At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
To determine which hyperbola has one focus point in common with the given hyperbola, we need to follow this step-by-step process:
### Step 1: Extract Parameters from the Given Hyperbola
The equation of the hyperbola is:
[tex]$ \frac{(y+11)^2}{15^2}-\frac{(x-7)^2}{8^2}=1 $[/tex]
This tells us:
- Center [tex]\((h, k) = (7, -11)\)[/tex]
- Semi-major axis [tex]\(a = 15\)[/tex]
- Semi-minor axis [tex]\(b = 8\)[/tex]
### Step 2: Calculate the Distance to the Foci (c) for the Given Hyperbola
For a hyperbola [tex]\( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)[/tex], the distance to each focus from the center, [tex]\(c\)[/tex], is given by:
[tex]\[ c = \sqrt{a^2 + b^2} \][/tex]
So,
[tex]\[ c = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \][/tex]
### Step 3: Determine the Foci of the Given Hyperbola
Since the given hyperbola is vertical:
- The foci are at:
[tex]\[ (7, -11 + 17) = (7, 6) \][/tex]
[tex]\[ (7, -11 - 17) = (7, -28) \][/tex]
### Step 4: Compare with Each Given Hyperbola
For each candidate hyperbola, we calculate the coordinates of the foci to check if they coincide with [tex]\((7, 6)\)[/tex] or [tex]\((7, -28)\)[/tex].
#### Hyperbola 1: [tex]\(\frac{(y+8)^2}{12^2} - \frac{(x-7)^2}{5^2} = 1\)[/tex]
- Center: [tex]\((7, -8)\)[/tex]
- [tex]\(a = 12\)[/tex]
- [tex]\(b = 5\)[/tex]
- [tex]\(c = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\)[/tex]
- Foci: [tex]\((7, -8 + 13) = (7, 5)\)[/tex] and [tex]\((7, -8 - 13) = (7, -21)\)[/tex]
No match with the given foci.
#### Hyperbola 2: [tex]\(\frac{(y+16)^2}{7^2} - \frac{(z-7)^2}{24^2} = 1\)[/tex]
- Invalid Variable (z instead of x), skip.
#### Hyperbola 3: [tex]\(\frac{(x-12)^2}{4^2} - \frac{(y+28)^2}{3^2} = 1\)[/tex]
- Horizontal Hyperbola, irrelevant.
#### Hyperbola 4: [tex]\(\frac{(z-15)^2}{6^2} - \frac{(y+28)^2}{8^2} = 1\)[/tex]
- Invalid Variable (z instead of x), skip.
#### Hyperbola 5: [tex]\(\frac{(y-21)^2}{9^2} - \frac{(x-7)^2}{12^2} = 1\)[/tex]
- Center: [tex]\((7, 21)\)[/tex]
- [tex]\(a = 9\)[/tex]
- [tex]\(b = 12\)[/tex]
- [tex]\(c = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\)[/tex]
- Foci: [tex]\((7, 21 + 15) = (7, 36)\)[/tex] and [tex]\((7, 21 - 15) = (7, 6)\)[/tex]
There is a matching focus at [tex]\((7, 6)\)[/tex].
### Conclusion
The hyperbola that shares one focus point, [tex]\((7, 6)\)[/tex], is:
[tex]\[ \frac{(y-21)^2}{9^2} - \frac{(x-7)^2}{12^2} = 1 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{5} \][/tex]
### Step 1: Extract Parameters from the Given Hyperbola
The equation of the hyperbola is:
[tex]$ \frac{(y+11)^2}{15^2}-\frac{(x-7)^2}{8^2}=1 $[/tex]
This tells us:
- Center [tex]\((h, k) = (7, -11)\)[/tex]
- Semi-major axis [tex]\(a = 15\)[/tex]
- Semi-minor axis [tex]\(b = 8\)[/tex]
### Step 2: Calculate the Distance to the Foci (c) for the Given Hyperbola
For a hyperbola [tex]\( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)[/tex], the distance to each focus from the center, [tex]\(c\)[/tex], is given by:
[tex]\[ c = \sqrt{a^2 + b^2} \][/tex]
So,
[tex]\[ c = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \][/tex]
### Step 3: Determine the Foci of the Given Hyperbola
Since the given hyperbola is vertical:
- The foci are at:
[tex]\[ (7, -11 + 17) = (7, 6) \][/tex]
[tex]\[ (7, -11 - 17) = (7, -28) \][/tex]
### Step 4: Compare with Each Given Hyperbola
For each candidate hyperbola, we calculate the coordinates of the foci to check if they coincide with [tex]\((7, 6)\)[/tex] or [tex]\((7, -28)\)[/tex].
#### Hyperbola 1: [tex]\(\frac{(y+8)^2}{12^2} - \frac{(x-7)^2}{5^2} = 1\)[/tex]
- Center: [tex]\((7, -8)\)[/tex]
- [tex]\(a = 12\)[/tex]
- [tex]\(b = 5\)[/tex]
- [tex]\(c = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\)[/tex]
- Foci: [tex]\((7, -8 + 13) = (7, 5)\)[/tex] and [tex]\((7, -8 - 13) = (7, -21)\)[/tex]
No match with the given foci.
#### Hyperbola 2: [tex]\(\frac{(y+16)^2}{7^2} - \frac{(z-7)^2}{24^2} = 1\)[/tex]
- Invalid Variable (z instead of x), skip.
#### Hyperbola 3: [tex]\(\frac{(x-12)^2}{4^2} - \frac{(y+28)^2}{3^2} = 1\)[/tex]
- Horizontal Hyperbola, irrelevant.
#### Hyperbola 4: [tex]\(\frac{(z-15)^2}{6^2} - \frac{(y+28)^2}{8^2} = 1\)[/tex]
- Invalid Variable (z instead of x), skip.
#### Hyperbola 5: [tex]\(\frac{(y-21)^2}{9^2} - \frac{(x-7)^2}{12^2} = 1\)[/tex]
- Center: [tex]\((7, 21)\)[/tex]
- [tex]\(a = 9\)[/tex]
- [tex]\(b = 12\)[/tex]
- [tex]\(c = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\)[/tex]
- Foci: [tex]\((7, 21 + 15) = (7, 36)\)[/tex] and [tex]\((7, 21 - 15) = (7, 6)\)[/tex]
There is a matching focus at [tex]\((7, 6)\)[/tex].
### Conclusion
The hyperbola that shares one focus point, [tex]\((7, 6)\)[/tex], is:
[tex]\[ \frac{(y-21)^2}{9^2} - \frac{(x-7)^2}{12^2} = 1 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{5} \][/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
