Graph Hyperbolas Not Centered at the Origin
-$$\frac { ( x - 3 ) ^ { 2 } } { 4 } - \frac { ( y - 1 ) ^ { 2 } } { 9 } = 1$$
A) Center: $( 3,1 )$; Vertices: $( 1,1 )$ and $( 5,1 )$; Foci: $( 3 - \sqrt { 13 } , 1 )$ and $( 3 + \sqrt { 13 } , 1 )$
B) Center: $( - 3 , - 1 )$; Vertices: $( - 5 , - 1 )$ and $( - 1 , - 1 )$; Foci: $( - 3 - \sqrt { 13 } , - 1 )$ and $( - 3 + \sqrt { 13 } , - 1 )$
C) Center: $( 3,1 )$; Vertices: $( 1 , - 1 )$ and $( 5 , - 1 )$; Foci: $( 3 - \sqrt { 13 } , - 1 )$ and $( 3 + \sqrt { 13 } , - 1 )$
D) Center: $( 3,1 )$; Vertices: $( 2,1 )$ and $( 6,1 )$; Foci: $( 4 + \sqrt { 13 } , 2 )$ and $( 2 + \sqrt { 13 } , 2 )$

Question

Quizplus · Accepted Answer

The center of the hyperbola is given by $(h, k)$ in the general equation $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, which is $(3, 1)$ for this equation. The vertices are found $\pm a$ units along the x-axis from the center, where $a^2 = 4$, so $a = 2$, making the vertices $(1, 1)$ and $(5, 1)$. The foci are found $\pm \sqrt{a^2 + b^2}$ units along the x-axis from the center, where $b^2 = 9$, so the foci are $(3 \pm \sqrt{13}, 1)$.

Graph Hyperbolas Not Centered at the Origin - (x−3)24−(y−1)29=1\frac { ( x - 3 ) ^ { 2 } } { 4 } - \frac { ( y - 1 ) ^ { 2 } } { 9 } = 14(x−3)2​−9(y−1)2​=1

Graph Hyperbolas Not Centered at the Origin
- $\frac { ( x - 3 ) ^ { 2 } } { 4 } - \frac { ( y - 1 ) ^ { 2 } } { 9 } = 1$