Solve the problem.
-Find the foci and asymptotes of the following hyperbola: $\frac { y ^ { 2 } } { 400 } - \frac { x ^ { 2 } } { 225 } = 1$

A) Foci: $( 0 , - 15 ) , ( 0,15 ) ;$ Asymptotes: $y = \frac { 4 } { 3 } x , y = - \frac { 4 } { 3 } x$
B) Foci: $( 0 , - 25 ) , ( 0,25 ) ;$ Asymptotes: $y = \frac { 4 } { 3 } x , y = - \frac { 4 } { 3 } x$
C) Foci: $( - 25,0 ) , ( 25,0 ) ;$ Asymptotes: $y = \frac { 3 } { 4 } x , y = - \frac { 3 } { 4 } x$
D) Foci: $( - 15,0 ) , ( 15,0 ) ;$ Asymptotes: $y = \frac { 3 } { 4 } x , y = - \frac { 3 } { 4 } x$

Question

Solve the problem.
-Find the foci and asymptotes of the following hyperbola: $\frac { y ^ { 2 } } { 400 } - \frac { x ^ { 2 } } { 225 } = 1$

A) Foci: $( 0 , - 15 ) , ( 0,15 ) ;$ Asymptotes: $y = \frac { 4 } { 3 } x , y = - \frac { 4 } { 3 } x$
B) Foci: $( 0 , - 25 ) , ( 0,25 ) ;$ Asymptotes: $y = \frac { 4 } { 3 } x , y = - \frac { 4 } { 3 } x$
C) Foci: $( - 25,0 ) , ( 25,0 ) ;$ Asymptotes: $y = \frac { 3 } { 4 } x , y = - \frac { 3 } { 4 } x$
D) Foci: $( - 15,0 ) , ( 15,0 ) ;$ Asymptotes: $y = \frac { 3 } { 4 } x , y = - \frac { 3 } { 4 } x$

Quizplus · Accepted Answer

The given equation is of a hyperbola with the y^2 term positive, indicating it opens up and down. The distance from the center to the foci (c) is found using the equation c^2 = a^2 + b^2, where a^2 = 400 and b^2 = 225. Solving for c gives c = sqrt(400 + 225) = sqrt(625) = 25. Therefore, the foci are at (0, ±25). The asymptotes for a hyperbola of this form are given by y = ±(a/b)x, where a = sqrt(400) = 20 and b = sqrt(225) = 15, so the asymptotes are y = ±(20/15)x = ±(4/3)x.

Solve the Problem y2400−x2225=1\frac { y ^ { 2 } } { 400 } - \frac { x ^ { 2 } } { 225 } = 1400y2​−225x2​=1

Solve the Problem $\frac { y ^ { 2 } } { 400 } - \frac { x ^ { 2 } } { 225 } = 1$