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Let $P=\left(\frac{a^2}{2},a\right)$ be a point on the parabola. First, we find the equation of the tangent to the parabola at $P.$



Since the tangent passes through $\left(\frac{a^2}{2},a\right),$ the equation of the tangent is of the form $$y-a=m\left(x-\frac{a^2}{2}\right)=mx-\frac{a^{2}m}{2}.$$Substituting $x=\frac{y^2}{2},$ we get $$y-a=\frac{m{y}^2}{2}-\frac{a^{2}m}{2}.$$This simplifies to $m{y}^2-2y+2a-a^{2}m=0.$ Since this is the equation of a tangent, the quadratic should have a double root of $y=a,$ which means its discriminant is 0, which gives us $$4-4m(2a-a^{2}m)=0.$$Then $4a^2{m}^2-8am+4=4(am-1{)}^2=0,$ so $m=\frac{1}{a}.$

Now, consider the point $P$ that is closest to $(6,12).$



Geometrically, the line connecting $P$ and $(6,12)$ is perpendicular to the tangent. In terms of slopes, this gives us $$\frac{a-12}{\frac{a^2}{2}-6}\cdot \frac{1}{a}=-1.$$This simplifies to $a^3-10a-24=0,$ which factors as $(a-4)(a^2+4a+6)=0.$ The quadratic factor has no real roots, so $a=4.$ Therefore, $P=(8,4),$ and the shortest distance is $\sqrt{(8-6{)}^2+(4-12{)}^2}=\boxed{2\sqrt{17}}.$