• question_answer The shortest distance between the parabolas ${{y}^{2}}=4x$and${{y}^{2}}-2x+6=0$ is A) $\sqrt{2}$        B) $\sqrt{5}$ C) $\sqrt{3}$                     D) None of these

[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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