A) \[a=1/2\]
B) \[a=1/4\]
C) \[a>1/2\]
D) None of these
Correct Answer: C
Solution :
| Equation of normal in slope form on \[{{y}^{2}}=4\] Ax is |
| \[y=mx-2Am-A{{m}^{3}}\] ...(i) |
| \[=mx-2\left( \frac{1}{4} \right)m-\left( \frac{1}{4} \right){{m}^{3}}\] \[\left[ \begin{align} & \because \,\,{{y}^{2}}=x \\ & \therefore \,\,A=\frac{1}{4} \\ \end{align} \right]\] |
| \[\Rightarrow \,\,4mx-4y-{{m}^{3}}-2m=0\] |
| \[\because \,\,(a,0)\]lies on the normal. |
| Then, \[4m\times a-4\times 0-{{m}^{3}}-2m=0\] |
| \[\Rightarrow \,\,m({{m}^{2}}+2-4a)=0\] |
| \[\Rightarrow \,\,m=0\] or \[{{m}^{2}}+2-4a=0\] |
| If \[m=0,\]then from (i), |
| \[y=0\]i.e., x-axis is one normal. |
| If \[{{m}^{2}}+2-4a=0\Rightarrow {{m}^{2}}=4a-2\] \[[\because \,{{m}^{2}}>0]\] |
| \[\Rightarrow \,\,4a-2>0\Rightarrow \,a>\frac{1}{2}\]. |
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