How do you find the center and vertices of the ellipse #x^2/(25/9)+y^2/(16/9)=1#?

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Answer 1 · 2017-01-14T14:14:56

The center is #(0,0)#
The vertices are A#=(h+a,k)=(5/3,0)#
A'#=(h-a,k)=(-5/3,0)#
B#=(h,k+b)=(0,4/3)#
B'#=(h,k-b)=(0,-4/3)#

The equation is #x^2/(25/9)+y^2/(16/9)=1#

We compare this to #(x-h)^2/a^2+(y-k)^2/b^2=1#

The center is O #=(h,k)=(0,0)#

The vertices are :

A#=(h+a,k)=(5/3,0)#

A'#=(h-a,k)=(-5/3,0)#

B#=(h,k+b)=(0,4/3)#

B'#=(h,k-b)=(0,-4/3)#

graph{x^2/(25/9)+y^2/(16/9)=1 [-3.898, 3.897, -1.947, 1.95]}