How do you find the center and vertices of the ellipse x^2/(25/9)+y^2/(16/9)=1x2259+y2169=1?

1 Answer
Narad T.
Jan 14, 2017

The center is (0,0)(0,0)
The vertices are A=(h+a,k)=(5/3,0)=(h+a,k)=(53,0)
A'=(h-a,k)=(-5/3,0)=(h−a,k)=(−53,0)
B=(h,k+b)=(0,4/3)=(h,k+b)=(0,43)
B'=(h,k-b)=(0,-4/3)=(h,k−b)=(0,−43)

Explanation:

The equation is x^2/(25/9)+y^2/(16/9)=1x2259+y2169=1

We compare this to (x-h)^2/a^2+(y-k)^2/b^2=1(x−h)2a2+(y−k)2b2=1

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

The vertices are :

A=(h+a,k)=(5/3,0)=(h+a,k)=(53,0)

A'=(h-a,k)=(-5/3,0)=(h−a,k)=(−53,0)

B=(h,k+b)=(0,4/3)=(h,k+b)=(0,43)

B'=(h,k-b)=(0,-4/3)=(h,k−b)=(0,−43)

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

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