NIMCET 2008 Question Paper with solutions

Tangent to parabola ${\mathit{y}}^2=4\mathit{a}\mathit{x}$ is
$\mathit{y}=\mathit{m}\mathit{x}+\frac{\mathit{a}}{\mathit{m}}$
⇒ Tangent to ${\mathit{y}}^2=4\mathit{x}$ will be
$\mathit{y}=\mathit{m}\mathit{x}+\frac{1}{\mathit{m}}$....(i)
It will be tangent to the circle
${\left(\mathit{x}-3\right)}^2+{\mathit{y}}^2=9={\left(3\right)}^2$
If length of perpendicular from (3,0) will be 3 .
$\Rightarrow \left|\frac{3\mathit{m}+\frac{1}{\mathit{m}}}{\sqrt{{\mathit{m}}^2+1}}\right|=3$
$\Rightarrow 3{\mathit{m}}^2+1=3\mathit{m}\sqrt{{\mathit{m}}^2+1}$
$\Rightarrow 9{\mathit{m}}^4+6{\mathit{m}}^2+1=9{\mathit{m}}^2\left({\mathit{m}}^2+1\right)$
$\Rightarrow \mathit{m}=\frac{1}{\sqrt{3}}$, for above x - axis
So, tangent is $\mathit{y}=\frac{1}{\sqrt{3}}\mathit{x}+\frac{1}{1∕\sqrt{3}}$
$\Rightarrow \sqrt{3}\mathit{y}=\mathit{x}+3$