Because the #x# coordinate of the foci is the coordinate that is changing, we know that the major axis of the ellipse is parallel to the #x# axis. Therefore, the standard Cartesian form of the equation of the ellipse is:
#(x - h)^2/a^2 + (y -k)^2/b^2 = 1##\ \ \ # #\cdots\ \cdots##\ \ \ ##[1]#
# #
# #
where #(h\ ,\ k)# is the center, #a# is half of the length of the major axis and #b# is half of the length of the minor axis.
We are given that the major axis is length 14, therefore, we substitute #7# for #a# into equation #[1]#:
#(x - h)^2/7^2 + (y -k)^2/b^2 = 1##\ \ \ # #\cdots\ \cdots##\ \ \ ##[2]#
# #
# #
The foci for this type of ellipse are located at:
#(h - sqrt(a^2 - b^2)\ ,\ k)# #" "#and #" "##(h + sqrt(a^2 - b^2)\ ,\ k)#
Using the foci, #(5 , -5 )# and #(1 , -5 )#, and #a = 7# we can write #3# equations that will help is to find the values of #h# , #k# , and #b#:
#h -\sqrt{7^2-b^2} = 5##\ \ \ # #\cdots\ \cdots##\ \ \ ##[3]#
#h +\sqrt{7^2-b^2} = 1##\ \ \ # #\ \ \ # #\cdots\ \cdots##\ \ \ ##[4]#
#k=-5##\ \ \ # #\ \ \ # #\cdots\ \cdots##\ \ \ ##[5]#
By adding [3] and [4] and solving for #h# we get:
#h=3#
By subtracting [4] from [3] and then solving for #b#, we get:
#b=3\sqrt{5}#
# #
# #
Substitute these values into equation #[2]#,
#(x - 3)^2/7^2 + (y +5)^2/(3\sqrt{5})^2 = 1#
# #
# #
That's it!
