Graph Hyperbolas Not Centered at the Origin
Find the location of the center, vertices, and foci for the hyperbola described by the equation.
-$$( y - 1 ) ^ { 2 } - 4 ( x - 3 ) ^ { 2 } = 4$$
A) Center: $( 3,1 )$; Vertices: $( 3 , - 1 )$ and $( 3,3 )$; Foci: $( 3,1 - \sqrt { 5 } )$ and $( 3,1 + \sqrt { 5 } )$
B) Center: $( - 3 , - 1 )$; Vertices: $( - 3 , - 3 )$ and $( - 3,1 )$; Foci: $( - 3 , - 1 - \sqrt { 5 } )$ and $( - 3 , - 1 + \sqrt { 5 } )$
C) Center: $( 3,1 )$; Vertices: $( - 3 , - 2 )$ and $( 3,2 ) ;$ Foci: $( 3 , - \sqrt { 5 } )$ and $( 3 , \sqrt { 5 } )$
D) Center: $( 3,1 )$; Vertices: $( 4,0 )$ and $( 4,4 )$; Foci: $( 4,2 - \sqrt { 5 } )$ and $( 4,2 + \sqrt { 5 } )$

Question

Quizplus · Accepted Answer

The given equation can be rewritten in standard form as:
$$\frac{(y-1)^2}{4} - \frac{(x-3)^2}{1} = 1$$
Comparing it with the standard equation of a hyperbola:
$$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$$
We get that the center is at $(h,k) = (3,1)$, semi-major axis $a=2$, and semi-minor axis $b=1$. The distance from the center to the vertices, which lie on the transverse axis, is equal to $a$. Therefore, the vertices are at $(3,-1)$ and $(3,3)$. Using the formula
$$c^2 = a^2 + b^2$$
where $c$ is the distance from the center to the foci, we get
$$c = \sqrt{a^2+b^2} = \sqrt{2^2+1^2} = \sqrt{5}$$
Since the foci lie on the transverse axis, their x-coordinate is the same as the center, i.e. $x=3$. The distance from the center to each focus is equal to $c$, so the foci are at $(3,1-\sqrt{5})$ and $(3,1+\sqrt{5})$. Therefore, the correct choice is $\boxed{\textbf{(A)}}$.

Graph Hyperbolas Not Centered at the Origin Find the Location (y−1)2−4(x−3)2=4( y - 1 ) ^ { 2 } - 4 ( x - 3 ) ^ { 2 } = 4(y−1)2−4(x−3)2=4

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