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.