A) \[\sqrt{3}y=3x+1\]
B) \[\sqrt{3}y=-(x+3)\]
C) \[\sqrt{3}y=x+3\]
D) \[\sqrt{3}y=-(3x+1)\]
Correct Answer: C
Solution :
Any tangent to \[{{y}^{2}}=4x\] is \[y=mx+\frac{1}{m}.\] It touches the circle, if \[3=\left| \frac{3m+\frac{1}{m}}{\sqrt{1+{{m}^{2}}}} \right|\] or \[9(1+{{m}^{2}})={{\left( 3m+\frac{1}{m} \right)}^{2}}\] or \[\frac{1}{{{m}^{2}}}=3\], \[\therefore \,m=\pm \frac{1}{\sqrt{3}}.\] For the common tangent to be above the x-axis, \[m=\frac{1}{\sqrt{3}}\] \Common tangent is, \[y=\frac{1}{\sqrt{3}}x+\sqrt{3}\] Þ \[\sqrt{3}y=x+3.\]You need to login to perform this action.
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