How do you find the center, foci and vertices of #(x-3)^2/(25/9)+(y-8)^2=1#?

1 Answer
Roy E.
Dec 28, 2016

Ellipse, center #(3,8)#,Vertices #(4/3,8)#, #(14/3,8)#, #(3,9)#, #(3,7)#, foci #(3+-4/3,8)#

Explanation:

The equation is already in a standard form #((x-x_c)/a)^2+((y-y_c)/b)^2#so the center #(x_c,y_c)# and semi-axes #a# and #b# can be read straight off it as #(3,8)#, #5/3# and #1#.

#a>b# so the line through the foci (the major axis) is parallel to the #x#-axis rather than the #y#.

The vertices are at the center #+-# semiaxes: #(3+-5/3,8+-1)#
The distance between each focus and the centre is #sqrt(a^2-b^2#=#sqrt((25/9)-1^2)=4/3#. So the foci are #(3+-4/3,8)#

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