A) \[x+4y=0\]and \[x+3y=0\]
B) \[2x-3y=0\] and \[x-4y=0\]
C) \[x-6y=0\]and \[x-3y=0\]
D) \[x+4y=0\]and \[x-3y=0\]
Correct Answer: D
Solution :
\[{{x}^{2}}+xy-12{{y}^{2}}=0\] Þ \[{{x}^{2}}+4xy-3xy-12{{y}^{2}}=0\] Þ \[(x+4y)(x-3y)=0\] Therefore separate equations for the lines are \[x+4y=0\] and\[x-3y=0\].You need to login to perform this action.
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