Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.