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(C) Step $1$: The equation of a tangent to the parabola $y^2 = 4ax$ (where $a=1$) is $y = mx + \frac{a}{m}$, which is $y = mx + \frac{1}{m}$.Step $2$:

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

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(C) Step $1$: The equation of a tangent to the parabola $y^2 = 4ax$ (where $a=1$) is $y = mx + \frac{a}{m}$, which is $y = mx + \frac{1}{m}$.Step $2$: Rewrite this as $mx - y + \frac{1}{m} = 0$. This line is also tangent to the circle $(x - 3)^2 + y^2 = 3^2$, so the perpendicular distance from the center $(3, 0)$ to the line must equal the radius $3$.Step $3$: Using the distance formula: $\frac{|m(3) - 0 + 1/m|}{\sqrt{m^2 + (-1)^2}} = 3$.Step $4$: $|3m + 1/m| = 3\sqrt{m^2 + 1} \implies |\frac{3m^2 + 1}{m}| = 3\sqrt{m^2 + 1}$.Step $5$: Squaring both sides: $\frac{(3m^2 + 1)^2}{m^2} = 9(m^2 + 1) \implies 9m^4 + 6m^2 + 1 = 9m^4 + 9m^2$.Step $6$: $3m^2 = 1 \implies m^2 = 1/3$. Since the tangent is above the $X$-axis, we take $m = 1/\sqrt{3}$.Step $7$: Substituting $m = 1/\sqrt{3}$ into $y = mx + 1/m$: $y = \frac{1}{\sqrt{3}}x + \sqrt{3} \implies \sqrt{3}y = x + 3$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

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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