Solve the problem.
-The hyperbola $\frac { y ^ { 2 } } { 11 } - \frac { x ^ { 2 } } { 5 } = 1$ is shifted down 7 units and right 5 units. Find an equation for the new hyperbola and find the new vertices.
A) $\frac { ( y + 7 ) ^ { 2 } } { 11 } - \frac { ( x - 5 ) ^ { 2 } } { 5 } = 1$; vertices: $( 5 \pm \sqrt { 5 } , - 7 )$
B) $\frac { ( y + 7 ) ^ { 2 } } { 11 } - \frac { ( x - 5 ) ^ { 2 } } { 5 } = 1$; vertices: $( 5 , - 7 \pm \sqrt { 11 } )$
C) $\frac { ( y - 7 ) ^ { 2 } } { 11 } - \frac { ( x + 5 ) ^ { 2 } } { 5 } = 1$; vertices: $( 5 , - 7 \pm \sqrt { 11 } )$
D) $\frac { ( y + 5 ) ^ { 2 } } { 11 } - \frac { ( x - 7 ) ^ { 2 } } { 5 } = 1$; vertices: $( 5 \pm \sqrt { 5 } , - 7 )$

Question

Quizplus · Accepted Answer

First, we need to shift the given hyperbola down 7 units and right 5 units. This means we need to replace $x$ with $x-5$ and $y$ with $y+7$ to obtain the equation of the new hyperbola. Doing so, we get: $$\frac{(y+7)^2}{11}-\frac{(x-5)^2}{5}=1$$Now, the vertices of the hyperbola are given by $(\pm\sqrt{a},0)$, where $a$ is the distance between the center of the hyperbola and its vertices. In this case, the center of the hyperbola is at the origin, so the distance between the center and vertices is $\sqrt{a^2+b^2}$, where $a=\sqrt{11}$ and $b=\sqrt{5}$. Therefore, the distance between the center and vertices is $\sqrt{11+5}=2\sqrt{11}$. So, the vertices are located at $(0,\pm2\sqrt{11})$.

However, we shifted the hyperbola down 7 units, so we need to subtract 7 from the $y$-coordinate of each vertex. Therefore, the vertices of the shifted hyperbola are at $(0,-2\sqrt{11}+7)$ and $(0,2\sqrt{11}+7)$.

Comparing the equation of the shifted hyperbola to the choices given, we see that only choice B matches both the equation and the coordinates of the vertices.

Solve the Problem y211−x25=1\frac { y ^ { 2 } } { 11 } - \frac { x ^ { 2 } } { 5 } = 111y2​−5x2​=1

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