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 :
\[{{x}^{2}}+xy-12{{y}^{2}}=0\] \[\Rightarrow \]\[{{x}^{2}}+4xy-3xy-12{{y}^{2}}=0\] \[\Rightarrow \]\[x(x+4y)-3y(x+4y)=0\] \[\Rightarrow \]\[(x+4y)(x-3y)=0\] \[x+4y=0\]and \[x-3y=0\]You need to login to perform this action.
You will be redirected in
3 sec