Solve the problem.
-The ellipse $\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 11 } = 1$ is shifted down 6 units and right 4 units. Find an equation for the new ellipse and find the new vertices.
A) $\frac { ( x + 4 ) ^ { 2 } } { 5 } + \frac { ( y - 6 ) ^ { 2 } } { 11 } = 1$; vertices: $( 4 , - 6 \pm \sqrt { 11 } )$
B) $\frac { ( x - 4 ) ^ { 2 } } { 5 } + \frac { ( y + 6 ) ^ { 2 } } { 11 } = 1$; vertices: $( 4 \pm \sqrt { 5 } , - 6 )$
C) $\frac { ( x - 4 ) ^ { 2 } } { 5 } + \frac { ( y + 6 ) ^ { 2 } } { 11 } = 1$; vertices: $( 4 , - 6 \pm \sqrt { 11 } )$
D) $\frac { ( x - 6 ) ^ { 2 } } { 5 } + \frac { ( y + 4 ) ^ { 2 } } { 11 } = 1$; vertices: $( 4 \pm \sqrt { 5 } , - 6 )$

Question

Quizplus · Accepted Answer

To shift the ellipse down 6 units and right 4 units, we need to subtract 6 from the y-coordinate of the center (which is 0) and add 4 to the x-coordinate of the center (which is also 0). Thus, the new equation is $$\frac{(x-4)^2}{5}+\frac{(y+6)^2}{11}=1.$$ The vertices are located at $(\pm\sqrt{a},0)$ and $(0,\pm\sqrt{b})$, where $a$ and $b$ are the coefficients of $x^2$ and $y^2$, respectively. Plugging in $a=5$ and $b=11$ into these formulas gives the vertices $(\pm\sqrt{5},0)$ and $(0,\pm\sqrt{11})$. After shifting, these vertices become $(\pm\sqrt{5}+4,-6)$ and $(4,\pm\sqrt{11}-6)$. Therefore, the correct answer is $\boxed{\textbf{(C)}}$.

Solve the Problem x25+y211=1\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 11 } = 15x2​+11y2​=1

Solve the Problem $\frac { x ^ { 2 } } { 5 } + \frac { y ^ { 2 } } { 11 } = 1$