Graph the ellipse and give the location of its foci. \[ 25(x+6)^{2}+9(y-4)^{2}=225 \]

We start with the given ellipse \[ 25(x+6)^{2}+9(y-4)^{2}=225. \] **Step 1. Write the equation in standard form.** Divide both sides by \(225\) to obtain \[ \frac{25(x+6)^2}{225}+\frac{9(y-4)^2}{225}=1. \] Simplify each fraction: \[ \frac{(x+6)^2}{9}+\frac{(y-4)^2}{25}=1. \] **Step 2. Identify the center and the lengths of the semi-axes.** The standard form of an ellipse is \[ \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1, \] where the center is \((h, k)\) and \(a\) is the semi-major axis while \(b\) is the semi-minor axis. Comparing with our equation, we have: - Center: \((h, k)=(-6,4)\). - Denominator under \((x+6)^2\) is \(9\) implying \(b^2=9\) so \(b=3\). - Denominator under \((y-4)^2\) is \(25\) implying \(a^2=25\) so \(a=5\). Since \(a > b\) and \(a^2\) is under the \(y\)-term, the major axis is vertical. **Step 3. Find the foci.** For an ellipse with a vertical major axis, the distance \(c\) from the center to each focus is given by \[ c=\sqrt{a^2-b^2}. \] Plug in the values: \[ c=\sqrt{25-9}=\sqrt{16}=4. \] The foci lie along the major (vertical) axis. Therefore, their coordinates are \[ (-6,\, 4+4)=(-6,8) \quad \text{and} \quad (-6,\, 4-4)=(-6,0). \] **Step 4. Graph the ellipse.** - **Center:** \((-6,4)\). - **Vertical Extent:** From \(4-5=-1\) to \(4+5=9\) (since \(a=5\)). - **Horizontal Extent:** From \(-6-3=-9\) to \(-6+3=-3\) (since \(b=3\)). - **Foci:** Plotted at \((-6,8)\) and \((-6,0)\). A rough sketch of the ellipse: ``` (x) ^ | (-9,?) (-3,?) | | (-6,9) • <-- Top of ellipse | | Focus at (-6,8) | • | Left end •-----• Right end (-9,4) (-3,4) | | • Focus at (-6,0) | (-6,-1) • <-- Bottom of ellipse | | ``` The ellipse is centered at \((-6,4)\) with its major axis vertical and minor axis horizontal. **Final Answer:** - The foci are located at \(\boxed{(-6,8)}\) and \(\boxed{(-6,0)}\).