Find the coordinates of foci, vertices, eccentricity of an ellipse and sketch the graph of equation:(x-2)^2\4+(y+3)^2\9=1

hint: find the center, then

a=2
b=3
c^2 = a^2-b^2

To find the coordinates of the foci, vertices, and the eccentricity of an ellipse, we'll need to identify the standard form of the equation first. The given equation is:

(x - 2)^2/4 + (y + 3)^2/9 = 1

To bring it to the standard form of an ellipse equation, let's start by isolating the denominators:

(x - 2)^2/2^2 + (y + 3)^2/3^2 = 1

Now, we can see that the center of the ellipse is (h, k) = (2, -3), and the horizontal radius is a = 2, and the vertical radius is b = 3.

To find the foci of the ellipse, we need to determine c, which is given by the equation c^2 = a^2 - b^2. Let's calculate c:

c^2 = 2^2 - 3^2 = 4 - 9 = -5

Since c^2 is negative, we know that this ellipse has no real foci. However, we can find the imaginary foci since they occur at (h ± ci, k).

The eccentricity (e) of the ellipse can then be found by the equation e = c/a. Let's plug the values in to calculate e:

e = √(-5)/2

Therefore, the eccentricity is e = √(-5)/2.

To find the vertices, we can use the formula (h ± a, k) and (h, k ± b), where h and k are the coordinates of the center and a and b are the semi-major and semi-minor axis values. Let's calculate them:

Vertice 1: (2 + 2, -3) = (4, -3)
Vertice 2: (2 - 2, -3) = (0, -3)
Vertice 3: (2, -3 + 3) = (2, 0)
Vertice 4: (2, -3 - 3) = (2, -6)

Now that we have all the information, we can sketch the graph of the ellipse, knowing the center, foci, vertices, and eccentricity.

(Note: The sketch is not possible here as the foci are imaginary, which means they aren't located on the graph of the ellipse.)

In summary:
- The center of the ellipse is (2, -3).
- The foci are imaginary and, therefore, cannot be plotted.
- The vertices are (4, -3), (0, -3), (2, 0), and (2, -6).
- The eccentricity is √(-5)/2.