Find the center and vertices of the ellipse. $$\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 4 } = 1$$
A) center: $( 7,2 ) \quad$ vertices: $( - 7 , - 2 ) , ( 7,2 )$
B) center: $( 0,0 ) \quad$ vertices: $( 0 , - 7 ) , ( 0,7 )$
C) center: $( 0,0 ) \quad$ vertices: $( - 7,0 ) , ( 7,0 )$
D) center: $( 7,0 ) \quad$ vertices: $( 0 , - 2 ) , ( 0,2 )$
E) center: $( 0,0 ) \quad$ vertices: $( - 2,0 ) , ( 2,0 )$

Question

Find the center and vertices of the ellipse. $$\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 4 } = 1$$ 
A) center: $( 7,2 ) \quad$ vertices: $( - 7 , - 2 ) , ( 7,2 )$
B) center: $( 0,0 ) \quad$ vertices: $( 0 , - 7 ) , ( 0,7 )$
C) center: $( 0,0 ) \quad$ vertices: $( - 7,0 ) , ( 7,0 )$
D) center: $( 7,0 ) \quad$ vertices: $( 0 , - 2 ) , ( 0,2 )$
E) center: $( 0,0 ) \quad$ vertices: $( - 2,0 ) , ( 2,0 )$

Quizplus · Accepted Answer

The given equation is in the standard form of an ellipse centered at the origin $(0,0)$. The square root of the denominator under $x^2$ gives the distance from the center to the vertices along the x-axis, which is 7, so the vertices are $(-7,0)$ and $(7,0)$.

Find the Center and Vertices of the Ellipse x249+y24=1\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 4 } = 149x2​+4y2​=1

Find the Center and Vertices of the Ellipse $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 4 } = 1$