Identify the center and vertices of the hyperbola. $x ^ { 2 } - 4 y ^ { 2 } + 40 y - 104 = 0$
A)center: (0,5) vertices: $( 0 , - 5 ) , ( 0,5 )$
B)center: (0,-5) vertices: $( - 2 , - 5 ) , ( 2 , - 5 )$
C)center: $( 0,5 )$ vertices: $( - 2,5 ) , ( 2,5 )$
D)center: $( 0 , - 5 )$ vertices: $( - 6 , - 5 ) , ( 6 , - 5 )$
E)center: $( 0 , - 5 )$ vertices: $( 0 , - 8 ) , ( 0 , - 2 )$

Question

Quizplus · Accepted Answer

First rewrite the equation in standard form by completing the square on y:
$$x^2-4y^2+40y-104=0$$
$$x^2-4(y^2-10y+25)-104+4(25)=0$$
$$x^2-4(y-5)^2+16=0$$
$$\frac{x^2}{16}-\frac{(y-5)^2}{4}=1$$
Therefore, the center is at $(0,5)$ and the distance between the center and vertices is equal to $\sqrt{a^2+b^2}=2$, so the vertices are at $(0,5\pm2)=(0,3)$ and $(0,7)$. The correct choice is therefore C.

Identify the Center and Vertices of the Hyperbola x2−4y2+40y−104=0x ^ { 2 } - 4 y ^ { 2 } + 40 y - 104 = 0x2−4y2+40y−104=0

Identify the Center and Vertices of the Hyperbola $x ^ { 2 } - 4 y ^ { 2 } + 40 y - 104 = 0$