Let's analyze the given equation step by step:
[tex]\[ \frac{y^2}{16} - \frac{x^2}{4} = 1 \][/tex]
This equation resembles the standard form of a hyperbola.
### Step 1: Identify the form of the hyperbola
The general form of a hyperbola centered at the origin can be written as:
[tex]\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \][/tex]
In this form, the hyperbola opens vertically because the [tex]\(y^2\)[/tex] term is positive.
### Step 2: Compare with the given equation
By comparing the given equation with the standard form:
[tex]\[ \frac{y^2}{16} - \frac{x^2}{4} = 1 \][/tex]
We can identify the denominators of [tex]\(y^2\)[/tex] and [tex]\(x^2\)[/tex].
### Step 3: Determine the values of [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex]
For the given hyperbola:
[tex]\[ a^2 = 16 \][/tex]
[tex]\[ b^2 = 4 \][/tex]
### Step 4: Calculate [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
To find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex], take the square roots of [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex]:
[tex]\[ a = \sqrt{16} = 4 \][/tex]
[tex]\[ b = \sqrt{4} = 2 \][/tex]
### Step 5: Summarize the key features of the hyperbola
1. Center: Since the equation is in standard form and there are no h or k terms (no shifts), the center is at the origin [tex]\((0, 0)\)[/tex].
2. Vertices: The vertices are along the y-axis at [tex]\((0, \pm a)\)[/tex]. Therefore, the vertices are at [tex]\((0, 4)\)[/tex] and [tex]\((0, -4)\)[/tex].
3. Co-vertices: Though not necessary for identifying the graph, the co-vertices are along the x-axis at [tex]\((\pm b, 0)\)[/tex], which are at [tex]\((2, 0)\)[/tex] and [tex]\((-2, 0)\)[/tex].
4. Asymptotes: The equations of the asymptotes for a vertically opening hyperbola are [tex]\(\pm \frac{a}{b} x\)[/tex]. Plugging in the values, the asymptotes are [tex]\(y = \pm 2x\)[/tex].
### Conclusion
The hyperbola represented by the equation [tex]\(\frac{y^2}{16} - \frac{x^2}{4} = 1\)[/tex] is centered at the origin, opens along the y-axis, and has vertices at [tex]\((0, 4)\)[/tex] and [tex]\((0, -4)\)[/tex]. The asymptotes are lines with slopes of [tex]\(\pm 2\)[/tex]. This set of features will correspond to the correct graph of the hyperbola.
