A) \[fh=cg\]
B) \[fg=ch\]
C) \[{{h}^{2}}=gf\]
D) \[fgh=c\]
Correct Answer: B
Solution :
\[hxy+gx+fy+c=0,\]Apply \[\Delta '=0\] i.e., \[a'b'c'+2f'g'h'-a'f{{'}^{2}}-b'g{{'}^{2}}-{c}'\,h{{'}^{2}}=0\] Here \[h'=\frac{h}{2},\ g'=\frac{g}{2},\ f'=\frac{f}{2},\ c'=c,\ a'=0,\ b'=0\] Hence, \[\frac{hgf}{4}-\frac{c{{h}^{2}}}{4}=0\]or\[fg=ch\].
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