Find the center, foci, and vertices of the ellipse.
-$$2 x ^ { 2 } + 5 y ^ { 2 } - 12 x + 60 y + 188 = 0$$
A) $\frac { ( x - 3 ) ^ { 2 } } { 2 } + \frac { ( y + 6 ) ^ { 2 } } { 5 } = 1$
center: $( - 3,6 )$; foci: $( - 1.3,6 ) , ( - 4.7,6 )$; vertices: $( - 5.2,6 ) , ( - 0.8,6 )$

B) $\frac { ( x - 3 ) ^ { 2 } } { 5 } + \frac { ( y + 6 ) ^ { 2 } } { 2 } = 1$
center: $( 3 , - 6 )$; foci: $( 4.7 , - 6 ) , ( 1.3 , - 6 )$; vertices: $( 5.2 , - 6 ) , ( 0.8 , - 6 )$

C) $\frac { ( x - 3 ) ^ { 2 } } { 5 } + \frac { ( y + 6 ) ^ { 2 } } { 2 } = 1$
center: $( - 3,6 )$; foci: $( - 1.3,6 ) , ( - 4.7,6 )$; vertices: $( - 5.2,6 ) , ( - 0.8,6 )$

D) $\frac { ( x - 3 ) ^ { 2 } } { 2 } + \frac { ( y + 6 ) ^ { 2 } } { 5 } = 1$
center: $( 3 , - 6 )$; foci: $( 4.7 , - 6 ) , ( 1.3 , - 6 )$; vertices: $( 5.2 , - 6 ) , ( 0.8 , - 6 )$

Question

Find the center, foci, and vertices of the ellipse.
-$$2 x ^ { 2 } + 5 y ^ { 2 } - 12 x + 60 y + 188 = 0$$
A) $\frac { ( x - 3 ) ^ { 2 } } { 2 } + \frac { ( y + 6 ) ^ { 2 } } { 5 } = 1$
center: $( - 3,6 )$; foci: $( - 1.3,6 ) , ( - 4.7,6 )$; vertices: $( - 5.2,6 ) , ( - 0.8,6 )$

B) $\frac { ( x - 3 ) ^ { 2 } } { 5 } + \frac { ( y + 6 ) ^ { 2 } } { 2 } = 1$
center: $( 3 , - 6 )$; foci: $( 4.7 , - 6 ) , ( 1.3 , - 6 )$; vertices: $( 5.2 , - 6 ) , ( 0.8 , - 6 )$

C) $\frac { ( x - 3 ) ^ { 2 } } { 5 } + \frac { ( y + 6 ) ^ { 2 } } { 2 } = 1$
center: $( - 3,6 )$; foci: $( - 1.3,6 ) , ( - 4.7,6 )$; vertices: $( - 5.2,6 ) , ( - 0.8,6 )$

D) $\frac { ( x - 3 ) ^ { 2 } } { 2 } + \frac { ( y + 6 ) ^ { 2 } } { 5 } = 1$
center: $( 3 , - 6 )$; foci: $( 4.7 , - 6 ) , ( 1.3 , - 6 )$; vertices: $( 5.2 , - 6 ) , ( 0.8 , - 6 )$

Quizplus · Accepted Answer

To find the center, foci, and vertices of the ellipse, we first complete the square for both $x$ and $y$ terms in the given equation $2x^2 + 5y^2 - 12x + 60y + 188 = 0$. Rearranging and completing the square:$$2(x^2 - 6x) + 5(y^2 + 12y) = -188$$$$2(x^2 - 6x + 9) + 5(y^2 + 12y + 36) = -188 + 18*2 + 180$$$$2(x - 3)^2 + 5(y + 6)^2 = 36$$Dividing by 36 to get the standard form of an ellipse equation:$$\frac{(x - 3)^2}{18} + \frac{(y + 6)^2}{36/5} = 1$$$$\frac{(x - 3)^2}{9/2} + \frac{(y + 6)^2}{36/5} = 1$$$$\frac{(x - 3)^2}{(3/\sqrt{2})^2} + \frac{(y + 6)^2}{(6/\sqrt{5})^2} = 1$$This matches the form of choice B, where the center of the ellipse is at $(3, -6)$, indicating that the $x$ and $y$ coordinates of the center are positive 3 and negative 6, respectively. The major axis is along the $y$-direction because the denominator under $y$ is larger, meaning the vertices and foci will be vertically aligned with the center. The distance from the center to the vertices along the major axis is $\sqrt{36/5}$, and the distance from the center to the foci is $\sqrt{a^2 - b^2} = \sqrt{36/5 - 9/2}$, which simplifies to distances that match the foci and vertices given in choice B.

Find the Center, Foci, and Vertices of the Ellipse 2x2+5y2−12x+60y+188=02 x ^ { 2 } + 5 y ^ { 2 } - 12 x + 60 y + 188 = 02x2+5y2−12x+60y+188=0

Find the Center, Foci, and Vertices of the Ellipse $2 x ^ { 2 } + 5 y ^ { 2 } - 12 x + 60 y + 188 = 0$