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Sagot :
Sure, let's match each hyperbola equation to its corresponding foci step-by-step:
1. [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]
- Foci calculated as [tex]\((3.0, 5)\)[/tex] and [tex]\((-7.0, 5)\)[/tex]
2. [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{15^2}=1\)[/tex]
- Foci calculated as [tex]\((21.0, -2)\)[/tex] and [tex]\((-13.0, -2)\)[/tex]
3. [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- Foci calculated as [tex]\((1, 12.0)\)[/tex] and [tex]\((1, -22.0)\)[/tex]
4. [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]
- Foci calculated as [tex]\((14.0, -2)\)[/tex] and [tex]\((-6.0, -2)\)[/tex]
5. [tex]\(\frac{(y-3)^2}{5^2}-\frac{(x-7)^2}{12^2}=1\)[/tex]
- Foci calculated as [tex]\((7, 16.0)\)[/tex] and [tex]\((7, -10.0)\)[/tex]
6. [tex]\(\frac{(y-3)^2}{5^2}-\frac{(x+7)^2}{12^2}=1\)[/tex]
- Foci calculated as [tex]\((7, 16.0)\)[/tex] and [tex]\((7, -10.0)\)[/tex]
7. [tex]\(\frac{(y+5)^2}{6^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- Foci calculated as [tex]\((1, 5.0)\)[/tex] and [tex]\((1, -15.0)\)[/tex]
Matching these results to the foci provided:
- [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1 \longleftrightarrow (1, -22) \text{ and } (1,12)\)[/tex]
- [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1 \longleftrightarrow (-7,5) \text{ and } (3,5)\)[/tex]
- [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1 \longleftrightarrow (-6,-2) \text{ and } (14,-2)\)[/tex]
- [tex]\(\frac{(y-3)^2}{5^2}-\frac{(x+7)^2}{12^2}=1 \longleftrightarrow (-7,-10) \text{ and } (-7,16)\)[/tex]
Therefore, the correct pairs are:
- The pair (1, -22) and (1,12) corresponds to the equation [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- The pair (-7,5) and (3,5) corresponds to the equation [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]
- The pair (-6,-2) and (14,-2) corresponds to the equation [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]
- The pair (-7,-10) and (-7,16) corresponds to the equation [tex]\(\frac{(y-3)^2}{5^2}-\frac{(x+7)^2}{12^2}=1\)[/tex]
Thus:
- [tex]\((1, -22) \text{ and } (1,12) \longleftrightarrow \frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- [tex]\((-7,5) \text{ and } (3,5) \longleftrightarrow \frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]
- [tex]\((-6,-2) \text{ and } (14,-2) \longleftrightarrow \frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]
- [tex]\((-7,-10) \text{ and } (-7,16) \longleftrightarrow \frac{(y-3)^2}{5^2}-\frac{(x+7)^2}{12^2}=1\)[/tex]
1. [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]
- Foci calculated as [tex]\((3.0, 5)\)[/tex] and [tex]\((-7.0, 5)\)[/tex]
2. [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{15^2}=1\)[/tex]
- Foci calculated as [tex]\((21.0, -2)\)[/tex] and [tex]\((-13.0, -2)\)[/tex]
3. [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- Foci calculated as [tex]\((1, 12.0)\)[/tex] and [tex]\((1, -22.0)\)[/tex]
4. [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]
- Foci calculated as [tex]\((14.0, -2)\)[/tex] and [tex]\((-6.0, -2)\)[/tex]
5. [tex]\(\frac{(y-3)^2}{5^2}-\frac{(x-7)^2}{12^2}=1\)[/tex]
- Foci calculated as [tex]\((7, 16.0)\)[/tex] and [tex]\((7, -10.0)\)[/tex]
6. [tex]\(\frac{(y-3)^2}{5^2}-\frac{(x+7)^2}{12^2}=1\)[/tex]
- Foci calculated as [tex]\((7, 16.0)\)[/tex] and [tex]\((7, -10.0)\)[/tex]
7. [tex]\(\frac{(y+5)^2}{6^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- Foci calculated as [tex]\((1, 5.0)\)[/tex] and [tex]\((1, -15.0)\)[/tex]
Matching these results to the foci provided:
- [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1 \longleftrightarrow (1, -22) \text{ and } (1,12)\)[/tex]
- [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1 \longleftrightarrow (-7,5) \text{ and } (3,5)\)[/tex]
- [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1 \longleftrightarrow (-6,-2) \text{ and } (14,-2)\)[/tex]
- [tex]\(\frac{(y-3)^2}{5^2}-\frac{(x+7)^2}{12^2}=1 \longleftrightarrow (-7,-10) \text{ and } (-7,16)\)[/tex]
Therefore, the correct pairs are:
- The pair (1, -22) and (1,12) corresponds to the equation [tex]\(\frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- The pair (-7,5) and (3,5) corresponds to the equation [tex]\(\frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]
- The pair (-6,-2) and (14,-2) corresponds to the equation [tex]\(\frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]
- The pair (-7,-10) and (-7,16) corresponds to the equation [tex]\(\frac{(y-3)^2}{5^2}-\frac{(x+7)^2}{12^2}=1\)[/tex]
Thus:
- [tex]\((1, -22) \text{ and } (1,12) \longleftrightarrow \frac{(y+5)^2}{15^2}-\frac{(x-1)^2}{8^2}=1\)[/tex]
- [tex]\((-7,5) \text{ and } (3,5) \longleftrightarrow \frac{(x+2)^2}{3^2}-\frac{(y-5)^2}{4^2}=1\)[/tex]
- [tex]\((-6,-2) \text{ and } (14,-2) \longleftrightarrow \frac{(x-4)^2}{8^2}-\frac{(y+2)^2}{6^2}=1\)[/tex]
- [tex]\((-7,-10) \text{ and } (-7,16) \longleftrightarrow \frac{(y-3)^2}{5^2}-\frac{(x+7)^2}{12^2}=1\)[/tex]
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