Graph Hyperbolas Not Centered at the Origin
-$\frac { ( y - 4 ) ^ { 2 } } { 49 } - \frac { ( x - 4 ) ^ { 2 } } { 9 } = 1$
A) Center: $( 4,4 )$; Vertices: $( 4 , - 3 )$ and $( 4,11 )$; Foci: $( 4,4 - \sqrt { 58 } )$ and $( 4,4 + \sqrt { 58 } )$
B) Center: $( - 4 , - 4 )$; Vertices: $( - 4 , - 11 )$ and $( - 4,3 )$; Foci: $( - 4 , - 4 - \sqrt { 58 } )$ and $( - 4 , - 4 + \sqrt { 58 } )$
C) Center: $( 4,4 )$; Vertices: $( 4,4 - \sqrt { 58 } )$ and $( 4,4 + \sqrt { 58 } )$; Foci: $( 4 , - 3 )$ and $( 4,11 )$
D) Center: $( 4,4 )$; Vertices: $( 3 , - 2 )$ and $( 5,12 )$; Foci: $( 3,5 - \sqrt { 58 } )$ and $( 5,5 + \sqrt { 58 } )$

Question

Quizplus · Accepted Answer

The center of the hyperbola is at $(4,4)$ because the equation is in the form $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$, where $(h,k)$ is the center. The vertices are along the y-axis from the center, at a distance of $\sqrt{49} = 7$, making them $(4,4-7)$ and $(4,4+7)$, or $(4,-3)$ and $(4,11)$. The foci are found using the formula $c^2 = a^2 + b^2$, giving $c = \sqrt{49 + 9} = \sqrt{58}$, so the foci are $(4,4-\sqrt{58})$ and $(4,4+\sqrt{58})$.

Graph Hyperbolas Not Centered at the Origin - (y−4)249−(x−4)29=1\frac { ( y - 4 ) ^ { 2 } } { 49 } - \frac { ( x - 4 ) ^ { 2 } } { 9 } = 149(y−4)2​−9(x−4)2​=1

Graph Hyperbolas Not Centered at the Origin
- $\frac { ( y - 4 ) ^ { 2 } } { 49 } - \frac { ( x - 4 ) ^ { 2 } } { 9 } = 1$