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Q.

If the equation of pair of asymptote of a hyperbola H is xy2xy+2=0 . Its conjugate hyperbola H'  passes through (3,0) then

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a

focus of H'  is  (2+12,2+12)

b

equation of transverse axis  of H is  xy+1=0

c

Eccentricity of H' is   2 

d

focus of  H is  (2,2)

answer is A, D.

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Detailed Solution

Equation of hyperbola is  xyy2x2=0
(x1)(y2)=4
XY=4
Conjugate hyperbola XY = -4
a22=4a=22
Focus of   with respect  to X-Y system =  (22,22)and(22,22)
So focus in x-y system  (122,2+22)and(1+22,222)
Equation of transverse axis Y= X
y2=x1xy+1=0
 

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