A) \[\sqrt{5}\]
B) \[\sqrt{21}\]
C) \[\sqrt{28}-\sqrt{5}\]
D) \[\sqrt{21}-\sqrt{5}\]
Correct Answer: A
Solution :
Centre and radius of the given circle is\[P(6,\,\,0)\]and\[\sqrt{5}\], respectively. Now minimum distance between two curves always occurs along a line which normal to both the curves. Equation of normal to\[{{y}^{2}}=4x\]at\[({{t}^{2}},\,\,2t)\]is If it is normal to circle also, then it must pass though\[(6,\,\,0)\]. \[\therefore \]\[0={{t}^{3}}-4t\] \[\Rightarrow \] \[t=0\] or \[t=\pm 2\] \[\Rightarrow \]\[A(4,\,\,4)\]and\[(4,\,\,-4)\]. \[\Rightarrow \]\[PA=PC=\sqrt{20}=2\sqrt{5}\] \[\Rightarrow \]Required minimum distance \[2\sqrt{5}-\sqrt{5}=\sqrt{5}\]You need to login to perform this action.
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