Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola.
-$$x ^ { 2 } - 16 y ^ { 2 } + 6 x - 32 y - 23 = 0$$
A) center at $( - 3 , - 1 )$
transverse axis is parallel to $x$-axis
vertices at $( - 7 , - 1 )$ and $( 1 , - 1 )$
foci at $( - 3 - \sqrt { 17 } , - 1 )$ and $( - 3 + \sqrt { 17 } , - 1 )$
asymptotes of $y + 1 = - \frac { 1 } { 4 } ( x + 3 )$ and $y + 1 = \frac { 1 } { 4 } ( x + 3 )$

B) center at $( - 3 , - 1 )$
transverse axis is parallel to $\mathrm { y }$-axis
vertices at $( - 3 , - 5 )$ and $( - 3,3 )$
foci at $( - 3 , - 1 - \sqrt { 17 } )$ and $( - 3 , - 1 + \sqrt { 17 } )$
asymptotes of $\mathrm { y } - 1 = - 4 ( \mathrm { x } - 3 )$ and $\mathrm { y } - 1 = 4 ( \mathrm { x } - 3 )$

C) center at $( - 3 , - 1 )$
transverse axis is parallel to $x$-axis
vertices at $( - 4 , - 1 )$ and $( - 2 , - 1 )$
foci at $( - 3 - \sqrt { 17 } , - 1 )$ and $( - 3 + \sqrt { 17 } , - 1 )$
asymptotes of $\mathrm { y } + 1 = - 4 ( \mathrm { x } + 3 )$ and $\mathrm { y } + 1 = 4 ( \mathrm { x } + 3 )$

D) center at $( - 1 , - 3 )$
transverse axis is parallel to $x$-axis
vertices at $( - 5 , - 3 )$ and $( 3 , - 3 )$
foci at $( - 1 - \sqrt { 17 } , - 3 )$ and $( - 1 + \sqrt { 17 } , - 3 )$
asymptotes of $y + 3 = - \frac { 1 } { 4 } ( x + 1 )$ and $y + 3 = \frac { 1 } { 4 } ( x + 1 )$

Question

Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola.
-$$x ^ { 2 } - 16 y ^ { 2 } + 6 x - 32 y - 23 = 0$$
A) center at $( - 3 , - 1 )$
transverse axis is parallel to $x$-axis
vertices at $( - 7 , - 1 )$ and $( 1 , - 1 )$
foci at $( - 3 - \sqrt { 17 } , - 1 )$ and $( - 3 + \sqrt { 17 } , - 1 )$
asymptotes of $y + 1 = - \frac { 1 } { 4 } ( x + 3 )$ and $y + 1 = \frac { 1 } { 4 } ( x + 3 )$

B) center at $( - 3 , - 1 )$
transverse axis is parallel to $\mathrm { y }$-axis
vertices at $( - 3 , - 5 )$ and $( - 3,3 )$
foci at $( - 3 , - 1 - \sqrt { 17 } )$ and $( - 3 , - 1 + \sqrt { 17 } )$
asymptotes of $\mathrm { y } - 1 = - 4 ( \mathrm { x } - 3 )$ and $\mathrm { y } - 1 = 4 ( \mathrm { x } - 3 )$

C) center at $( - 3 , - 1 )$
transverse axis is parallel to $x$-axis
vertices at $( - 4 , - 1 )$ and $( - 2 , - 1 )$
foci at $( - 3 - \sqrt { 17 } , - 1 )$ and $( - 3 + \sqrt { 17 } , - 1 )$
asymptotes of $\mathrm { y } + 1 = - 4 ( \mathrm { x } + 3 )$ and $\mathrm { y } + 1 = 4 ( \mathrm { x } + 3 )$

D) center at $( - 1 , - 3 )$
transverse axis is parallel to $x$-axis
vertices at $( - 5 , - 3 )$ and $( 3 , - 3 )$
foci at $( - 1 - \sqrt { 17 } , - 3 )$ and $( - 1 + \sqrt { 17 } , - 3 )$
asymptotes of $y + 3 = - \frac { 1 } { 4 } ( x + 1 )$ and $y + 3 = \frac { 1 } { 4 } ( x + 1 )$

Quizplus · Accepted Answer

To find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola, we first complete the square for both $x$ and $y$ terms in the given equation $x^2 - 16y^2 + 6x - 32y - 23 = 0$. Completing the square:- For $x^2 + 6x$, we add and subtract $(6/2)^2 = 9$, getting $x^2 + 6x + 9$.- For $-16(y^2 + 2y)$, we add and subtract $-16(2/2)^2 = -16$, getting $-16(y^2 + 2y + 1)$.Adding these adjustments to both sides, we get $x^2 + 6x + 9 - 16(y^2 + 2y + 1) = 23 + 9 - 16$, simplifying to $(x + 3)^2 - 16(y + 1)^2 = 16$. Dividing through by 16 gives $\frac{(x + 3)^2}{16} - \frac{(y + 1)^2}{1} = 1$, which is a standard form of a hyperbola centered at $(-3, -1)$ with a transverse axis parallel to the $x$-axis. The vertices are found by adding and subtracting the square root of the $x$-term denominator from the $x$-coordinate of the center, giving vertices at $(-7, -1)$ and $(1, -1)$. The foci are found by calculating $c = \sqrt{a^2 + b^2} = \sqrt{16 + 1} = \sqrt{17}$, giving foci at $(-3 - \sqrt{17}, -1)$ and $(-3 + \sqrt{17}, -1)$. The asymptotes are given by the equation of the hyperbola in its standard form, leading to $y + 1 = \pm \frac{1}{4}(x + 3)$.

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