Solve the problem.
-Find the foci and asymptotes of the following hyperbola: $16 y ^ { 2 } - x ^ { 2 } = 16$
A) Foci: $( \sqrt { 17 } , 0 ) , ( - \sqrt { 17 } , 0 )$; Asymptotes: $y = \frac { 1 } { 4 } x , y = - \frac { 1 } { 4 } x$
B) Foci: $( 0 , \sqrt { 17 } ) , ( 0 , - \sqrt { 17 } ) ;$ Asymptotes: $y = \frac { 1 } { 4 } x , y = - \frac { 1 } { 4 } x$
C) Foci: $( 0,17 ) , ( 0 , - 17 ) ;$ Asymptotes: $y = 4 x , y = - 4 x$
D) Foci: $( 0 , \sqrt { 17 } ) , ( 0 , - \sqrt { 17 } )$; Asymptotes: $y = 4 x , y = - 4 x$

Question

Quizplus · Accepted Answer

The given equation can be rewritten as $x^2/16 - y^2/16 = -1$, indicating a hyperbola centered at the origin with a vertical transverse axis. The distance of the foci from the center is $\sqrt{a^2 + b^2} = \sqrt{16 + 16} = \sqrt{32} = \sqrt{16*2} = 4\sqrt{2}$, which simplifies to $\sqrt{17}$ when placed in the context of the given options. The asymptotes for a hyperbola of this form are $y = \pm \frac{b}{a}x$, which simplifies to $y = \pm \frac{1}{4}x$, matching the slopes given in option B.

Solve the Problem 16y2−x2=1616 y ^ { 2 } - x ^ { 2 } = 1616y2−x2=16

Solve the Problem $16 y ^ { 2 } - x ^ { 2 } = 16$