A) \[g(a'-b')=g'(a+b)\]
B) \[g(a'+b')=g'(a+b)\]
C) \[g(a'+b')=g'(a-b)\]
D) \[g(a'-b')=g'(a-b)\]
Correct Answer: B
Solution :
The family of lines passing through point of intersection of the given curves will be \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+\lambda (a'{{x}^{2}}+2h'xy+b'{{y}^{2}}+2g'x)=0\]\[\Rightarrow (a+a'\lambda ){{x}^{2}}+(2h+2h'\lambda )xy\]\[+(b+b'\lambda ){{y}^{2}}+(2g+2g'\lambda )x=0\] Now the condition for perpendicularity is \[\Delta =0\]and\[a+b=0\]. \[\Rightarrow a+a'\lambda +b+b'\lambda =0\Rightarrow \lambda =-\frac{a+b}{a'+b'}\] and\[\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\] \[\Rightarrow 0+0-0-(b+b'\lambda ){{(2g+2g'\lambda )}^{2}}-0=0\] \[\Rightarrow 4(b+b'\lambda ){{(g+g'\lambda )}^{2}}=0\] Now on putting the value ofl, we get \[g(a'+b')=g'(a+b)\].
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