Answer :
To determine which graph represents the hyperbola given by the equation
[tex]\[ \frac{y^2}{6^2} - \frac{x^2}{2^2} = 1 \][/tex]
let's break down the equation step by step.
1. Standard Form of a Hyperbola:
The equation given is
[tex]\[ \frac{y^2}{6^2} - \frac{x^2}{2^2} = 1. \][/tex]
2. Identify the components:
- The term \(\frac{y^2}{6^2}\) indicates that the \(y\)-term is associated with the positive component of the hyperbola.
- The term \(\frac{x^2}{2^2}\) is associated with the negative component.
3. General Form of Vertically Oriented Hyperbola:
This particular form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) tells us it is a vertical hyperbola centered at the origin \((0,0)\). Here, \(a^2 = 36\) and \(b^2 = 4\).
4. Identify \(a\) and \(b\):
- \(a = \sqrt{36} = 6\)
- \(b = \sqrt{4} = 2\)
These values help to understand the vertices and the asymptotes of the hyperbola.
5. Vertices:
The vertices of this hyperbola are found at \((0, \pm a)\), thus they are at \((0, 6)\) and \((0, -6)\).
6. Asymptotes:
The equations of the asymptotes for this hyperbola are given by the lines:
[tex]\[ y = \pm \frac{a}{b} x = \pm \frac{6}{2} x = \pm 3x \][/tex]
7. Shape and Orientation:
Since the hyperbola is vertical:
- It opens upwards and downwards along the \(y\)-axis.
- The branches of the hyperbola extend away from the center (0,0) along the \(y\)-axis while approaching the asymptotes.
By synthesizing all of this information, we can determine that the correct graph should display a hyperbola with:
- Vertices at \((0, 6)\) and \((0, -6)\)
- Asymptotes along the lines \(y = 3x\) and \(y = -3x\)
- An overall structure where the two branches of the hyperbola open in the vertical directions (up and down) from the center at (0, 0).
When you examine your graphing tool or visual reference, look for the graph that matches this description. This is the graph of the hyperbola described by the given equation.
[tex]\[ \frac{y^2}{6^2} - \frac{x^2}{2^2} = 1 \][/tex]
let's break down the equation step by step.
1. Standard Form of a Hyperbola:
The equation given is
[tex]\[ \frac{y^2}{6^2} - \frac{x^2}{2^2} = 1. \][/tex]
2. Identify the components:
- The term \(\frac{y^2}{6^2}\) indicates that the \(y\)-term is associated with the positive component of the hyperbola.
- The term \(\frac{x^2}{2^2}\) is associated with the negative component.
3. General Form of Vertically Oriented Hyperbola:
This particular form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) tells us it is a vertical hyperbola centered at the origin \((0,0)\). Here, \(a^2 = 36\) and \(b^2 = 4\).
4. Identify \(a\) and \(b\):
- \(a = \sqrt{36} = 6\)
- \(b = \sqrt{4} = 2\)
These values help to understand the vertices and the asymptotes of the hyperbola.
5. Vertices:
The vertices of this hyperbola are found at \((0, \pm a)\), thus they are at \((0, 6)\) and \((0, -6)\).
6. Asymptotes:
The equations of the asymptotes for this hyperbola are given by the lines:
[tex]\[ y = \pm \frac{a}{b} x = \pm \frac{6}{2} x = \pm 3x \][/tex]
7. Shape and Orientation:
Since the hyperbola is vertical:
- It opens upwards and downwards along the \(y\)-axis.
- The branches of the hyperbola extend away from the center (0,0) along the \(y\)-axis while approaching the asymptotes.
By synthesizing all of this information, we can determine that the correct graph should display a hyperbola with:
- Vertices at \((0, 6)\) and \((0, -6)\)
- Asymptotes along the lines \(y = 3x\) and \(y = -3x\)
- An overall structure where the two branches of the hyperbola open in the vertical directions (up and down) from the center at (0, 0).
When you examine your graphing tool or visual reference, look for the graph that matches this description. This is the graph of the hyperbola described by the given equation.
