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

Foci:
[tex]\[
\begin{array}{l}
(1,-22) \text{ and } (1,12) \\
(-7,5) \text{ and } (3,5) \\
(-6,-2) \text{ and } (14,-2) \\
(-7,-10) \text{ and } (-7,16) \\
\end{array}
\][/tex]



Answer :

GinnyAnswer
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]

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