Solve the problem.
-The ellipse $\frac { x ^ { 2 } } { 256 } + \frac { y ^ { 2 } } { 400 } = 1$ is shifted horizontally and vertically to obtain the ellipse $\frac { ( x + 4 ) ^ { 2 } } { 256 } + \frac { ( y + 4 ) ^ { 2 } } { 400 } = 1$ Find the foci of the new ellipse.
A) (-4, -20), (-4, 12)
B) (-20, -4), (12, -4)
C) (-4, -16), (-4, 8)
D) (-16, -4), (8, -4)

Question

Quizplus · Accepted Answer

The foci of an ellipse $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ are given by $(h, k \pm c)$ or $(h \pm c, k)$, where $c = \sqrt{a^2 - b^2}$ for a horizontal ellipse and $c = \sqrt{b^2 - a^2}$ for a vertical ellipse. For the given ellipse, $a^2 = 400$ and $b^2 = 256$, making it a vertical ellipse since $a > b$. Thus, $c = \sqrt{400 - 256} = \sqrt{144} = 12$. The center of the new ellipse is at $(-4, -4)$, so the foci are $(-4, -4 \pm 12)$, which simplifies to $(-4, -16)$ and $(-4, 8)$.

Solve the Problem x2256+y2400=1\frac { x ^ { 2 } } { 256 } + \frac { y ^ { 2 } } { 400 } = 1256x2​+400y2​=1

Solve the Problem $\frac { x ^ { 2 } } { 256 } + \frac { y ^ { 2 } } { 400 } = 1$