SOLUTION: What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, (y-4)^2/16

SOLUTION: What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, (y-4)^2/16 - (x+2)^2/9 = 1?

Question 28333: What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, (y-4)^2/16 - (x+2)^2/9 = 1?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING AND YOU SHOULD BE ABLE TO SOLVE YOUR PROBLEM BY YOUR SELF.....
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What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, y^2 - 4x^2 - 2y - 16x + 1 = 0?
(Y^2-2*Y*1+1^2)-{(2X)^2+2*(2X)*4+4^2}-1^2+4^2+1=0
(Y-1)^2-(2X+4)^2=-16
4(X+2)^2-(Y-1)^2=16
{4(X+2)^2}/16-{(Y-1)^2}/16=1
(X+2)^2/2^2-(Y-1)^2/4^2=1...
COMPARING WITH STANDARD EQN.
(X-H)^2/A^2-(Y-K)^2/B^2=1....WE HAVE
VERTICES ARE {(H-A),K} AND {(H+A),K}=(-2-2,1) AND (-2+2,1)=(-4,1) AND (0,1)
FOCI ARE {(H-AE),K} AND {(H+AE),K}...WHERE E IS
ECCENTRICITY =SQRT{(A^2+B^2)/A^2}=SQRT((4+16)/4)=SQRT(5)
SO FOCI ARE }=(-2-2SQRT(5),1) AND (-2+2SQRT(5),1)
SLOPE OF ASYMPTOTE IS GIVEN BY DIFFERENTIATION.HAVE YOU BEEN TAUGHT?PLEASE INFORM.I SHALL COME BACK ON HEARING FROM YOU.
or you can take this proposition as proved formula
the pair of asymptotes for a conic is given by the same equation as the conic except for the constant term which has to be found using the condition for the equation to represent a pair of straight lines.
HENCE EQN OF ASYMPTOTES IS GIVEN BY
y^2 - 4x^2 - 2y - 16x + K=0 , WHERE K IS DETERMINED using the condition for the equation to represent a pair of straight lines.
SINCE WE ARE TO FIND ONLY SLOPES ,WE NEED NOT DETERMINE THE CONSTANT BUT ASSUME THAT THIS EQN REPRESENTS A PAIR OF STRAIGHT LINES.SO
y^2 - 4x^2 - 2y - 16x + K=0 = (Y+2X+A)(Y-2X+B)
HENCE SLOPES ARE +2 AND -2

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