Find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola.
-

\frac { ( x - 4 ) ^ { 2 } } { 25 } - \frac { ( y + 1 ) ^ { 2 } } { 16 } = 1

A) center at

( 4 , - 1 )

transverse axis is parallel to

x

-axis
vertices at

( 0 , - 1 )

and

( 8 , - 1 )

foci at

( 4 - \sqrt { 41 } , - 1 )

and

( 4 + \sqrt { 41 } , - 1 )

asymptotes of

y + 1 = - \frac { 5 } { 4 } ( x - 4 )

and

y + 1 = \frac { 5 } { 4 } ( x - 4 )

B) center at

( - 1,4 )

transverse axis is parallel to

x

-axis
vertices at

( - 6,4 )

and

( 4,4 )

foci at

( - 1 - \sqrt { 41 } , 4 )

and

( - 1 + \sqrt { 41 } , 4 )

asymptotes of

y - 4 = - \frac { 4 } { 5 } ( x + 1 )

and

y - 4 = \frac { 4 } { 5 } ( x + 1 )

C) center at

( 4 , - 1 )

transverse axis is parallel to

\mathrm { y }

-axis
vertices at

( 4 , - 6 )

and

( 4,4 )

foci at

( 4 , - 1 - \sqrt { 41 } )

and

( 4 , - 1 + \sqrt { 41 } )

asymptotes of

y - 1 = - \frac { 5 } { 4 } ( x + 4 )

and

y - 1 = \frac { 5 } { 4 } ( x + 4 )

D) center at

( 4 , - 1 )

transverse axis is parallel to

x

-axis
vertices at

( - 1 , - 1 )

and

( 9 , - 1 )

foci at

( 4 - \sqrt { 41 } , - 1 )

and

( 4 + \sqrt { 41 } , - 1 )

asymptotes of

y + 1 = - \frac { 4 } { 5 } ( x - 4 )

and

y + 1 = \frac { 4 } { 5 } ( x - 4 )

Convert the Polar Equation to a Rectangular Equation r=4sec⁡θsec⁡θ+2r = \frac { 4 \sec \theta } { \sec \theta + 2 }r=secθ+24secθ​

Multiple Choice