How do you find the center, vertices, foci and eccentricity of #(X^2/25)+(y^2/64)=1#?

1 Answer
Narad T.
Oct 28, 2016

The center is #(0,0)#
The vertices are #(5,0)#, #(-5,0)#, #(0,8)# and #(0,-8)#
The foci are #(sqrt0,sqrt39)# and #(0,-sqrt39)#
The eccentricity is #=sqrt39/8#

Explanation:

This is the equation of an ellipse with a vertical major axis

#(x-x_0)^2/a^2+(y-y_0)^2/b^2=1#

Here #(x_0,y_0)=(0,0)#
So the center of the ellipse is #(0,0)#
#a=5# and #b=8# are the axes

The vertices are #(5,0)#, #(-5,0)#, #(0,8)# and #(0,-8)#

The distance of the center to the focus is #sqrt(b^2-a^2)=sqrt(64-25)=sqrt39#

So the foci are #(sqrt0,sqrt39)# and #(0,-sqrt39)#

The eccentricity is #=sqrt39/8#

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