Answer :
To determine which hyperbola has one focus point in common with the hyperbola [tex]\(\frac{(y+11)^2}{15^2}-\frac{(x-7)^2}{8^2}=1\)[/tex], we need to follow these steps:
1. Identify the center and the foci of the given hyperbola:
- The given hyperbola [tex]\(\frac{(y+11)^2}{15^2}-\frac{(x-7)^2}{8^2}=1\)[/tex] has its center at [tex]\((7, -11)\)[/tex].
- Since the equation is of the form [tex]\(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)[/tex], it is a vertical hyperbola.
- For a vertical hyperbola, the distance to the foci, [tex]\(c\)[/tex], is found using [tex]\(c = \sqrt{a^2 + b^2}\)[/tex].
Given [tex]\(a^2 = 15^2 = 225\)[/tex] and [tex]\(b^2 = 8^2 = 64\)[/tex]:
[tex]\[ c = \sqrt{225 + 64} = \sqrt{289} = 17 \][/tex]
- The foci are located at [tex]\((h, k + c)\)[/tex] and [tex]\((h, k - c)\)[/tex]:
[tex]\[ (7, -11 + 17) = (7, 6) \quad \text{and} \quad (7, -11 - 17) = (7, -28) \][/tex]
2. Analyze each given hyperbola and find their foci:
- For [tex]\(\frac{(y+8)^2}{12^2}-\frac{(x-7)^2}{5^2}=1\)[/tex]:
- Center: [tex]\((7, -8)\)[/tex]
- [tex]\(a^2 = 12^2 = 144\)[/tex] and [tex]\(b^2 = 5^2 = 25\)[/tex], thus
[tex]\[ c = \sqrt{144 + 25} = \sqrt{169} = 13 \][/tex]
- Foci: [tex]\((7, -8 + 13) = (7, 5)\)[/tex] and [tex]\((7, -8 - 13) = (7, -21)\)[/tex]
- For [tex]\(\frac{(y+16)^2}{7^2}-\frac{(x-7)^2}{24^2}=1\)[/tex]:
- Center: [tex]\((7, -16)\)[/tex]
- [tex]\(a^2 = 7^2 = 49\)[/tex] and [tex]\(b^2 = 24^2 = 576\)[/tex], thus
[tex]\[ c = \sqrt{49 + 576} = \sqrt{625} = 25 \][/tex]
- Foci: [tex]\((7, -16 + 25) = (7, 9)\)[/tex] and [tex]\((7, -16 - 25) = (7, -41)\)[/tex]
- For [tex]\(\frac{(x-12)^2}{4^2}-\frac{(y+28)^2}{3^2}=1\)[/tex] (horizontal hyperbola):
- Center: [tex]\((12, -28)\)[/tex]
- [tex]\(a^2 = 4^2 = 16\)[/tex] and [tex]\(b^2 = 3^2 = 9\)[/tex], thus
[tex]\[ c = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
- Foci: [tex]\((12 + 5, -28) = (17, -28)\)[/tex] and [tex]\((12 - 5, -28) = (7, -28)\)[/tex]
- For [tex]\(\frac{(z-15)^2}{6^2}-\frac{(y+28)^2}{8^2}=1\)[/tex] (z-plane):
- This hyperbola is not in the xy-plane and is thus invalid for comparison.
- For [tex]\(\frac{(y-21)^2}{9^2}-\frac{(x-7)^2}{12^2}=1\)[/tex]:
- Center: [tex]\((7, 21)\)[/tex]
- [tex]\(a^2 = 9^2 = 81\)[/tex] and [tex]\(b^2 = 12^2 = 144\)[/tex], thus
[tex]\[ c = \sqrt{81 + 144} = \sqrt{225} = 15 \][/tex]
- Foci: [tex]\((7, 21 + 15) = (7, 36)\)[/tex] and [tex]\((7, 21 - 15) = (7, 6)\)[/tex]
3. Identify common foci:
- The foci of the given hyperbola are [tex]\((7, 6)\)[/tex] and [tex]\((7, -28)\)[/tex].
- For hyperbola 1: the foci are [tex]\((7, 5)\)[/tex] and [tex]\((7, -21)\)[/tex], no common focus.
- For hyperbola 2: the foci are [tex]\((7, 9)\)[/tex] and [tex]\((7, -41)\)[/tex], no common focus.
- For hyperbola 3: the foci are [tex]\((17, -28)\)[/tex] and [tex]\((7, -28)\)[/tex], one common focus at [tex]\((7, -28)\)[/tex].
- For hyperbola 5: the foci are [tex]\((7, 36)\)[/tex] and [tex]\((7, 6)\)[/tex], one common focus at [tex]\((7, 6)\)[/tex].
Thus, the hyperbolas that share a focus with the given hyperbola are:
[tex]\[ \boxed{5} \][/tex]
1. Identify the center and the foci of the given hyperbola:
- The given hyperbola [tex]\(\frac{(y+11)^2}{15^2}-\frac{(x-7)^2}{8^2}=1\)[/tex] has its center at [tex]\((7, -11)\)[/tex].
- Since the equation is of the form [tex]\(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)[/tex], it is a vertical hyperbola.
- For a vertical hyperbola, the distance to the foci, [tex]\(c\)[/tex], is found using [tex]\(c = \sqrt{a^2 + b^2}\)[/tex].
Given [tex]\(a^2 = 15^2 = 225\)[/tex] and [tex]\(b^2 = 8^2 = 64\)[/tex]:
[tex]\[ c = \sqrt{225 + 64} = \sqrt{289} = 17 \][/tex]
- The foci are located at [tex]\((h, k + c)\)[/tex] and [tex]\((h, k - c)\)[/tex]:
[tex]\[ (7, -11 + 17) = (7, 6) \quad \text{and} \quad (7, -11 - 17) = (7, -28) \][/tex]
2. Analyze each given hyperbola and find their foci:
- For [tex]\(\frac{(y+8)^2}{12^2}-\frac{(x-7)^2}{5^2}=1\)[/tex]:
- Center: [tex]\((7, -8)\)[/tex]
- [tex]\(a^2 = 12^2 = 144\)[/tex] and [tex]\(b^2 = 5^2 = 25\)[/tex], thus
[tex]\[ c = \sqrt{144 + 25} = \sqrt{169} = 13 \][/tex]
- Foci: [tex]\((7, -8 + 13) = (7, 5)\)[/tex] and [tex]\((7, -8 - 13) = (7, -21)\)[/tex]
- For [tex]\(\frac{(y+16)^2}{7^2}-\frac{(x-7)^2}{24^2}=1\)[/tex]:
- Center: [tex]\((7, -16)\)[/tex]
- [tex]\(a^2 = 7^2 = 49\)[/tex] and [tex]\(b^2 = 24^2 = 576\)[/tex], thus
[tex]\[ c = \sqrt{49 + 576} = \sqrt{625} = 25 \][/tex]
- Foci: [tex]\((7, -16 + 25) = (7, 9)\)[/tex] and [tex]\((7, -16 - 25) = (7, -41)\)[/tex]
- For [tex]\(\frac{(x-12)^2}{4^2}-\frac{(y+28)^2}{3^2}=1\)[/tex] (horizontal hyperbola):
- Center: [tex]\((12, -28)\)[/tex]
- [tex]\(a^2 = 4^2 = 16\)[/tex] and [tex]\(b^2 = 3^2 = 9\)[/tex], thus
[tex]\[ c = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
- Foci: [tex]\((12 + 5, -28) = (17, -28)\)[/tex] and [tex]\((12 - 5, -28) = (7, -28)\)[/tex]
- For [tex]\(\frac{(z-15)^2}{6^2}-\frac{(y+28)^2}{8^2}=1\)[/tex] (z-plane):
- This hyperbola is not in the xy-plane and is thus invalid for comparison.
- For [tex]\(\frac{(y-21)^2}{9^2}-\frac{(x-7)^2}{12^2}=1\)[/tex]:
- Center: [tex]\((7, 21)\)[/tex]
- [tex]\(a^2 = 9^2 = 81\)[/tex] and [tex]\(b^2 = 12^2 = 144\)[/tex], thus
[tex]\[ c = \sqrt{81 + 144} = \sqrt{225} = 15 \][/tex]
- Foci: [tex]\((7, 21 + 15) = (7, 36)\)[/tex] and [tex]\((7, 21 - 15) = (7, 6)\)[/tex]
3. Identify common foci:
- The foci of the given hyperbola are [tex]\((7, 6)\)[/tex] and [tex]\((7, -28)\)[/tex].
- For hyperbola 1: the foci are [tex]\((7, 5)\)[/tex] and [tex]\((7, -21)\)[/tex], no common focus.
- For hyperbola 2: the foci are [tex]\((7, 9)\)[/tex] and [tex]\((7, -41)\)[/tex], no common focus.
- For hyperbola 3: the foci are [tex]\((17, -28)\)[/tex] and [tex]\((7, -28)\)[/tex], one common focus at [tex]\((7, -28)\)[/tex].
- For hyperbola 5: the foci are [tex]\((7, 36)\)[/tex] and [tex]\((7, 6)\)[/tex], one common focus at [tex]\((7, 6)\)[/tex].
Thus, the hyperbolas that share a focus with the given hyperbola are:
[tex]\[ \boxed{5} \][/tex]
