A) \[c=-2am-a{{m}^{3}}\]
B) \[c=-\frac{a}{m}\]
C) \[c=\frac{a}{m}\]
D) \[c=2am+a{{m}^{3}}\]
Correct Answer: A
Solution :
Given the, equation of parabola \[{{y}^{2}}=4ax,\] let the parametric coordinate is \[(a{{m}^{2}},2am).\] \[\Rightarrow \] \[2y\frac{dy}{dx}=4a\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{2a}{y}\] Slope of normal \[=\left( \frac{-y}{2a} \right)\] At \[(a{{m}^{2}},2am)=\frac{-2am}{2a}=-m\] Now, the equation of normal to the parabola is \[(y-2am)=(-m)(x-a{{m}^{2}})\] \[y-2am=-mx+a{{m}^{3}}\] \[mx+y-(2am+a{{m}^{3}})=0\] ….(i) Also, given the line \[y=mx+c\] or \[mx-y+c=0\] …..(ii) Is normal to parabola, then On comparing \[c=-2am-a{{m}^{3}}\]
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