How do you find the equation of an ellipse with vertices #(0,+-8)# and foci #(0, +-4)#?

1 Answer
Dec 30, 2016

#x^2/48 +y^2/64=1#

Explanation:

Find the equation of an ellipse with vertices #(0, +-8)# and foci #(0,+-4)#.

The equation of an ellipse is #(x-h)^2/a^2 +(y-k)^2/b^2=1# for a horizontally oriented ellipse and #(x-h)^2/b^2 +(y-k)^2/a^2 =1# for a vertically oriented ellipse.

#(h,k)# is the center and the distance #c# from the center to the foci is given by #a^2-b^2=c^2#. #a# is the distance from the center to the vertices and #b# is the distance from the center to the co-vertices.

The center of the ellipse is half way between the vertices. Thus, the center #(h,k)# of the ellipse is #(0,0)# and the ellipse is vertically oriented.

#a# is the distance between the center and the vertices, so #a=8#.
#c# is the distance between the center and the foci, so #c=4#

#a^2-b^2=c^2=> b^2=a^2-c^2#

#b^2=8^2-4^2=64-16=48#

The equation is:

#(x-0)^2/48 +(y-0)^2/64=1# or #x^2/48 +y^2/64=1#