Find the center, foci, vertices, and eccentricity of the ellipse. $$\frac { ( x - 3 ) ^ { 2 } } { 25 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$$
A) center $( 3 , - 2 )$; vertices: $( - 2 , - 2 ) , ( 8 , - 2 ) ;$ foci $( 3 , - 6 ) , ( 7,2 )$; eccentricity: $e = \frac { 5 } { 4 }$
B) center: $( 3 , - 2 )$; vertices: $( - 2 , - 2 ) , ( 8 , - 2 )$; foci: $( - 1 , - 2 ) , ( 7 , - 2 )$; eccentricity: $e = \frac { 2 } { 5 }$
C) center $( 3 , - 2 )$; vertices: $( 3 , - 7 ) , ( 3,3 )$; foci: $( 3 , - 6 ) , ( 37,2 )$; eccentricity:e $= \frac { 4 } { 5 }$
D) center: $( 3 , - 2 )$; vertices: $( - 2 , - 2 ) , ( 8 , - 2 )$; foci: $( - 1 , - 2 ) , ( 7 , - 2 )$; eccentricity: $e = \frac { 5 } { 2 }$
E) center: $( 3 , - 2 )$; vertices: $( - 2 , - 2 ) , ( 8 , - 2 )$; foci: $( - 1 , - 2 ) , ( 7 , - 2 )$; eccentricity: $e = \frac { 4 } { 5 }$

Question

Quizplus · Accepted Answer

The Answer is E

Find the Center, Foci, Vertices, and Eccentricity of the Ellipse (x−3)225+(y+2)29=1\frac { ( x - 3 ) ^ { 2 } } { 25 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 125(x−3)2​+9(y+2)2​=1

Find the Center, Foci, Vertices, and Eccentricity of the Ellipse $\frac { ( x - 3 ) ^ { 2 } } { 25 } + \frac { ( y + 2 ) ^ { 2 } } { 9 } = 1$