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.
