Let's find the vertices and foci of the given hyperbola:
[tex]\[
\frac{y^2}{25} - \frac{x^2}{144} = 1
\][/tex]
This equation represents a vertical hyperbola, because the term with [tex]\( y^2 \)[/tex] is positive. The standard form for a vertical hyperbola is:
[tex]\[
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
\][/tex]
By comparing the given equation with the standard form, we can identify that:
[tex]\[
a^2 = 25 \quad \text{and} \quad b^2 = 144
\][/tex]
First, let's find [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
a = \sqrt{25} = 5
\][/tex]
[tex]\[
b = \sqrt{144} = 12
\][/tex]
The vertices of the hyperbola are located at [tex]\( (0, \pm a) \)[/tex]. Therefore, the coordinates of the vertices are:
[tex]\[
(0, \pm 5) \quad \rightarrow \quad \text{Vertices: } (0, 5) \text{ and } (0, -5)
\][/tex]
Next, we need to find the coordinates of the foci. The distance from the center to each focus is given by [tex]\( c \)[/tex], where [tex]\( c \)[/tex] can be found using the formula:
[tex]\[
c^2 = a^2 + b^2
\][/tex]
So, we calculate:
[tex]\[
c^2 = 25 + 144 = 169
\][/tex]
Then,
[tex]\[
c = \sqrt{169} = 13
\][/tex]
The foci of the hyperbola are located at [tex]\( (0, \pm c) \)[/tex]. Therefore, the coordinates of the foci are:
[tex]\[
(0, \pm 13) \quad \rightarrow \quad \text{Foci: } (0, 13) \text{ and } (0, -13)
\][/tex]
To summarize:
- The vertices of the hyperbola are at [tex]\( (0, 5) \)[/tex] and [tex]\( (0, -5) \)[/tex].
- The foci of the hyperbola are at [tex]\( (0, 13) \)[/tex] and [tex]\( (0, -13) \)[/tex].
Thus, we have:
[tex]\[
\begin{array}{l}
\qquad \frac{y^2}{25} - \frac{x^2}{144} = 1 \\
\text{Vertices: } (0, 5) \text{ and } (0, -5) \\
\text{Foci: } (0, 13) \text{ and } (0, -13) \\
\end{array}
\][/tex]
