Answer :
Sure, let's pair the equations of the hyperbolas to their corresponding foci:
1. For the equation [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]:
- The foci are [tex]\((-7, -10)\)[/tex] and [tex]\((-7, 16)\)[/tex].
2. For the equation [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{15^2}=1\)[/tex]:
- The foci are [tex]\((-6, -2)\)[/tex] and [tex]\((14, -2)\)[/tex].
3. For the equation [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]:
- The foci are [tex]\((1, -22)\)[/tex] and [tex]\((1, 12)\)[/tex].
4. For the equation [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]:
- The foci are [tex]\((-7, 5)\)[/tex] and [tex]\((3, 5)\)[/tex].
Let's form the correct pairs:
- [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]
- Corresponds to foci: [tex]\((-7, -10)\)[/tex] and [tex]\((-7, 16)\)[/tex]
- [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{15^2}=1\)[/tex]
- Corresponds to foci: [tex]\((-6, -2)\)[/tex] and [tex]\((14, -2)\)[/tex]
- [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- Corresponds to foci: [tex]\((1, -22)\)[/tex] and [tex]\((1, 12)\)[/tex]
- [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]
- Corresponds to foci: [tex]\((-7, 5)\)[/tex] and [tex]\((3, 5)\)[/tex]
Therefore, the correct matches are:
1. [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex] with foci [tex]\((-7, -10)\)[/tex] and [tex]\((-7, 16)\)[/tex]
2. [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{15^2}=1\)[/tex] with foci [tex]\((-6, -2)\)[/tex] and [tex]\((14, -2)\)[/tex]
3. [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex] with foci [tex]\((1, -22)\)[/tex] and [tex]\((1, 12)\)[/tex]
4. [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex] with foci [tex]\((-7, 5)\)[/tex] and [tex]\((3, 5)\)[/tex]
1. For the equation [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]:
- The foci are [tex]\((-7, -10)\)[/tex] and [tex]\((-7, 16)\)[/tex].
2. For the equation [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{15^2}=1\)[/tex]:
- The foci are [tex]\((-6, -2)\)[/tex] and [tex]\((14, -2)\)[/tex].
3. For the equation [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]:
- The foci are [tex]\((1, -22)\)[/tex] and [tex]\((1, 12)\)[/tex].
4. For the equation [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]:
- The foci are [tex]\((-7, 5)\)[/tex] and [tex]\((3, 5)\)[/tex].
Let's form the correct pairs:
- [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]
- Corresponds to foci: [tex]\((-7, -10)\)[/tex] and [tex]\((-7, 16)\)[/tex]
- [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{15^2}=1\)[/tex]
- Corresponds to foci: [tex]\((-6, -2)\)[/tex] and [tex]\((14, -2)\)[/tex]
- [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- Corresponds to foci: [tex]\((1, -22)\)[/tex] and [tex]\((1, 12)\)[/tex]
- [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]
- Corresponds to foci: [tex]\((-7, 5)\)[/tex] and [tex]\((3, 5)\)[/tex]
Therefore, the correct matches are:
1. [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex] with foci [tex]\((-7, -10)\)[/tex] and [tex]\((-7, 16)\)[/tex]
2. [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{15^2}=1\)[/tex] with foci [tex]\((-6, -2)\)[/tex] and [tex]\((14, -2)\)[/tex]
3. [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex] with foci [tex]\((1, -22)\)[/tex] and [tex]\((1, 12)\)[/tex]
4. [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex] with foci [tex]\((-7, 5)\)[/tex] and [tex]\((3, 5)\)[/tex]