Answer :
To determine the equations of the asymptotes of the given hyperbola, we need to identify the general form of the hyperbola equation and utilize specific elements of it.
The given hyperbola equation is:
[tex]\[ \dfrac{(y-2)^2}{9} - (x+2)^2 = 1 \][/tex]
This equation is in the standard form of a vertical hyperbola given by:
[tex]\[ \dfrac{(y - k)^2}{a^2} - \dfrac{(x - h)^2}{b^2} = 1 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the hyperbola, [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the lengths of the semi-major and semi-minor axes, respectively.
In our equation:
- [tex]\(h = -2\)[/tex] (notice the form [tex]\((x + 2)\)[/tex])
- [tex]\(k = 2\)[/tex] (notice the form [tex]\((y - 2)\)[/tex])
- [tex]\(a^2 = 9 \Rightarrow a = 3\)[/tex]
- Since [tex]\(b^2\)[/tex] is not explicitly stated, but we know that the coefficient for [tex]\(x\)[/tex] term needs to be [tex]\(1\)[/tex].
The equations of the asymptotes for the hyperbola are:
[tex]\[ y - k = \pm \dfrac{a}{b} (x - h) \][/tex]
Plugging in the values we obtained:
[tex]\[ y - 2 = \pm \dfrac{3}{1} (x + 2) \][/tex]
This simplifies to:
[tex]\[ y - 2 = \pm 3 (x + 2) \][/tex]
Thus, we have two asymptote equations:
1. [tex]\[ y - 2 = 3(x + 2) \][/tex]
2. [tex]\[ y - 2 = -3(x + 2) \][/tex]
So, the correct answers for the drop-down menus would be:
- [tex]\[ y - 2 = 3(x + 2) \][/tex]
- [tex]\[ y - 2 = -3(x + 2) \][/tex]
The given hyperbola equation is:
[tex]\[ \dfrac{(y-2)^2}{9} - (x+2)^2 = 1 \][/tex]
This equation is in the standard form of a vertical hyperbola given by:
[tex]\[ \dfrac{(y - k)^2}{a^2} - \dfrac{(x - h)^2}{b^2} = 1 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the hyperbola, [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the lengths of the semi-major and semi-minor axes, respectively.
In our equation:
- [tex]\(h = -2\)[/tex] (notice the form [tex]\((x + 2)\)[/tex])
- [tex]\(k = 2\)[/tex] (notice the form [tex]\((y - 2)\)[/tex])
- [tex]\(a^2 = 9 \Rightarrow a = 3\)[/tex]
- Since [tex]\(b^2\)[/tex] is not explicitly stated, but we know that the coefficient for [tex]\(x\)[/tex] term needs to be [tex]\(1\)[/tex].
The equations of the asymptotes for the hyperbola are:
[tex]\[ y - k = \pm \dfrac{a}{b} (x - h) \][/tex]
Plugging in the values we obtained:
[tex]\[ y - 2 = \pm \dfrac{3}{1} (x + 2) \][/tex]
This simplifies to:
[tex]\[ y - 2 = \pm 3 (x + 2) \][/tex]
Thus, we have two asymptote equations:
1. [tex]\[ y - 2 = 3(x + 2) \][/tex]
2. [tex]\[ y - 2 = -3(x + 2) \][/tex]
So, the correct answers for the drop-down menus would be:
- [tex]\[ y - 2 = 3(x + 2) \][/tex]
- [tex]\[ y - 2 = -3(x + 2) \][/tex]
