A) \[\sqrt{2}\]
B) \[\sqrt{5}\]
C) \[\sqrt{3}\]
D) None of these
Correct Answer: B
Solution :
[b]: Equation of normal to the curve \[{{y}^{2}}=4x\] at \[({{m}^{2}},2m)\]is taken as \[y-{{y}_{1}}=-\frac{1}{\left( \frac{dy}{dx} \right)}(x-{{x}_{1}})\] \[\Rightarrow \]\[(y-2m)=-m(x-{{m}^{2}})\] \[\Rightarrow \]\[y+mx-2m-{{m}^{3}}=0\] ?(i) Similarly normal to \[{{y}^{2}}-2x+6=0\]at \[\left( \frac{1}{2}{{t}^{2}}+3,t \right)\]is \[y+t(x-3)-t-\frac{1}{2}{{t}^{3}}=0\] ...(ii) Shortest distance between two curves exist along the common normal. Let (i) and (ii) are same \[\therefore \]\[-2m-{{m}^{3}}=-4m-\frac{1}{2}{{m}^{3}}\Rightarrow m=0,m=\pm 2\] Points on the parabolas\[({{m}^{2}},2m)=(4,4)\]and\[\left( \frac{1}{2}{{m}^{2}}+3,m \right)=(5,2)\] \[\therefore \]Shortest distance \[=\sqrt{{{(5-4)}^{2}}+{{(4-2)}^{2}}}=\sqrt{5}\]
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