A) \[p{{x}^{2}}+2xy+p{{y}^{2}}=0\]
B) \[p{{x}^{2}}-2xy+p{{y}^{2}}=0\]
C) \[p{{x}^{2}}+2xy-p{{y}^{2}}=0\]
D) None of these
Correct Answer: C
Solution :
The bisectors of the angles between the lines in the new position are same as the bisectors of the angles between their old positions. Therefore, required equation is \[\frac{{{x}^{2}}-{{y}^{2}}}{1-(-1)}=\frac{xy}{-p}\] \[\Rightarrow \] \[-p{{x}^{2}}+p{{y}^{2}}=2xy\] \[\Rightarrow \] \[p{{x}^{2}}+2xy-p{{y}^{2}}=0\]You need to login to perform this action.
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