What are the vertices and co vertices of the ellipse x^2/4 + y^2/25=1

I actually have no idea how to even start to solve this problem, so any help will be appreciated.
Steps on how this problem is solved are also appreciated.

the vertices are at the ends of the major axis, and the covertices are on the minor axis. So, since 25 > 4, the vertices are on the y-axis.

when x=0, y=5
when y=0, x=2

So, the vertices are at (0,±5)
the covertices are at (±2,0)

Looks like you need to review the section where this was explained. With graphs, I'm sure.

To find the vertices and co-vertices of an ellipse, you need to identify the equation of the ellipse in standard form. The standard form of the equation of an ellipse is:

(x - h)^2/a^2 + (y - k)^2/b^2 = 1

Where (h, k) represents the center of the ellipse, and 'a' and 'b' are the semi-major and semi-minor axes, respectively.

In your given equation x^2/4 + y^2/25 = 1, we can see that:

a^2 = 4
b^2 = 25

To find the values of 'a' and 'b', we take the square root:

a = √4 = 2
b = √25 = 5

Since the ellipse is centered at the origin (0, 0), we can determine that the center (h, k) is (0, 0).

Now we have all the information we need to find the vertices and co-vertices of the ellipse.

The vertices of an ellipse are located on the major axis and are positioned a distance 'a' units away from the center. In this case, the major axis is along the x-axis.

So the vertices are: (±a, 0) = (±2, 0)

The co-vertices, on the other hand, are located on the minor axis and positioned a distance 'b' units away from the center. In this case, the minor axis is along the y-axis.

So, the co-vertices are: (0, ±b) = (0, ±5)

Therefore, the vertices of the ellipse are (±2, 0) and the co-vertices are (0, ±5).

By following these steps, you can find the vertices and co-vertices of any ellipse given its equation in standard form.