SOLUTION: Hyperbola Determine the center,vertices, foci,conjugat, and the latus rectum of the (y+3)^2/16

SOLUTION: Hyperbola Determine the center,vertices, foci,conjugat, and the latus rectum of the (y+3)^2/16 - (x-2)^2/25=1

Question 992302: Hyperbola
Determine the center,vertices, foci,conjugat, and the latus rectum of the
(y+3)^2/16 - (x-2)^2/25=1
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
....if you compare to you see that ,, ,
since the part of the equation is added, then the center, foci, and vertices will be above and below the center (on a line paralleling the y-axis)
so,
the center is at (,)= (,)
semi-major axis length
semi-minor axis length

since and , the equation tells me that , so , and
the eccentricity is

the vertices and foci are above and below the center,
so the foci are at
(,)=(,) and
(,)= (,)
or, approximately at (, ) and (,)

the vertices are at (,) and (, )
(,) =>(,)
and
(, )=>(,)

The length of the Latus Rectum:
In a hyperbola, it is twice the square of the length of the transverse axis divided by the length of the conjugate axis.
the length of the transverse axis is =>=>
the length of the conjugate axis is =>=>
the Latus Rectum is

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