Answer :
To find the center and foci of the given ellipse equation:
[tex]\[ 25x^2 + 16y^2 + 50x - 64y - 311 = 0 \][/tex]
we can follow these steps:
### 1. Completing the Square
First, we need to rearrange the equation and complete the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
#### For [tex]\(x\)[/tex]:
Starting with the [tex]\(x\)[/tex] terms:
[tex]\[ 25x^2 + 50x \][/tex]
Factor out the 25 from the x terms:
[tex]\[ 25(x^2 + 2x) \][/tex]
Complete the square inside the parentheses. Add and subtract 1 inside the parentheses, then multiply by 25:
[tex]\[ 25((x + 1)^2 - 1) \][/tex]
[tex]\[ = 25(x + 1)^2 - 25 \][/tex]
#### For [tex]\(y\)[/tex]:
Starting with the [tex]\(y\)[/tex] terms:
[tex]\[ 16y^2 - 64y \][/tex]
Factor out the 16 from the y terms:
[tex]\[ 16(y^2 - 4y) \][/tex]
Complete the square inside the parentheses. Add and subtract 4 inside the parentheses, then multiply by 16:
[tex]\[ 16((y - 2)^2 - 4) \][/tex]
[tex]\[ = 16(y - 2)^2 - 64 \][/tex]
### 2. Rewriting the Equation
Combine these results into the original equation:
[tex]\[ 25((x + 1)^2 - 1) + 16((y - 2)^2 - 4) - 311 = 0 \][/tex]
Simplify by combining constant terms:
[tex]\[ 25(x + 1)^2 - 25 + 16(y - 2)^2 - 64 - 311 = 0 \][/tex]
[tex]\[ 25(x + 1)^2 + 16(y - 2)^2 - 400 = 0 \][/tex]
[tex]\[ 25(x + 1)^2 + 16(y - 2)^2 = 400 \][/tex]
### 3. Standard Form
Divide the entire equation by 400 to obtain the standard form of the ellipse equation:
[tex]\[ \frac{(x + 1)^2}{16} + \frac{(y - 2)^2}{25} = 1 \][/tex]
From this standard form, we determine:
- The center [tex]\((h, k) = (-1, 2)\)[/tex]
- [tex]\(a^2 = 25 \implies a = 5\)[/tex]
- [tex]\(b^2 = 16 \implies b = 4\)[/tex]
### 4. Finding the Foci
To find the foci, we need to calculate the distance [tex]\(c\)[/tex] from the center to each focus where:
[tex]\[ c^2 = a^2 - b^2 \][/tex]
[tex]\[ c^2 = 25 - 16 \][/tex]
[tex]\[ c^2 = 9 \][/tex]
[tex]\[ c = 3 \][/tex]
The foci are located at [tex]\((h, k \pm c)\)[/tex], given that the major axis is vertical. Hence, the coordinates for the foci will be:
[tex]\[ (-1, 2 - 3) \][/tex]
[tex]\[ (-1, 2 + 3) \][/tex]
So, the foci are:
[tex]\[ (-1, -1) \; \text{and} \; (-1, 5) \][/tex]
Thus, the final answer for finding the center and foci is:
- Center: [tex]\((-1, 2)\)[/tex]
- Foci: [tex]\((-1, -1)\)[/tex] and [tex]\((-1, 5)\)[/tex]
### Final Answer:
Enter the smallest [tex]\(y\)[/tex] first:
[tex]\[ (-1, -1) \; \text{and} \; (-1, 5) \][/tex]
[tex]\[ 25x^2 + 16y^2 + 50x - 64y - 311 = 0 \][/tex]
we can follow these steps:
### 1. Completing the Square
First, we need to rearrange the equation and complete the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
#### For [tex]\(x\)[/tex]:
Starting with the [tex]\(x\)[/tex] terms:
[tex]\[ 25x^2 + 50x \][/tex]
Factor out the 25 from the x terms:
[tex]\[ 25(x^2 + 2x) \][/tex]
Complete the square inside the parentheses. Add and subtract 1 inside the parentheses, then multiply by 25:
[tex]\[ 25((x + 1)^2 - 1) \][/tex]
[tex]\[ = 25(x + 1)^2 - 25 \][/tex]
#### For [tex]\(y\)[/tex]:
Starting with the [tex]\(y\)[/tex] terms:
[tex]\[ 16y^2 - 64y \][/tex]
Factor out the 16 from the y terms:
[tex]\[ 16(y^2 - 4y) \][/tex]
Complete the square inside the parentheses. Add and subtract 4 inside the parentheses, then multiply by 16:
[tex]\[ 16((y - 2)^2 - 4) \][/tex]
[tex]\[ = 16(y - 2)^2 - 64 \][/tex]
### 2. Rewriting the Equation
Combine these results into the original equation:
[tex]\[ 25((x + 1)^2 - 1) + 16((y - 2)^2 - 4) - 311 = 0 \][/tex]
Simplify by combining constant terms:
[tex]\[ 25(x + 1)^2 - 25 + 16(y - 2)^2 - 64 - 311 = 0 \][/tex]
[tex]\[ 25(x + 1)^2 + 16(y - 2)^2 - 400 = 0 \][/tex]
[tex]\[ 25(x + 1)^2 + 16(y - 2)^2 = 400 \][/tex]
### 3. Standard Form
Divide the entire equation by 400 to obtain the standard form of the ellipse equation:
[tex]\[ \frac{(x + 1)^2}{16} + \frac{(y - 2)^2}{25} = 1 \][/tex]
From this standard form, we determine:
- The center [tex]\((h, k) = (-1, 2)\)[/tex]
- [tex]\(a^2 = 25 \implies a = 5\)[/tex]
- [tex]\(b^2 = 16 \implies b = 4\)[/tex]
### 4. Finding the Foci
To find the foci, we need to calculate the distance [tex]\(c\)[/tex] from the center to each focus where:
[tex]\[ c^2 = a^2 - b^2 \][/tex]
[tex]\[ c^2 = 25 - 16 \][/tex]
[tex]\[ c^2 = 9 \][/tex]
[tex]\[ c = 3 \][/tex]
The foci are located at [tex]\((h, k \pm c)\)[/tex], given that the major axis is vertical. Hence, the coordinates for the foci will be:
[tex]\[ (-1, 2 - 3) \][/tex]
[tex]\[ (-1, 2 + 3) \][/tex]
So, the foci are:
[tex]\[ (-1, -1) \; \text{and} \; (-1, 5) \][/tex]
Thus, the final answer for finding the center and foci is:
- Center: [tex]\((-1, 2)\)[/tex]
- Foci: [tex]\((-1, -1)\)[/tex] and [tex]\((-1, 5)\)[/tex]
### Final Answer:
Enter the smallest [tex]\(y\)[/tex] first:
[tex]\[ (-1, -1) \; \text{and} \; (-1, 5) \][/tex]
