A) \[\frac{1}{n}+\frac{1}{m}=0\]
B) \[\frac{1}{n}-\frac{1}{m}=0\]
C) \[nm-1=0\]
D) \[nm+1=0\]
E) \[\frac{1}{m}-\frac{1}{n}=0\]
Correct Answer: D
Solution :
Equation of bisectors of angles between \[{{x}^{2}}-2nxy-{{y}^{2}}=0\]is \[\frac{{{x}^{2}}-{{y}^{2}}}{1+1}=\frac{xy}{-n}\] \[{{x}^{2}}+\frac{2}{n}xy+{{y}^{2}}=0\] ...(i) This equation is identical to \[{{x}^{2}}-2mxy-{{y}^{2}}=0\] \[\therefore \] \[\frac{\frac{2}{n}}{-2m}=1\] \[\Rightarrow \] \[\frac{1}{n}=-m\] \[\Rightarrow \] \[mn+1=0\]You need to login to perform this action.
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