Find the center, foci, and vertices of the hyperbola. $\frac { ( x - 1 ) ^ { 2 } } { 16 } - \frac { ( y + 2 ) ^ { 2 } } { 1 } = 1$
A) center: $( 1 , - 2 )$; vertices: $( - 3 , - 2 ) , ( 5 , - 2 )$; foci: $( 1 - \sqrt { 17 } , - 2 ) , ( 1 + \sqrt { 17 } , - 2 )$.
B) center: $( 1 , - 2 )$; vertices: $( - 3 , - 2 ) , ( 5 , - 2 )$; foci: $( 1 , - 2 - \sqrt { 17 } ) , ( 1 , - 2 + \sqrt { 17 } )$.
C) center: $( 1 , - 2 )$; vertices: $( - 3 , - 2 ) , ( - 5 , - 2 )$; foci: $( 1 , - 2 - \sqrt { 17 } ) , ( 1 , - 2 + \sqrt { 17 } )$.
D) center: $( 1 , - 2 )$; vertices: $( 1 , - 3 ) , ( 1 , - 1 )$; foci $( 1 , - 2 - \sqrt { 17 } ) , ( 1 , - 2 + \sqrt { 17 } )$.
E) center: $( 1 , - 2 )$; vertices: $( 1 , - 3 ) , ( 1 , - 1 )$; foci: $( 1 - \sqrt { 17 } , - 2 ) , ( 1 + \sqrt { 17 } , - 2 )$.

Question

Quizplus · Accepted Answer

- The given equation is in the standard form $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ where $(h,k)$ is the center of the hyperbola, $a$ is the distance from the center to each vertex along the x-axis, $b$ is the distance from the center to each vertex along the y-axis.
- Comparing the given equation with standard form, we get: $(h,k)=(1,-2)$, $a=4$, and $b=1$.
- The distance between the center and each focus is $c=\sqrt{a^2+b^2}=\sqrt{16+1}=\sqrt{17}$.
- The distance between the center and the vertices along the y-axis is $b=1$.
- Using the above information, we can determine that the vertices are at $(1\pm a,-2)=(5,-2)$ and $(-3,-2)$.
- The foci are located along the major axis which is the x-axis in this case. So, the foci are at $(1\pm c,-2)=(1\pm\sqrt{17},-2)$.
- Therefore, the correct choice is $	extbf{(A)}$.

Find the Center, Foci, and Vertices of the Hyperbola (x−1)216−(y+2)21=1\frac { ( x - 1 ) ^ { 2 } } { 16 } - \frac { ( y + 2 ) ^ { 2 } } { 1 } = 116(x−1)2​−1(y+2)2​=1

Find the Center, Foci, and Vertices of the Hyperbola $\frac { ( x - 1 ) ^ { 2 } } { 16 } - \frac { ( y + 2 ) ^ { 2 } } { 1 } = 1$