Use the point $(-3,4)$ to find $b$:** Substitute $x = -3$ and $y = 4$ into the ellipse equation: $$\frac{(-3)^2}{5^2} + \frac{4^2}{b^2} = 1$$ $$\frac{9}{25} + \frac{16}{b^2} = 1$$ 5. **Solve for $b^2$:** $$\frac{16}{b^2} = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25}$$ 6. **Isolate $b^2$:** $$\frac{16}{b^2} = \frac{16}{25}$$ Cross-multiply: $$16 \times 25 = 16 \times b^2$$ $$400 = 16 b^2$$ Divide both sides by 16: $$\cancel{16} \times 25 = \cancel{16} b^2$$ $$25 = b^2$$ 7. **Write the final equation:** $$\frac{x^2}{25} + \frac{y^2}{25} = 1$$ This simplifies to: $$\frac{x^2}{25} + \frac{y^2}{25} = 1$$ which is a circle of radius 5, but since the problem states an ellipse with vertices at $\pm 5$ on x-axis and passing through $(-3,4)$, this is consistent. **Final answer:** $$\frac{x^2}{25} + \frac{y^2}{25} = 1$$