help plz

 the smallest distance

$y=0.5(x^2-18)\\ l^2=x^2+y^2\\ l^2=x^2+0.5^2\cdot (x^2-18)^2\\ \dfrac{d\ l^2}{dx}=2x+0.25\cdot 2\cdot (x^2-18)\cdot 2x=0\\ 2x+x(x^2-18)=0\\ 2+x^2-18=0\\ x^2=16\\ x\in \{-4,4\}$

$l=\sqrt{x^2+y^2}=\sqrt{4^2+(-1)^2}\\ \color{blue}l=\sqrt{17}=4.123$

Tthe smallest distance is 4.123.

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