Find the vertices of the ellipse defined by the \( \frac{(x-4)^{2}}{9}+\frac{(y-2)^{2}}{16}=1 \). If necessary, round to

To find the vertices of the ellipse defined by the equation \[ \frac{(x-4)^{2}}{9} + \frac{(y-2)^{2}}{16} = 1, \] follow these steps: ### 1. Identify the Center The general form of an ellipse is: \[ \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1, \] where \((h, k)\) is the center of the ellipse. For the given equation: - \(h = 4\) - \(k = 2\) **Center:** \((4, 2)\) ### 2. Determine the Orientation Compare the denominators under \(x\) and \(y\): - \(a^{2} = 9\) → \(a = 3\) - \(b^{2} = 16\) → \(b = 4\) Since \(b > a\), the major axis is vertical. ### 3. Find the Vertices The vertices lie along the major axis. Since the major axis is vertical: - **Top Vertex:** \((h, k + b) = (4, 2 + 4) = (4, 6)\) - **Bottom Vertex:** \((h, k - b) = (4, 2 - 4) = (4, -2)\) ### **Vertices of the Ellipse** \[ (4,\ 6) \quad \text{and} \quad (4,\ -2) \]