The equation of the common tangent touching t | vedclass

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(C) The equation of a tangent to the parabola $y^{2}=4x$ is given by $y=mx+\frac{a}{m}$,where $a=1$. Thus,$y=mx+\frac{1}{m}$.This line is also tangent

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$\sqrt{3}y=x+3$

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(C) The equation of a tangent to the parabola $y^{2}=4x$ is given by $y=mx+\frac{a}{m}$,where $a=1$. Thus,$y=mx+\frac{1}{m}$.This line is also tangent to the circle $(x-3)^{2}+y^{2}=9$,which has center $(3,0)$ and radius $r=3$.The perpendicular distance from the center $(3,0)$ to the line $mx-y+\frac{1}{m}=0$ must equal the radius $3$:$\frac{|m(3)-0+\frac{1}{m}|}{\sqrt{m^{2}+(-1)^{2}}}=3$$|3m+\frac{1}{m}|=3\sqrt{m^{2}+1}$Squaring both sides:$(3m+\frac{1}{m})^{2}=9(m^{2}+1)$$9m^{2}+6+\frac{1}{m^{2}}=9m^{2}+9$$\frac{1}{m^{2}}=3$ $\Rightarrow m^{2}=\frac{1}{3}$ $\Rightarrow m=\pm\frac{1}{\sqrt{3}}$.Since the tangent touches the parabola above the $x$-axis,we choose $m=\frac{1}{\sqrt{3}}$.Substituting $m$ into the tangent equation:$y=\frac{1}{\sqrt{3}}x+\sqrt{3}$Multiplying by $\sqrt{3}$ gives $\sqrt{3}y=x+3$.

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

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