Identify the vertices and co-vertices of the ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ .

A) vertices: $(-7,0),(7,0)$ co-vertices: $(0,-2),(0,2)$
B) vertices: $(-49,0),(49,0)$ co-vertices: $(0,-4),(0,4)$
C) vertices: $(-2,0),(2,0)$ co-vertices: $(0,-7),(0,7)$
D) vertices: $(0,-49),(0,49)$ co-vertices: $(-4,0),(4,0)$
E) vertices: $(0,-2),(0,2)$ co-vertices: $(-7,0),(7,0)$

The standard form of the equation of an ellipse is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$, where $(h,k)$ is the center of the ellipse, $a$ is the semi-major axis (half the length of the major axis), and $b$ is the semi-minor axis (half the length of the minor axis). Comparing this to $\frac{x^2}{49}+\frac{y^2}{4}=1$, we can see that $h=k=0$, $a=7$, and $b=2$. The vertices are located at $(\pm a,0)=(\pm 7,0)$ and the co-vertices are located at $(0,\pm b)=(0,\pm 2)$, giving us the answer choice of $\boxed{\textbf{(A)}}$.