identify the graph of $\frac{y^{2}}{25}-\frac{x^{2}}{4}=1$.

Explanation:

Step1: Recall hyperbola standard - form

The standard form of a hyperbola is $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1$ (opens up and down) or $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ (opens left and right). Given $\frac{y^{2}}{25}-\frac{x^{2}}{4}=1$, where $a^{2}=25$ (so $a = 5$) and $b^{2}=4$ (so $b = 2$). This is a hyperbola that opens up and down.

Step2: Analyze the vertices

For a hyperbola of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$, the vertices are at $(0,a)$ and $(0, - a)$. Here, $a = 5$, so the vertices are at $(0,5)$ and $(0,-5)$.

Step3: Match with the graph

The hyperbola that opens up - and - down with vertices at $(0,5)$ and $(0, - 5)$ is graph C.

Answer:

C