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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]
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