Find center, vertices, co-vertices and foci for the following;

(x-3(^2/49 + (y-4)^2/4 = 1
Center would be (3,4)
A=7
b=2
Vertices would be found this way:
(h+-a,k)
(3-7,4)
(3+7,4)
Vertices = (-4,4)
(10,4)
Co-vertices
(h,k+-b)
(3,4-2)
(3,4+2)
Co-vertices=(3,2)
(3,6)

If ound "c" by 49-4 = c^2
45=c^2
3sqrt(5)=c
6.7=c
Foci =(h+-c,k)
(3-6.7,4)
(3+6.7,4)
Co-vertices=(-3.7,4)
(9.7,4)or left in radical it would be: 3- (3sqrt5),4
3+(3sqrt5),4

I agree with all your answers.

Thank you

To find the center, vertices, co-vertices, and foci for the equation (x-3)^2/49 + (y-4)^2/4 = 1, follow these steps:

1. Center: The equation is in the form (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) represents the center of the ellipse. In this case, (h,k) = (3,4). So the center of the ellipse is (3,4).

2. Semi-major axis (a): The value of "a" can be found by taking the square root of the denominator of (x-h)^2, which is 49. So a = √49 = 7.

3. Semi-minor axis (b): The value of "b" can be found by taking the square root of the denominator of (y-k)^2, which is 4. So b = √4 = 2.

4. Vertices: The vertices of the ellipse can be found by adding or subtracting the value of "a" from the x-coordinate of the center, while keeping the y-coordinate constant. So the vertices are (3-7,4) = (-4,4) and (3+7,4) = (10,4).

5. Co-vertices: The co-vertices of the ellipse can be found by adding or subtracting the value of "b" from the y-coordinate of the center, while keeping the x-coordinate constant. So the co-vertices are (3,4-2) = (3,2) and (3,4+2) = (3,6).

6. Foci: The foci of the ellipse can be found using the formula c = √(a^2 - b^2), where c represents the distance from the center to each focus. In this case, c = √(7^2 - 2^2) = √45 = 6.7.

The foci can be obtained by adding or subtracting the value of "c" from the x-coordinate of the center, while keeping the y-coordinate constant. So the foci are (3-6.7,4) ≈ (-3.7,4) and (3+6.7,4) ≈ (9.7,4). Alternatively, the foci can be expressed in radical form as (3-√(5),4) and (3+√(5),4).