identify the vertices, the co-vertixes, and the foci of the ellipse
x^2/4+y^2/9=1
A. the vertexes are (0,2) and (0,-2). The co-vertexes are (3,0) and (-3,0). The foci are ((0, 3√5)
B. The vertex ices are (0,3) and 0.-3). The co-vertex are (2,0) and (0,-2). the foci are (0,5) and (0,-5)
C. The vertices are (3,0) and (-3,0) The co-vertices are (0,2) and (0,-2). The foci are (√5,0) and (-√5,0)
D the vertices are (0,3), and (0,-3). The co vertex are (2,0) and (-2,0). The foci are (0,√5) and (0,√5)
E none
To identify the vertices, co-vertices, and foci of the ellipse given by the equation x^2/4 + y^2/9 = 1, we can use the standard form of an ellipse:
[(x - h)^2 / a^2] + [(y - k)^2 / b^2] = 1
Where (h, k) represents the center of the ellipse, a represents the semi-major axis, and b represents the semi-minor axis.
Comparing the given equation to the standard form, we can see that the center of the ellipse is (h, k) = (0, 0) since the x and y terms are squared and there is no added constant.
The square of 'a' is 4, so 'a' is 2. This means that the length of the semi-major axis is 2.
The square of 'b' is 9, so 'b' is 3. This means that the length of the semi-minor axis is 3.
Using this information, we can now identify the vertices, co-vertices, and foci of the ellipse:
The vertices are located at (±a, k) = (±2, 0) = (2, 0) and (-2, 0).
The co-vertices are located at (h, ±b) = (0, ±3) = (0, 3) and (0, -3).
To find the foci, we can use the formula:
c = √(a^2 - b^2)
Since a^2 = 4 and b^2 = 9, we have:
c = √(4 - 9) = √(-5)
Since the term under the square root is negative, this means that the ellipse does not have any real foci.
Therefore, the correct answer is option E: none.