Answer :
To analyze the hyperbola given by the equation:
[tex]\[ \frac{x^2}{9} - \frac{y^2}{9} = 1 \][/tex]
We start by noting the standard form of a hyperbola:
[tex]\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \][/tex]
Here, we identify [tex]\(a^2 = 9\)[/tex] and [tex]\(b^2 = 9\)[/tex].
### Center
The center of the hyperbola is straightforward since the equation is in the form of [tex]\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)[/tex], where [tex]\(h = 0\)[/tex] and [tex]\(k = 0\)[/tex]. Therefore, the center of the hyperbola is:
[tex]\[ (0, 0) \][/tex]
### Vertices
The vertices of the hyperbola are located at [tex]\(\pm a\)[/tex] units away from the center along the x-axis, because the term with [tex]\(x^2\)[/tex] comes first and represents the direction in which the hyperbola opens. Since:
[tex]\[ a^2 = 9 \implies a = \sqrt{9} = 3 \][/tex]
Thus, the vertices are at:
[tex]\[ (-a, 0) = (-3, 0) \][/tex]
[tex]\[ (a, 0) = (3, 0) \][/tex]
Therefore, the vertices are:
[tex]\[ (-3, 0) \][/tex] and [tex]\[ (3, 0) \][/tex]
### Foci
The foci of the hyperbola are determined by the distance [tex]\(c\)[/tex] from the center, calculated using the relationship:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Given that [tex]\(a^2 = 9\)[/tex] and [tex]\(b^2 = 9\)[/tex], we find:
[tex]\[ c^2 = 9 + 9 = 18 \implies c = \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \approx 4.242640687119285 \][/tex]
So, the foci are at:
[tex]\[ (-c, 0) = (-4.242640687119285, 0) \][/tex]
[tex]\[ (c, 0) = (4.242640687119285, 0) \][/tex]
Therefore, the foci are:
[tex]\[ (-4.242640687119285, 0) \][/tex] and [tex]\[ (4.242640687119285, 0) \][/tex]
### Asymptotes
The asymptotes for a hyperbola of this form are given by the equations:
[tex]\[ y = \pm \frac{b}{a} x \][/tex]
Given [tex]\(a = 3\)[/tex] and [tex]\(b = 3\)[/tex], we compute:
[tex]\[ \frac{b}{a} = \frac{3}{3} = 1 \][/tex]
Thus, the equations of the asymptotes are:
[tex]\[ y = x \][/tex]
[tex]\[ y = -x \][/tex]
So the asymptotes are:
[tex]\[ y = 1.0 x \][/tex] and [tex]\[ y = -1.0 x \][/tex]
### Conclusion
Summarizing all the information:
- The center of the hyperbola is [tex]\((0, 0)\)[/tex].
- The vertices are [tex]\((-3.0, 0)\)[/tex] and [tex]\((3.0, 0)\)[/tex].
- The foci are [tex]\((-4.242640687119285, 0)\)[/tex] and [tex]\((4.242640687119285, 0)\)[/tex].
- The asymptotes are [tex]\(y = 1.0 x\)[/tex] and [tex]\(y = -1.0 x\)[/tex].
[tex]\[ \frac{x^2}{9} - \frac{y^2}{9} = 1 \][/tex]
We start by noting the standard form of a hyperbola:
[tex]\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \][/tex]
Here, we identify [tex]\(a^2 = 9\)[/tex] and [tex]\(b^2 = 9\)[/tex].
### Center
The center of the hyperbola is straightforward since the equation is in the form of [tex]\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)[/tex], where [tex]\(h = 0\)[/tex] and [tex]\(k = 0\)[/tex]. Therefore, the center of the hyperbola is:
[tex]\[ (0, 0) \][/tex]
### Vertices
The vertices of the hyperbola are located at [tex]\(\pm a\)[/tex] units away from the center along the x-axis, because the term with [tex]\(x^2\)[/tex] comes first and represents the direction in which the hyperbola opens. Since:
[tex]\[ a^2 = 9 \implies a = \sqrt{9} = 3 \][/tex]
Thus, the vertices are at:
[tex]\[ (-a, 0) = (-3, 0) \][/tex]
[tex]\[ (a, 0) = (3, 0) \][/tex]
Therefore, the vertices are:
[tex]\[ (-3, 0) \][/tex] and [tex]\[ (3, 0) \][/tex]
### Foci
The foci of the hyperbola are determined by the distance [tex]\(c\)[/tex] from the center, calculated using the relationship:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Given that [tex]\(a^2 = 9\)[/tex] and [tex]\(b^2 = 9\)[/tex], we find:
[tex]\[ c^2 = 9 + 9 = 18 \implies c = \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \approx 4.242640687119285 \][/tex]
So, the foci are at:
[tex]\[ (-c, 0) = (-4.242640687119285, 0) \][/tex]
[tex]\[ (c, 0) = (4.242640687119285, 0) \][/tex]
Therefore, the foci are:
[tex]\[ (-4.242640687119285, 0) \][/tex] and [tex]\[ (4.242640687119285, 0) \][/tex]
### Asymptotes
The asymptotes for a hyperbola of this form are given by the equations:
[tex]\[ y = \pm \frac{b}{a} x \][/tex]
Given [tex]\(a = 3\)[/tex] and [tex]\(b = 3\)[/tex], we compute:
[tex]\[ \frac{b}{a} = \frac{3}{3} = 1 \][/tex]
Thus, the equations of the asymptotes are:
[tex]\[ y = x \][/tex]
[tex]\[ y = -x \][/tex]
So the asymptotes are:
[tex]\[ y = 1.0 x \][/tex] and [tex]\[ y = -1.0 x \][/tex]
### Conclusion
Summarizing all the information:
- The center of the hyperbola is [tex]\((0, 0)\)[/tex].
- The vertices are [tex]\((-3.0, 0)\)[/tex] and [tex]\((3.0, 0)\)[/tex].
- The foci are [tex]\((-4.242640687119285, 0)\)[/tex] and [tex]\((4.242640687119285, 0)\)[/tex].
- The asymptotes are [tex]\(y = 1.0 x\)[/tex] and [tex]\(y = -1.0 x\)[/tex].