A) \[y=mx-2am-a{{m}^{3}}\]
B) \[y=m\,(x+a)-2am-a{{m}^{3}}\]
C) \[y=m\,(x-a)+\frac{a}{m}\]
D) \[y=m\,(x-a)-2am-a{{m}^{3}}\]
Correct Answer: D
Solution :
Let normal at \[(h,k)\]be \[y=mx+c\] then, \[k=mh+c\]also \[{{k}^{2}}=4a(h-a)\] slope of tangent at \[(h,k)\]is \[{{m}_{1}}\]then on differentiating equation of parabola. \[2y{{m}_{1}}=4a\]Þ \[{{m}_{1}}=\frac{2a}{k}\] also \[m{{m}_{1}}=-1\] Þ \[m=-\frac{k}{2a},\] solving and replacing \[h,k)\] by \[(x,y)\] Þ \[y=m(x-a)-2am-a{{m}^{3}}\].
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