To solve for the center, vertices, foci, and the asymptotes of the hyperbola given by the equation [tex]\( x^2 - 16y^2 = 144 \)[/tex], we will need to rewrite the equation in its standard form and identify key parameters.
### Step 1: Rewrite the Equation in Standard Form
Given equation of the hyperbola:
[tex]\[ x^2 - 16y^2 = 144 \][/tex]
First, divide the entire equation by 144 to normalize it:
[tex]\[ \frac{x^2}{144} - \frac{16y^2}{144} = 1 \][/tex]
Simplify the terms:
[tex]\[ \frac{x^2}{144} - \frac{y^2}{9} = 1 \][/tex]
The standard form of a hyperbola centered at the origin [tex]\((0,0)\)[/tex] with a horizontal transverse axis is:
[tex]\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \][/tex]
By comparing, we get:
[tex]\[ a^2 = 144 \][/tex]
[tex]\[ b^2 = 9 \][/tex]
### Step 2: Find the Center
Since the given equation already fits the form [tex]\((x - h)^2 - (y - k)^2 = 1\)[/tex] with [tex]\(h = 0\)[/tex] and [tex]\(k = 0\)[/tex], the center of the hyperbola is:
[tex]\[ \text{Center} = (0, 0) \][/tex]
### Step 3: Find the Vertices
Vertices are [tex]\(a\)[/tex] units away from the center along the transverse axis (horizontal axis for this hyperbola).
Calculate [tex]\(a\)[/tex]:
[tex]\[ a = \sqrt{144} = 12 \][/tex]
Thus, the vertices are:
[tex]\[ \text{Vertices} = (-a, 0) \text{ and } (a, 0) = (-12, 0) \text{ and } (12, 0) \][/tex]
### Step 4: Find the Foci
The distance from the center to each focus is given by [tex]\(c\)[/tex] where [tex]\(c^2 = a^2 + b^2\)[/tex].
Calculate [tex]\(c\)[/tex]:
[tex]\[ c^2 = 144 + 9 = 153 \][/tex]
[tex]\[ c = \sqrt{153} \approx 12.37 \][/tex]
Thus, the foci are:
[tex]\[ \text{Foci} = (-c, 0) \text{ and } (c, 0) \approx (-12.37, 0) \text{ and } (12.37, 0) \][/tex]
### Step 5: Find the Asymptotes
The equations of the asymptotes for a hyperbola in the form [tex]\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)[/tex] are given by:
[tex]\[ y = \pm \frac{b}{a} x \][/tex]
Given [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ \frac{b}{a} = \frac{\sqrt{9}}{\sqrt{144}} = \frac{3}{12} = \frac{1}{4} \][/tex]
Thus, the asymptote equations are:
[tex]\[ y = \frac{x}{4} \][/tex]
[tex]\[ y = -\frac{x}{4} \][/tex]
### Summary
For the hyperbola [tex]\( x^2 - 16y^2 = 144 \)[/tex]:
- Center: [tex]\( (0, 0) \)[/tex]
- Vertices: [tex]\( (-12, 0) \)[/tex] and [tex]\( (12, 0) \)[/tex]
- Foci: [tex]\( (-12.37, 0) \)[/tex] and [tex]\( (12.37, 0) \)[/tex]
- Asymptotes: [tex]\( y = \frac{x}{4} \)[/tex] and [tex]\( y = -\frac{x}{4} \)[/tex]
