A) \[\frac{3\sqrt{2}}{8}\]
B) \[\frac{8}{3\sqrt{2}}\]
C) \[\frac{4}{\sqrt{3}}\]
D) \[\frac{\sqrt{3}}{4}\]
Correct Answer: A
Solution :
Let \[P({{y}^{2}},y)\] \[\therefore \]Perpendicular distance from P to \[x-y+1=0\] \[=\frac{|{{y}^{2}}-y+1|}{\sqrt{2}}\] As, \[{{y}^{2}}-y+1>0\,\forall \,y\,\in \,R\] \[\therefore \] Min value \[=\frac{1}{\sqrt{2}}\left( \frac{4ac-{{b}^{2}}}{4a} \right)=\frac{3}{4\sqrt{2}}\]
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