he foci of the ellipse defined by the equation $$ \frac{(x-5)^{2}}{25}+\frac{(y+1)}{4} $$ essary, round to the nearest tenth.

Question

UpStudy AI · Accepted Answer

To find the foci of the ellipse defined by the equation:

$$
\frac{(x - 5)^2}{25} + \frac{(y + 1)^2}{4} = 1
$$

we can follow these steps:

### 1. Identify the Standard Form
The given equation is in the standard form of an ellipse:

$$
\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
$$

Where:
- $$(h, k)$$ is the center of the ellipse.
- $$a$$ is the semi-major axis length.
- $$b$$ is the semi-minor axis length.

### 2. Extract Parameters
From the given equation:
- Center $$(h, k) = (5, -1)$$
- $$a^2 = 25 \Rightarrow a = 5$$
- $$b^2 = 4 \Rightarrow b = 2$$

### 3. Determine the Orientation
Since $$a^2 > b^2$$, the major axis is horizontal.

### 4. Calculate the Distance to the Foci ($$c$$)
The distance from the center to each focus is given by:

$$
c = \sqrt{a^2 - b^2} = \sqrt{25 - 4} = \sqrt{21} \approx 4.6
$$

### 5. Locate the Foci
Since the major axis is horizontal, the foci are located to the left and right of the center:

$$
\text{Foci} = (h \pm c, k) = (5 \pm 4.6, -1)
$$

Calculating the coordinates:
- Right Focus: $$5 + 4.6 = 9.6$$, so $$(9.6, -1)$$
- Left Focus: $$5 - 4.6 = 0.4$$, so $$(0.4, -1)$$

### **Final Answer**
The foci of the ellipse are at the points ( 0.4, –1 ) and ( 9.6, –1 ).
The foci of the ellipse are at (0.4, -1) and (9.6, -1).

he foci of the ellipse defined by the equation \( \frac{(x-5)^{2}}{25}+\frac{(y+1)}{4} \) essary, round to the nearest tenth.