Certainly! Let's work through finding the center, the vertices, the foci, and the asymptotes for the hyperbola given by the equation:
[tex]\[ y^2 - x^2 = \frac{9}{25} \][/tex]
### 1. Center
The given equation is of the form [tex]\( y^2 - x^2 = C \)[/tex]. This form indicates that the hyperbola is centered at the origin [tex]\((0,0)\)[/tex].
### 2. Vertices
To find the vertices, we determine the values of [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex].
[tex]\[ y^2 = \frac{9}{25} \Rightarrow y = \pm\sqrt{\frac{9}{25}} = \pm\frac{3}{5} \][/tex]
Therefore, the vertices of the hyperbola are at:
[tex]\[ (0, \frac{3}{5}) \][/tex] and [tex]\[ (0, -\frac{3}{5}) \][/tex]
### 3. Foci
The formula for the distance from the center to the foci of a hyperbola of this form is given by [tex]\(\sqrt{a^2 + b^2}\)[/tex], where [tex]\( a^2 \)[/tex] is the value under the positive term (here, [tex]\( y^2 \)[/tex]) and [tex]\( b^2 \)[/tex] is the value under the negative term (here, [tex]\( x^2 \)[/tex]).
In our equation:
[tex]\[ a^2 = \frac{9}{25} \][/tex]
[tex]\[ b^2 = \frac{9}{25} \][/tex] (because it’s the same as [tex]\( a^2 \)[/tex])
Then:
[tex]\[ a = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
[tex]\[ b = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
The distance to the foci is:
[tex]\[ \sqrt{a^2 + b^2} = \sqrt{\frac{9}{25} + \frac{9}{25}} = \sqrt{\frac{18}{25}} = \frac{\sqrt{18}}{5} = \frac{3\sqrt{2}}{5} \approx 0.848528137423857 \][/tex]
Thus, the foci are located at:
[tex]\[ (0, \frac{3\sqrt{2}}{5}) \][/tex] and [tex]\[ (0, -\frac{3\sqrt{2}}{5}) \][/tex]
### 4. Asymptotes
The asymptotes of a hyperbola of the form [tex]\( y^2/a^2 - x^2/b^2 = 1 \)[/tex] are given by:
[tex]\[ y = \pm\frac{a}{b}x \][/tex]
In our case, since both [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are equal:
[tex]\[ y = \pm\frac{\frac{3}{5}}{\frac{3}{5}}x = \pm x \][/tex]
Therefore, the asymptotes are:
[tex]\[ y = x \][/tex]
[tex]\[ y = -x \][/tex]
### Summary
- Center: [tex]\((0,0)\)[/tex]
- Vertices: [tex]\((0, \frac{3}{5})\)[/tex] and [tex]\((0, -\frac{3}{5})\)[/tex]
- Foci: [tex]\((0, 0.848528137423857)\)[/tex] and [tex]\((0, -0.848528137423857)\)[/tex]
- Asymptotes: [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex]
This concludes the detailed breakdown for the given hyperbola.
