Drag the tiles to the boxes to form correct pairs. Not all tiles will be used.

Match the hyperbolas represented by the equations to their foci.

[tex]\[
\begin{array}{l}
\frac{(x+2)^2}{3^2} - \frac{(y-5)^2}{4^2} = 1 \quad \frac{(x-4)^2}{8^2} - \frac{(y+2)^2}{15^2} = 1 \quad \frac{(y+5)^2}{15^2} - \frac{(x-1)^2}{8^2} = 1 \\
\frac{(x-4)^2}{8^2} - \frac{(y+2)^2}{6^2} = 1 \quad \frac{(y-3)^2}{5^2} - \frac{(x-7)^2}{12^2} = 1 \quad \frac{(y-3)^2}{5^2} - \frac{(x+7)^2}{12^2} = 1 \\
\frac{(y+5)^2}{6^2} - \frac{(x-1)^2}{8^2} = 1 \\
\end{array}
\][/tex]

- [tex]\((1, -22)\)[/tex] and [tex]\((1, 12) \longleftrightarrow \square\)[/tex]
- [tex]\((-7, 5)\)[/tex] and [tex]\((3, 5) \longleftrightarrow \square\)[/tex]
- [tex]\((-6, -2)\)[/tex] and [tex]\((14, -2) \longleftrightarrow \square\)[/tex]
- [tex]\((-7, -10)\)[/tex] and [tex]\((-7, 16) \longleftrightarrow \square\)[/tex]

Question

20-06-2024
Mathematics

Answered

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used.

Match the hyperbolas represented by the equations to their foci.

[tex]\[
\begin{array}{l}
\frac{(x+2)^2}{3^2} - \frac{(y-5)^2}{4^2} = 1 \quad \frac{(x-4)^2}{8^2} - \frac{(y+2)^2}{15^2} = 1 \quad \frac{(y+5)^2}{15^2} - \frac{(x-1)^2}{8^2} = 1 \\
\frac{(x-4)^2}{8^2} - \frac{(y+2)^2}{6^2} = 1 \quad \frac{(y-3)^2}{5^2} - \frac{(x-7)^2}{12^2} = 1 \quad \frac{(y-3)^2}{5^2} - \frac{(x+7)^2}{12^2} = 1 \\
\frac{(y+5)^2}{6^2} - \frac{(x-1)^2}{8^2} = 1 \\
\end{array}
\][/tex]

- [tex]\((1, -22)\)[/tex] and [tex]\((1, 12) \longleftrightarrow \square\)[/tex]
- [tex]\((-7, 5)\)[/tex] and [tex]\((3, 5) \longleftrightarrow \square\)[/tex]
- [tex]\((-6, -2)\)[/tex] and [tex]\((14, -2) \longleftrightarrow \square\)[/tex]
- [tex]\((-7, -10)\)[/tex] and [tex]\((-7, 16) \longleftrightarrow \square\)[/tex]

Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2024-06-20T16:08:19-04:00

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]