Solve the problem.
-The ellipse $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1$ is shifted up 6 units and left 8 units. Find an equation for the new ellipse and find the new center.
A) $\frac { ( x + 8 ) ^ { 2 } } { 16 } + \frac { ( y - 6 ) ^ { 2 } } { 9 } = 1$; center: $( - 8,6 )$
B) $\frac { ( x + 6 ) ^ { 2 } } { 7 } + \frac { ( y - 8 ) ^ { 2 } } { 3 } = 1 ;$ center: $( 6 , - 8 )$
C) $\frac { ( x + 8 ) ^ { 2 } } { 7 } + \frac { ( y - 6 ) ^ { 2 } } { 3 } = 1 ;$ center: $( 8,6 )$
D) $\frac { ( x - 8 ) ^ { 2 } } { 16 } + \frac { ( y + 6 ) ^ { 2 } } { 9 } = 1 ;$ center: $( - 8,6 )$

Question

Quizplus · Accepted Answer

The center of the original ellipse is at the point (0,0). Shifting it up 6 units and left 8 units means that the new center is at (-8,6).
To find the new equation, we just replace x with (x+8) and y with (y-6) in the equation of the original ellipse:
\begin{align*}
\frac{(x+8)^2}{16}+\frac{(y-6)^2}{9}&=1\
\frac{(x+8)^2}{(4\cdot4)}+\frac{(y-6)^2}{(3\cdot3)}&=1\
\frac{(x+8)^2}{4^2}+\frac{(y-6)^2}{3^2}&=1
\end{align*}
Therefore, the equation of the new ellipse is $\frac{(x+8)^2}{16}+\frac{(y-6)^2}{9}=1$ and its center is at $(-8,6)$. The answer is (A).

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Solve the Problem $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1$