State the center, the vertices, the foci, the lengths of the major and minor axis for the following ellipse. ((x-1)^2/25)+((y+3)^2/9=1

We'll recall the general equation of the
ellipse:

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

The center of the ellipse is no longer (0,0), but
(h,k).

Now, we'll check the given equation of the ellipse.
We notice that the larger of the values a and b is 25 and it is the denominator of x
fraction, therefore the major axis of the ellipse is
horizontal.

In our case, the center of ellipse is C(1 ,
-3).

To determine the horizontal vertices of the ellipse,
we'll add and subtract 5 from x value: A(6 , -3) and A'(-4 ,
-3).

To determine the vertical vertices of the ellipse,
we'll add and subtract 3 from x value: B(1 , 0) and B'(1 ,
-6).

Since a = 5 and b = 3, we'll determine the lengths of
the major and minor axis.

The major axis length is given by
2a = 2*5 = 10

The minor axis length is given by 2b = 2*3 =
6

To find the coordinates of the foci, we need to determine
the value of c first.

c = sqrt(a^2 -
b^2)

c = sqrt(25 - 9)

c =
sqrt16

c = +/-4

Now, we'll
determine the foci adding and subtracting 4 from x value, since the foci are located on
the major axis, that is x axis, in this case.

The
coordinates of the foci are: F(5,-3) and F'(-3 ,
-3).

The center of ellipse: C(1 , -3). The
vertices of ellipse: A(6 , -3),  A'(-4 , -3), B(1 , 0) and B'(1 , -6). The coordinates
of the foci are: F(5,-3) and F'(-3 , -3). The lengths of the major and minor axis: 2a=10
and 2b=6.