Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
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]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
