question_answer

The equation of the common tangent touching the circle \[{{(x-3)}^{2}}+{{y}^{2}}=9\] and the parabola \[{{y}^{2}}=4x\] above the x-axis, is [IIT Screening 2001]

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 \[\sqrt{3}y=x+3.\]