A) \[\sqrt{3}y=x+3\]
B) \[\sqrt{3}y=3x+3\]
C) \[2\sqrt{3}y=\,\,-x-12\]
D) \[2\sqrt{3}\,y=12x+1\]
Correct Answer: A
Solution :
\[{{y}^{2}}\,\,=\,\,4x\] \[y\,\,=\,\,mx\,+\frac{1}{m}\,\,is\,\,\operatorname{tangent}\,\,to\,\,{{x}^{2}}+{{y}^{2}}\,\,=\,6x\] \[{{x}^{2}}\,+{{\left( mx+\frac{1}{m} \right)}^{2}}\,=\,\,6x\] \[{{x}^{2}}\,+\,{{m}^{2}}{{x}^{2}}\,\,+\,\frac{1}{{{m}^{2}}}\,+\,2mx\,\,.\,\,\frac{1}{m}\,\,=\,\,6x\] \[{{x}^{2}}\,(1+{{m}^{2}})\,-\,\,4x\,\,+\,\,\frac{1}{{{m}^{2}}}\,\,=\,\,0\,\,\] \[D\,\,=\,\,0\] \[16\,\,=\,\,4(1\,\,+\,\,{{m}^{2}})\,\,.\,\,\frac{1}{{{m}^{2}}}\] \[4{{m}^{2}}=1\text{ }+{{m}^{2}}\] \[3{{m}^{2}}=\text{ }1\] \[m\,\,=\,\,\pm \,\,\frac{1}{\sqrt{3}}\] if \[m\,\,=\,\,+\,\frac{1}{\sqrt{3}}\] then equation of tangent is \[y\,\,=\,\,\frac{1}{\sqrt{3}}x\,\,+\,\sqrt{3}\,\] \[\sqrt{3}y\,\,=\,\,x\,\,+\,\,3\]
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