A) \[{{a}^{2}}+{{b}^{2}}=1\]
B) \[{{a}^{2}}+{{b}^{2}}=2\]
C) \[2({{a}^{2}}+{{b}^{2}})=1\]
D) none of these
Correct Answer: B
Solution :
[b] Lines \[x\cos a+y\sin a=p\]and \[x\sin \alpha -y\cos \alpha =0\] are mutually perpendicular, thus , \[ax+by+p=0\] will be equally inclined to these lines and would be the angle bisector of these lines. Now, the equations of angle bisectors are \[x\sin \alpha -y\cos \alpha =\pm (xcos\alpha +ysin\alpha -p)\] Or \[x(cos\alpha -sin\alpha )+y(sin\alpha +cos\alpha )=p\] Or \[x(sin\alpha +cos\alpha )-y(cos\alpha -sin\alpha )=p\] Comparing th4ese lines with \[ax+by+p=0\], we get \[\frac{a}{\cos \alpha -\sin \alpha }=\frac{b}{\sin \alpha +\cos \alpha }=1\] Or \[{{a}^{2}}+{{b}^{2}}=2\] Or \[\frac{a}{\sin \alpha +\cos \alpha }=\frac{b}{\cos \alpha -\sin \alpha }=1\] Or \[{{a}^{2}}+{{b}^{2}}=2\]
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