Find the center, vertices, and foci of the ellipse with the given equation.
-$$5 x ^ { 2 } + 7 y ^ { 2 } = 35$$
A) Center: $( 0,0 )$; Vertices: $( - \sqrt { 7 } , 0 ) , ( - \sqrt { 7 } , 0 )$; Foci: $( - \sqrt { 2 } , 0 ) , ( \sqrt { 2 } , 0 )$
B) Center: $( 0,0 )$; Vertices: $( 0 , - 7 ) , ( 0,7 ) ;$ Foci: $( 0 , - 2 \sqrt { 6 } ) , ( 0,2 \sqrt { 6 } )$
C) Center: $( 0,0 )$; Vertices: $( 0 , - \sqrt { 7 } ) , ( 0 , - \sqrt { 7 } )$; Foci: $( 0 , - \sqrt { 2 } ) , ( 0 , \sqrt { 2 } )$
D) Center: $( 0,0 )$; Vertices: $( - 7,0 ) , ( 7,0 )$; Foci: $( - 2 \sqrt { 6 } , 0 ) , ( 2 \sqrt { 6 } , 0 )$

Question

Quizplus · Accepted Answer

The given equation is in the standard form of an ellipse, $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $a^2$ and $b^2$ are the denominators under $x^2$ and $y^2$ respectively. By dividing the entire equation by 35, we get $ \frac{x^2}{7} + \frac{y^2}{5} = 1 $, indicating that $a^2 = 7$ and $b^2 = 5$, with $a > b$. This means the ellipse is stretched more along the x-axis, making the center (0,0), vertices $(\pm\sqrt{7}, 0)$, and foci $(\pm\sqrt{a^2-b^2}, 0)$ = $(\pm\sqrt{7-5}, 0)$ = $(\pm\sqrt{2}, 0)$. However, there seems to be a mistake in the choices provided as none of them accurately reflect the correct vertices and foci based on the calculations. The correct vertices should indeed be $(\pm\sqrt{7}, 0)$, but the correct foci should be $(\pm\sqrt{2}, 0)$, and none of the options match this description accurately. Given the options, none perfectly match the correct analysis, suggesting a potential error in the interpretation of the choices or a mistake in the question's options.

Find the Center, Vertices, and Foci of the Ellipse with the Given