At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To find the center and foci of the given ellipse, we need to rewrite the given equation in its standard form, which typically looks like:
[tex]\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the ellipse, and [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the lengths of the semi-major and semi-minor axes, respectively.
Given the equation:
[tex]\[ 9x^2 + 16y^2 + 126x + 96y + 441 = 0 \][/tex]
We need to complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
### Completing the Square
First, we group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ 9x^2 + 126x + 16y^2 + 96y + 441 = 0 \][/tex]
#### For [tex]\(x\)[/tex]-terms:
[tex]\[ 9(x^2 + \frac{126}{9}x) = 9(x^2 + 14x) \][/tex]
To complete the square for [tex]\(x\)[/tex]:
[tex]\[ x^2 + 14x \][/tex]
[tex]\[ = (x + 7)^2 - 49 \][/tex]
So,
[tex]\[ 9(x + 7)^2 - 9 \cdot 49 \][/tex]
#### For [tex]\(y\)[/tex]-terms:
[tex]\[ 16(y^2 + \frac{96}{16}y) = 16(y^2 + 6y) \][/tex]
To complete the square for [tex]\(y\)[/tex]:
[tex]\[ y^2 + 6y \][/tex]
[tex]\[ = (y + 3)^2 - 9 \][/tex]
So,
[tex]\[ 16(y + 3)^2 - 16 \cdot 9 \][/tex]
### Substitute Back:
[tex]\[ 9(x + 7)^2 - 9 \cdot 49 + 16(y + 3)^2 - 16 \cdot 9 + 441 = 0 \][/tex]
Simplify the constants:
[tex]\[ 9(x + 7)^2 - 441 + 16(y + 3)^2 - 144 + 441 = 0 \][/tex]
Combine the constants:
[tex]\[ 9(x + 7)^2 + 16(y + 3)^2 - 144 = 0 \][/tex]
### Standard Form:
[tex]\[ 9(x + 7)^2 + 16(y + 3)^2 = 144 \][/tex]
Divide through by 144:
[tex]\[ \frac{(x + 7)^2}{16} + \frac{(y + 3)^2}{9} = 1 \][/tex]
Now we have the standard form of the ellipse equation:
[tex]\[ \frac{(x + 7)^2}{16} + \frac{(y + 3)^2}{9} = 1 \][/tex]
From this, we can identify:
- The center [tex]\((h, k)\)[/tex] of the ellipse is [tex]\((-7, -3)\)[/tex].
- The lengths of the semi-major axis [tex]\(a\)[/tex] and the semi-minor axis [tex]\(b\)[/tex]:
- [tex]\(a^2 = 16\)[/tex] so [tex]\(a = 4\)[/tex]
- [tex]\(b^2 = 9\)[/tex] so [tex]\(b = 3\)[/tex]
### Determining the Foci:
The distance [tex]\(c\)[/tex] from the center to each focus is found using:
[tex]\[ c = \sqrt{a^2 - b^2} \][/tex]
Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ c = \sqrt{16 - 9} = \sqrt{7} \][/tex]
### Foci Coordinates:
The foci are along the [tex]\(x\)[/tex]-axis (horizontal) because [tex]\(a > b\)[/tex], so we add and subtract [tex]\(c\)[/tex] from the [tex]\(x\)[/tex]-coordinate of the center:
The foci are at:
[tex]\[ (-7 + \sqrt{7}, -3) \text{ and } (-7 - \sqrt{7}, -3) \][/tex]
### Answers:
- Center: [tex]\((-7, -3)\)[/tex]
- Foci: [tex]\((-7 \pm \sqrt{7}, -3)\)[/tex]
[tex]\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the ellipse, and [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the lengths of the semi-major and semi-minor axes, respectively.
Given the equation:
[tex]\[ 9x^2 + 16y^2 + 126x + 96y + 441 = 0 \][/tex]
We need to complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
### Completing the Square
First, we group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ 9x^2 + 126x + 16y^2 + 96y + 441 = 0 \][/tex]
#### For [tex]\(x\)[/tex]-terms:
[tex]\[ 9(x^2 + \frac{126}{9}x) = 9(x^2 + 14x) \][/tex]
To complete the square for [tex]\(x\)[/tex]:
[tex]\[ x^2 + 14x \][/tex]
[tex]\[ = (x + 7)^2 - 49 \][/tex]
So,
[tex]\[ 9(x + 7)^2 - 9 \cdot 49 \][/tex]
#### For [tex]\(y\)[/tex]-terms:
[tex]\[ 16(y^2 + \frac{96}{16}y) = 16(y^2 + 6y) \][/tex]
To complete the square for [tex]\(y\)[/tex]:
[tex]\[ y^2 + 6y \][/tex]
[tex]\[ = (y + 3)^2 - 9 \][/tex]
So,
[tex]\[ 16(y + 3)^2 - 16 \cdot 9 \][/tex]
### Substitute Back:
[tex]\[ 9(x + 7)^2 - 9 \cdot 49 + 16(y + 3)^2 - 16 \cdot 9 + 441 = 0 \][/tex]
Simplify the constants:
[tex]\[ 9(x + 7)^2 - 441 + 16(y + 3)^2 - 144 + 441 = 0 \][/tex]
Combine the constants:
[tex]\[ 9(x + 7)^2 + 16(y + 3)^2 - 144 = 0 \][/tex]
### Standard Form:
[tex]\[ 9(x + 7)^2 + 16(y + 3)^2 = 144 \][/tex]
Divide through by 144:
[tex]\[ \frac{(x + 7)^2}{16} + \frac{(y + 3)^2}{9} = 1 \][/tex]
Now we have the standard form of the ellipse equation:
[tex]\[ \frac{(x + 7)^2}{16} + \frac{(y + 3)^2}{9} = 1 \][/tex]
From this, we can identify:
- The center [tex]\((h, k)\)[/tex] of the ellipse is [tex]\((-7, -3)\)[/tex].
- The lengths of the semi-major axis [tex]\(a\)[/tex] and the semi-minor axis [tex]\(b\)[/tex]:
- [tex]\(a^2 = 16\)[/tex] so [tex]\(a = 4\)[/tex]
- [tex]\(b^2 = 9\)[/tex] so [tex]\(b = 3\)[/tex]
### Determining the Foci:
The distance [tex]\(c\)[/tex] from the center to each focus is found using:
[tex]\[ c = \sqrt{a^2 - b^2} \][/tex]
Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ c = \sqrt{16 - 9} = \sqrt{7} \][/tex]
### Foci Coordinates:
The foci are along the [tex]\(x\)[/tex]-axis (horizontal) because [tex]\(a > b\)[/tex], so we add and subtract [tex]\(c\)[/tex] from the [tex]\(x\)[/tex]-coordinate of the center:
The foci are at:
[tex]\[ (-7 + \sqrt{7}, -3) \text{ and } (-7 - \sqrt{7}, -3) \][/tex]
### Answers:
- Center: [tex]\((-7, -3)\)[/tex]
- Foci: [tex]\((-7 \pm \sqrt{7}, -3)\)[/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.