IDNLearn.com is your go-to resource for finding expert answers and community support. Our community is here to provide detailed and trustworthy answers to any questions you may have.

Find the center, the vertices, the foci, and the asymptotes of the hyperbola.

[tex]\[
\frac{x^2}{9}-\frac{y^2}{9}=1
\][/tex]

The center is [tex]\((0,0)\)[/tex].


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].