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Select the correct equations. Identify all the hyperbolas which open horizontally.

[tex]\[
\begin{array}{ll}
\frac{(x+2)^2}{3^2}-\frac{(2y-10)^2}{8^2}=1 & \frac{(x-1)^2}{6^2}-\frac{(2y+6)^2}{5^2}=1 \\
\frac{(2y-6)^2}{3^2}-\frac{(x+\sqrt{5})^2}{2^2}=1 & \frac{(y+4)^2}{8^2}-\frac{(2x-3)^2}{30^2}=1 \\
\frac{(2y-10)^2}{10^2}-\frac{(x+3)^2}{12^2}=1 & \frac{(y+1)^2}{4^2}-\frac{(2x-8)^2}{15^2}=1
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
To identify which hyperbolas open horizontally, we need to understand the general forms of hyperbola equations:

1. A hyperbola that opens horizontally has the general form:
[tex]\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \][/tex]
In this form, the term with [tex]\(x\)[/tex] is positive, and the term with [tex]\(y\)[/tex] is negative.

2. A hyperbola that opens vertically has the general form:
[tex]\[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \][/tex]
In this form, the term with [tex]\(y\)[/tex] is positive, and the term with [tex]\(x\)[/tex] is negative.

Let's examine each equation provided:

1. [tex]\(\frac{(x+2)^2}{3^2} - \frac{(2y-10)^2}{8^2} = 1\)[/tex]

Here, the [tex]\( (x+2)^2 \)[/tex] term is positive, and [tex]\(\frac{(2y-10)^2}{8^2}\)[/tex] term is negative. This matches the form [tex]\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)[/tex], indicating the hyperbola opens horizontally.

2. [tex]\(\frac{(x-1)^2}{6^2} - \frac{(2y+6)^2}{5^2} = 1\)[/tex]

Here, the [tex]\( (x-1)^2 \)[/tex] term is positive, and [tex]\(\frac{(2y+6)^2}{5^2}\)[/tex] term is negative. This matches the form [tex]\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)[/tex], indicating the hyperbola opens horizontally.

3. [tex]\(\frac{(2y-6)^2}{3^2} - \frac{(x+\sqrt{5})^2}{2^2} = 1\)[/tex]

Here, the [tex]\( (2y-6)^2 \)[/tex] term is positive, and [tex]\(\frac{(x+\sqrt{5})^2}{2^2}\)[/tex] term is negative. This matches the form [tex]\(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)[/tex], indicating the hyperbola opens vertically rather than horizontally.

4. [tex]\(\frac{(y+4)^2}{8^2} - \frac{(2x-3)^2}{30^2} = 1\)[/tex]

Here, the [tex]\( (y+4)^2 \)[/tex] term is positive, and [tex]\(\frac{(2x-3)^2}{30^2}\)[/tex] term is negative. This matches the form [tex]\(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)[/tex], indicating the hyperbola opens vertically rather than horizontally.

5. [tex]\(\frac{(2y-10)^2}{10^2} - \frac{(x+3)^2}{12^2} = 1\)[/tex]

Here, the [tex]\( (2y-10)^2 \)[/tex] term is positive, and [tex]\(\frac{(x+3)^2}{12^2}\)[/tex] term is negative. This matches the form [tex]\(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)[/tex], indicating the hyperbola opens vertically rather than horizontally.

6. [tex]\(\frac{(y+1)^2}{4^2} - \frac{(2x-8)^2}{15^2} = 1\)[/tex]

Here, the [tex]\( (y+1)^2 \)[/tex] term is positive, and [tex]\(\frac{(2x-8)^2}{15^2}\)[/tex] term is negative. This matches the form [tex]\(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)[/tex], indicating the hyperbola opens vertically rather than horizontally.

Thus, the hyperbolas that open horizontally are given by the equations:
[tex]\[ \frac{(x+2)^2}{3^2} - \frac{(2y-10)^2}{8^2} = 1 \quad \text{and} \quad \frac{(x-1)^2}{6^2} - \frac{(2y+6)^2}{5^2} = 1 \][/tex]

So the correct selections are equations 1 and 2.
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