document.write( "Question 28335: What is the vertices, foci, and slope of the asymptote for the hyperbola whose equation is, y^2 - 4x^2 - 2y - 16x + 1 = 0? \n" ); document.write( "
Algebra.Com's Answer #15409 by venugopalramana(3286)\"\" \"About 
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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?
\n" ); document.write( "(Y^2-2*Y*1+1^2)-{(2X)^2+2*(2X)*4+4^2}-1^2+4^2+1=0
\n" ); document.write( "(Y-1)^2-(2X+4)^2=-16
\n" ); document.write( "4(X+2)^2-(Y-1)^2=16
\n" ); document.write( "{4(X+2)^2}/16-{(Y-1)^2}/16=1
\n" ); document.write( "(X+2)^2/2^2-(Y-1)^2/4^2=1...
\n" ); document.write( "COMPARING WITH STANDARD EQN.
\n" ); document.write( "(X-H)^2/A^2-(Y-K)^2/B^2=1....WE HAVE
\n" ); document.write( "VERTICES ARE {(H-A),K} AND {(H+A),K}=(-2-2,1) AND (-2+2,1)=(-4,1) AND (0,1)
\n" ); document.write( "FOCI ARE {(H-AE),K} AND {(H+AE),K}...WHERE E IS
\n" ); document.write( "ECCENTRICITY =SQRT{(A^2+B^2)/A^2}=SQRT((4+16)/4)=SQRT(5)
\n" ); document.write( "SO FOCI ARE }=(-2-2SQRT(5),1) AND (-2+2SQRT(5),1)
\n" ); document.write( "SLOPE OF ASYMPTOTE IS GIVEN BY DIFFERENTIATION.HAVE YOU BEEN TAUGHT?PLEASE INFORM.I SHALL COME BACK ON HEARING FROM YOU.
\n" ); document.write( "or you can take this proposition as proved formula
\n" ); document.write( "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.
\n" ); document.write( "HENCE EQN OF ASYMPTOTES IS GIVEN BY
\n" ); document.write( "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.
\n" ); document.write( "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
\n" ); document.write( "y^2 - 4x^2 - 2y - 16x + K=0 = (Y+2X+A)(Y-2X+B)
\n" ); document.write( "HENCE SLOPES ARE +2 AND -2\r
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