the locus of the midpoint of the portion of a line of constant slope m between two branches of the rectangular hyperbola xy=1 is

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xy=1Letequation of line be y=mx+cPut in hyperbolaxmx+c=1mx2+cx-1=0Use Shridharacharya's formulax=-c±c2+4m2mx1=-c+c2+4m2m and x2=-c-c2+4m2mPoint lie on hyperbola , hence y1=1x1  and y2=1x2y1=2m-c+c2+4m and y2=2m-c-c2+4mx1,y1=-c+c2+4m2m,2m-c+c2+4mx2,y2=-c-c2+4m2m,2m-c-c2+4mLet mid point be h,kh=x1+x22=-c+c2+4m2m+-c-c2+4m2m2h=-c2m...ik=y1+y22=2m-c+c2+4m+2m-c-c2+4m2=2m21c2+4m-c-1c2+4m+c=mc2+4m+c-c2+4m+cc2+4m-cc2+4m+c=m2cc2+4m-c2=m2c4m=k=c2c=2kPut in ih=-2k2mmh=-kk+mh=0Hence locus isy+mx=0 Answer

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