Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. 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 appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
