Solve the problem.
-The ellipse $\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 16 } = 1$ is shifted down 3 units and right 8 units. Find an equation for the new ellipse and find the new center.
A) $\frac { ( x + 3 ) ^ { 2 } } { 5 } + \frac { ( y - 8 ) ^ { 2 } } { 11 } = 1$; center: $( 3 , - 8 )$
B) $\frac { ( x - 3 ) ^ { 2 } } { 5 } + \frac { ( y + 8 ) ^ { 2 } } { 11 } = 1$; center: $( - 3,8 )$
C) $\frac { ( x - 8 ) ^ { 2 } } { 5 } + \frac { ( y + 3 ) ^ { 2 } } { 16 } = 1 ;$ center: $( 8 , - 3 )$
D) $\frac { ( x + 8 ) ^ { 2 } } { 5 } + \frac { ( y - 3 ) ^ { 2 } } { 11 } = 1$; center: $( 8 , - 3 )$

Question

Quizplus · Accepted Answer

To shift an ellipse down 3 units and right 8 units, we need to subtract 3 from the y-coordinate and add 8 to the x-coordinate of the center. This gives us the new center as (8,-3). 
Next, we need to adjust the equation of the ellipse accordingly. Whenever an ellipse is shifted in the x or y direction, we only need to adjust the corresponding numerator. If the center is shifted (h,k) units, the adjusted equation is: 
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
In this case, we replace $x$ with $x-8$ and $y$ with $y+3$ to get:
$\frac{(x-8)^2}{5}+\frac{(y+3)^2}{16}=1$
Simplifying gives us the equation in choice (C).

Solve the Problem x25+y216=1\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 16 } = 15x2​+16y2​=1

Solve the Problem $\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 16 } = 1$