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

A) Foci: $( - 25,0 ) , ( 25,0 ) ;$ Asymptotes: $y = \frac { 4 } { 3 } x , y = - \frac { 4 } { 3 } x$
B) Foci: $( 0 , - 20 ) , ( 0,20 ) ;$ Asymptotes: $y = \frac { 3 } { 4 } x , y = - \frac { 3 } { 4 } x$
C) Foci: $( 0 , - 25 ) , ( 0,25 ) ;$ Asymptotes: $y = \frac { 4 } { 3 } x , y = - \frac { 4 } { 3 } x$
D) Foci: $( - 25,0 ) , ( 25,0 ) ;$ Asymptotes: $y = \frac { 3 } { 4 } x , y = - \frac { 3 } { 4 } x$

Question

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

A) Foci: $( - 25,0 ) , ( 25,0 ) ;$ Asymptotes: $y = \frac { 4 } { 3 } x , y = - \frac { 4 } { 3 } x$
B) Foci: $( 0 , - 20 ) , ( 0,20 ) ;$ Asymptotes: $y = \frac { 3 } { 4 } x , y = - \frac { 3 } { 4 } x$
C) Foci: $( 0 , - 25 ) , ( 0,25 ) ;$ Asymptotes: $y = \frac { 4 } { 3 } x , y = - \frac { 4 } { 3 } x$
D) Foci: $( - 25,0 ) , ( 25,0 ) ;$ Asymptotes: $y = \frac { 3 } { 4 } x , y = - \frac { 3 } { 4 } x$

Quizplus · Accepted Answer

The given equation $\frac{x^2}{225} - \frac{y^2}{400} = 1$ is of a hyperbola centered at the origin with a horizontal transverse axis. The distance of the foci from the center along the x-axis is given by $c = \sqrt{a^2 + b^2}$, where $a^2 = 225$ and $b^2 = 400$. Calculating $c$ gives $c = \sqrt{225 + 400} = 25$, so the foci are at $(-25, 0)$ and $(25, 0)$. The asymptotes of a hyperbola of this form are given by $y = \pm \frac{b}{a}x$, where $a = 15$ and $b = 20$, so the equations of the asymptotes are $y = \pm \frac{4}{3}x$.

Solve the Problem x2225−y2400=1\frac { x ^ { 2 } } { 225 } - \frac { y ^ { 2 } } { 400 } = 1225x2​−400y2​=1

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