A) \[{{(a{b}'-{a}'b)}^{2}}=4(a{h}'-{a}'h)\,(h{b}'-{h}'b)\]
B) \[{{(a{b}'+{a}'b)}^{2}}=4(a{h}'-{a}'h)\,(h{b}'-{h}'b)\]
C) \[{{(a{b}'-{a}'b)}^{2}}=(a{h}'-{a}'h)\,(h{b}'-{h}'b)\]
D) None of these
Correct Answer: A
Solution :
The equation of given lines are \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] .....(i) \[a'{{x}^{2}}+2h'xy+b'{{y}^{2}}=0\] .....(ii) Let common line to both is \[y=mx\], then it will satisfy both the above equations. Hence, \[a+2mh+b{{m}^{2}}=0\] .....(iii) and \[a'\,+\,2mh'\,+\,b'{{m}^{2}}=0\] .....(iv) Now eliminating 'm' from the equation (iii) and (iv), we get \[\frac{{{m}^{2}}}{2ha'-2h'a}=\frac{-m}{ba'-b'a}=\frac{1}{2bh'-2b'h}\] \[\Rightarrow {{m}^{2}}=\frac{ha'-h'a}{bh'-b'h}\] .....(v) and \[{{m}^{2}}=\frac{{{(ab'-ba')}^{2}}}{4{{(bh'-b'h)}^{2}}}\] .....(vi) From (v) and (vi), we get the required condition.
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