A) \[c=\frac{1}{4}\]
B) \[c=\frac{1}{2}\]
C) \[c>\frac{1}{2}\]
D) None of these
Correct Answer: C
Solution :
Key Idea: The slope form of the normal equation to the parabola is \[{{y}^{2}}=4ax\]is \[y=mx-2am-a{{m}^{3}}\] The slope form of the normal to the parabola\[{{y}^{2}}=4ax\]is Since, the given curve is \[{{y}^{2}}=x\] Here, If it passes through \[(c,0),\] then \[0=mc-\frac{1}{2}m-\frac{1}{4}{{m}^{3}}\] \[\Rightarrow \]\[m=0\]or \[c-\frac{1}{2}-\frac{1}{4}{{m}^{2}}=0\] \[\Rightarrow \]\[m=\pm \,2\sqrt{c-\frac{1}{2}}\] For three normal values of \[m,\]it should be real. \[\Rightarrow \] \[m=\pm \,2\sqrt{c-\frac{1}{2}}\] For three normal value of \[m,\]it should be real. \[\therefore \] \[c>\frac{1}{2}.\]
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