question_answer

Minimum distance between the curves\[{{y}^{2}}=4ax\]and\[{{x}^{2}}+{{y}^{2}}-12x+31=0\]is

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 \[2\sqrt{5}-\sqrt{5}=\sqrt{5}\]