Find the center, foci, and vertices of the ellipse.
-$$16 ( x + 2 ) ^ { 2 } + 9 ( y - 1 ) ^ { 2 } = 144$$
A) center at $ (-1,1) $
foci at $ (-1,1-\sqrt{7}),(-1,1+\sqrt{7}) $
vertices at $ (-1,5),(-1,-3) $

B)center at $ (1,-2) $
foci at $ (1,-2-\sqrt{7}),(1,-2+\sqrt{7}) $
vertices at $ (1,5),(1,-3) $

C) center at $ (2,1) $
foci at $ (2,1-\sqrt{7}),(2,1+\sqrt{7}) $
vertices at $ (2,5),(2,-3) $

D) center at $ (-2,1) $
foci at $ (-2,1-\sqrt{7}),(-2,1+\sqrt{7}) $
vertices at $ (-2,5),(-2,-3) $

Question

Find the center, foci, and vertices of the ellipse.
-$$16 ( x + 2 ) ^ { 2 } + 9 ( y - 1 ) ^ { 2 } = 144$$
A) center at $ (-1,1) $
foci at $ (-1,1-\sqrt{7}),(-1,1+\sqrt{7}) $
vertices at $ (-1,5),(-1,-3) $

B)center at $ (1,-2) $
foci at $ (1,-2-\sqrt{7}),(1,-2+\sqrt{7}) $
vertices at $ (1,5),(1,-3) $

C) center at $ (2,1) $
foci at $ (2,1-\sqrt{7}),(2,1+\sqrt{7}) $
vertices at $ (2,5),(2,-3) $

D) center at $ (-2,1) $
foci at $ (-2,1-\sqrt{7}),(-2,1+\sqrt{7}) $
vertices at $ (-2,5),(-2,-3) $

Quizplus · Accepted Answer

The equation of the ellipse is given in the form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $(h,k)$ is the center of the ellipse. By comparing, we find the center at $(-2,1)$. The lengths of the semi-major axis $a$ and semi-minor axis $b$ are found from the coefficients of the ellipse equation, with $a^2 = 144/16 = 9$ and $b^2 = 144/9 = 16$. The foci are located along the major axis, which is vertical in this case, and their distance from the center is $\sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}$. Therefore, the foci are at $(-2, 1 \pm \sqrt{7})$, and the vertices, being on the major axis and a distance $a$ from the center, are at $(-2, 1 \pm 4)$, which simplifies to $(-2, 5)$ and $(-2, -3)$.

Find the Center, Foci, and Vertices of the Ellipse 16(x+2)2+9(y−1)2=14416 ( x + 2 ) ^ { 2 } + 9 ( y - 1 ) ^ { 2 } = 14416(x+2)2+9(y−1)2=144

Find the Center, Foci, and Vertices of the Ellipse $16 ( x + 2 ) ^ { 2 } + 9 ( y - 1 ) ^ { 2 } = 144$