Graph Hyperbolas Not Centered at the Origin
Find the location of the center, vertices, and foci for the hyperbola described by the equation.
-$$\frac { ( x - 1 ) ^ { 2 } } { 49 } - \frac { ( y + 4 ) ^ { 2 } } { 36 } = 1$$
A) Center: $( 1 , - 4 )$; Vertices: $( - 6 , - 4 )$ and $( 8 , - 4 )$; Foci: $( 1 - \sqrt { 85 } , - 4 )$ and $( 1 + \sqrt { 85 } , - 4 )$
B) Center: $( - 1,4 )$; Vertices: $( - 8,4 )$ and $( 6,4 )$; Foci: $( - 1 - \sqrt { 85 } , 4 )$ and $( - 1 + \sqrt { 85 } , 4 )$
C) Center: $( 1 , - 4 )$; Vertices: $( - 6,4 )$ and $( 8,4 )$; Foci: $( 1 - \sqrt { 85 } , 4 )$ and $( 1 + \sqrt { 85 } , 4 )$
D) Center: $( 1 , - 4 )$; Vertices: $( - 5 , - 4 )$ and $( 9 , - 4 )$; Foci: $( 2 + \sqrt { 85 } , - 3 )$ and $( - 3 + \sqrt { 85 } , - 3 )$

Question

Quizplus · Accepted Answer

First, we can identify that this is a hyperbola because of the subtraction of the $y$ term. 
To find the center, we look at the signs in front of the $(x-1)^2$ and $(y+4)^2$ terms. They are positive and negative, respectively, which means that the center is $(1,-4)$. 
To find the vertices, we look at the denominator of the $(x-1)^2$ term and the $(y+4)^2$ term. They are 49 and 36, respectively, which means that $a=7$ and $b=6$. From the center, we move $a$ units to the left and right to get the two vertices: $(-6,-4)$ and $(8,-4)$. 
To find the foci, we use the formula $c=\sqrt{a^2+b^2}$, where $c$ is the distance from the center to each focus. Plugging in $a=7$ and $b=6$, we get $c=\sqrt{85}$. Then, we add/subtract $c$ from the $x$-coordinate of the center to get the two foci: $(1-\sqrt{85},-4)$ and $(1+\sqrt{85},-4)$. 
Therefore, the answer is A.

Graph Hyperbolas Not Centered at the Origin Find the Location (x−1)249−(y+4)236=1\frac { ( x - 1 ) ^ { 2 } } { 49 } - \frac { ( y + 4 ) ^ { 2 } } { 36 } = 149(x−1)2​−36(y+4)2​=1

Graph Hyperbolas Not Centered at the Origin
Find the Location $\frac { ( x - 1 ) ^ { 2 } } { 49 } - \frac { ( y + 4 ) ^ { 2 } } { 36 } = 1$