If $(\alpha,\beta)$ is a point on $y^2=6x$, that is closest to $(3,\frac{3}{2})$, then find the value 0f $2(\alpha+\beta)$.

Question

If $(\alpha,\beta)$ is a point on $y^2=6x$, that is closest to $(3,\frac{3}{2})$, then find the value 0f $2(\alpha+\beta)$.

There are many method of doing this problem so I am using the following method and got stuck at the end.

${y^2} = 6x \Rightarrow {y^2} = 4 \times \frac{3}{2} \times x$

$y = mx + \frac{3}{{2m}} \Rightarrow mx - y + \frac{3}{{2m}} = 0$ represent the equation of tangent

Writing in line form we get the followng equation $\frac{{x - 3}}{m} = \frac{{y - \frac{3}{2}}}{{ - 1}} = - \frac{{\left( {3m - \frac{3}{2} + \frac{3}{{2m}}} \right)}}{{{m^2} + 1}} \Rightarrow x - 3 = - \frac{{m\left( {3m - \frac{3}{2} + \frac{3}{{2m}}} \right)}}{{{m^2} + 1}}$

$ \Rightarrow y - \frac{3}{2} = \frac{{\left( {3m - \frac{3}{2} + \frac{3}{{2m}}} \right)}}{{{m^2} + 1}}$

${D^2} = {\left( {x - 3} \right)^2} + {\left( {y - \frac{3}{2}} \right)^2} = \frac{{{{\left( {3m - \frac{3}{2} + \frac{3}{{2m}}} \right)}^2}}}{{{m^2} + 1}}$

$D^2$ needs to be minimum , how do we get the minimum value of $m$.

I have already solved this problem and was experimenting this method to arrive at the answer but not able to proceed