A) \[-2am-a{{m}^{3}}\]
B) \[2am+a{{m}^{3}}\]
C) \[-\frac{2a}{m}-\frac{a}{{{m}^{3}}}\]
D) \[\frac{2a}{m}+\frac{a}{{{m}^{3}}}\]
Correct Answer: A
Solution :
The equation of the normal to \[{{x}^{2}}=4ay\]is of the form \[x=my-2am-a{{m}^{3}}.\] Therefore\[c=-2am-a{{m}^{3.}}\].
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