The condition that one of the straight line given by the equation ax2+2hxy+by2=0 may coincide with one of those given by the equation a'x2+2h'xy+b'y2=0 is

Let the common line be  y=mx. Then it must satisfy both the equations. Therefore, we have bm2+2hm+a=0______(1)b'm2+2h' m+a'=0_____(2)Solved (1) and (2), we getm22(ha'−h'a)=mab'−a'b=12(bh'−b'h)Eliminating m we get  [ab'−a'b2(2bh'−b'h)]2=ha'−h'abh'−b'hor (ab'−a'b)2=4(ha'−h'a)(bh'−b'h)