Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola.
-$$( x - 1 ) ^ { 2 } - 4 ( y - 2 ) ^ { 2 } = 4$$
A) center at $( 1,2 )$
transverse axis is parallel to $\mathrm { y }$-axis
vertices at $( 1,0 )$ and $( 1,4 )$,
foci at $( 1,2 - \sqrt { 5 } )$ and $( 1,2 + \sqrt { 5 } )$,
asymptotes of $y + 2 = - 2 ( x + 1 )$ and $y + 2 = 2 ( x + 1 )$
B) center at $( 1,2 )$
transverse axis is parallel to $x$-axis
vertices at $( 0,2 )$ and $( 2,2 )$
foci at $( 1 - \sqrt { 5 } , 2 )$ and $( 1 + \sqrt { 5 } , 2 )$
asymptotes of $y - 2 = - 2 ( x - 1 )$ and $y - 2 = 2 ( x - 1 )$
C) center at $( 1,2 )$
transverse axis is parallel to $x$-axis
vertices at $( - 1,2 )$ and $( 3,2 )$
foci at $( 1 - \sqrt { 5 } , 2 )$ and $( 1 + \sqrt { 5 } , 2 )$
asymptotes of $y - 2 = - \frac { 1 } { 2 } ( x - 1 )$ and $y - 2 = \frac { 1 } { 2 } ( x - 1 )$
D) center at $( 2,1 )$
transverse axis is parallel to $x$-axis
vertices at $( 0,1 )$ and $( 4,1 )$
foci at $( 2 - \sqrt { 5 } , 1 )$ and $( 2 + \sqrt { 5 } , 1 )$
asymptotes of $y - 1 = - \frac { 1 } { 2 } ( x - 2 )$ and $y - 1 = \frac { 1 } { 2 } ( x - 2 )$

Question

Quizplus · Accepted Answer

The given equation $(x-1)^2 - 4(y-2)^2 = 4$ can be rewritten as $\frac{(x-1)^2}{4} - \frac{(y-2)^2}{1} = 1$, which is a hyperbola centered at $(1,2)$ with a transverse axis parallel to the x-axis. The vertices are $(1 \pm 2, 2)$ or $(-1, 2)$ and $(3, 2)$. The distance to the foci is $\sqrt{4+1} = \sqrt{5}$, so the foci are $(1 \pm \sqrt{5}, 2)$. The asymptotes are given by $y - 2 = \pm \frac{1}{2}(x - 1)$.

Find the Center, Transverse Axis, Vertices, Foci, and Asymptotes of the Hyperbola