A) \[4al+n=0\]
B) \[4al+4am+n=0\]
C) \[4am+n=0\]
D) \[al+n=0\]
Correct Answer: A
Solution :
[a] Make homogeneous and put A+B=0 \[{{y}^{2}}=4ax\left( \frac{lx+my}{-n} \right)\] \[\therefore 4al+n=0\]
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