Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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].
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
