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]}

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