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\(y=\frac{1}{\sqrt2}-(x^2-3)\\ y=-x^2+3+\frac{1}{\sqrt2}\\ y=-x^2+\frac{6+\sqrt2}{2}\\ \)

This is a concave down parabola with y intercept     \(\frac{6+\sqrt2}{2}\approx 3.707\)

And the y axis is the axis of symmetry.

A general point on this parabola is    \((x,\left[-x^2+\frac{6+\sqrt2}{2}\right])\\\)

The distance square from (0,0) to the parabola for any given x value will be

\(d^2=D=(x-0)^2+(-x^2+\frac{6+\sqrt2}{2}-0)^2\\ D=x^2+(-x^2+\frac{6+\sqrt2}{2})^2\\ D=x^2+x^4+(-2*\frac{6+\sqrt2}{2})x^2+(\frac{6+\sqrt2}{2})^2\\ D=x^2+x^4+(-6-\sqrt2)x^2+(\frac{36+2+12\sqrt2}{4})\\ D=x^4+(1-6-\sqrt2)x^2+(\frac{19}{2}+3\sqrt2)\\ D=x^4+(-5-\sqrt2)x^2+(\frac{19}{2}+3\sqrt2)\\~\\ D'=4x^3+2(-5-\sqrt2)x\\ \text{D will be minimum when }D'=0\\ 0=4x^3+2(-5-\sqrt2)x\\ \text{I know just from thinking about the graph that }\;x\ne0\\ 0=4x^2+2(-5-\sqrt2)\\ 0=4x^2-10-2\sqrt2\\ 4x^2=10+2\sqrt2\\ 2x^2=5+\sqrt2\\ x^2=\frac{5+\sqrt2}{2}\\ x=\pm\sqrt{\frac{5+\sqrt2}{2}}\\\)

So the distance will be minimum when     \(x=\pm\sqrt{\frac{5+\sqrt2}{2}}\\\)

Sub to get y

\(y=-\left[\pm\sqrt{\frac{5+\sqrt2}{2}}\right]^2+\frac{6+\sqrt2}{2}\\ y=-\left[\frac{5+\sqrt2}{2}\right]+\frac{6+\sqrt2}{2}\\ y=\frac{-5-\sqrt2}{2}+\frac{6+\sqrt2}{2}\\ y=\frac{1}{2}\)

So the closest points on the parabola to (0,0)     are      \(\left(\sqrt{\frac{5+\sqrt2}{2}},\frac{1}{2}\right)\quad and \quad \left(-\sqrt{\frac{5+\sqrt2}{2}},\frac{1}{2}\right)\)

So the distance from either both these points to the origin will be    

\(d=\sqrt{          \left(\frac{5+\sqrt2}{2}\right)   + \frac{1}{4}       }\\ d=\sqrt{          \left(\frac{10+2\sqrt2+1}{4}\right) }\\ d=\sqrt{          \left(\frac{11+2\sqrt2}{4}\right) }\\ smallest\;\; distance=\frac{\sqrt{  11+2\sqrt2  }}{2}\\ smallest\;\; distance\approx 1.86\;units\)

Here is the related graph:

LaTex:

y=\frac{1}{\sqrt2}-(x^2-3)\\
y=-x^2+3+\frac{1}{\sqrt2}\\
y=-x^2+\frac{6+\sqrt2}{2}\\

d^2=D=(x-0)^2+(-x^2+\frac{6+\sqrt2}{2}-0)^2\\
D=x^2+(-x^2+\frac{6+\sqrt2}{2})^2\\
D=x^2+x^4+(-2*\frac{6+\sqrt2}{2})x^2+(\frac{6+\sqrt2}{2})^2\\
D=x^2+x^4+(-6-\sqrt2)x^2+(\frac{36+2+12\sqrt2}{4})\\
D=x^4+(1-6-\sqrt2)x^2+(\frac{19}{2}+3\sqrt2)\\
D=x^4+(-5-\sqrt2)x^2+(\frac{19}{2}+3\sqrt2)\\~\\
D'=4x^3+2(-5-\sqrt2)x\\
\text{D will be minimum when }D'=0\\
0=4x^3+2(-5-\sqrt2)x\\
\text{I know just from thinking about the graph that  }\;x\ne0\\
0=4x^2+2(-5-\sqrt2)\\
0=4x^2-10-2\sqrt2\\
4x^2=10+2\sqrt2\\
2x^2=5+\sqrt2\\
x^2=\frac{5+\sqrt2}{2}\\
x=\pm\sqrt{\frac{5+\sqrt2}{2}}\\

y=-\left[\pm\sqrt{\frac{5+\sqrt2}{2}}\right]^2+\frac{6+\sqrt2}{2}\\
y=-\left[\frac{5+\sqrt2}{2}\right]+\frac{6+\sqrt2}{2}\\
y=\frac{-5-\sqrt2}{2}+\frac{6+\sqrt2}{2}\\
y=\frac{1}{2}

\left(\sqrt{\frac{5+\sqrt2}{2}},\frac{1}{2}\right)\quad and \quad \left(-\sqrt{\frac{5+\sqrt2}{2}},\frac{1}{2}\right)

d=\sqrt{          \left(\frac{5+\sqrt2}{2}\right)   + \frac{1}{4}       }\\
d=\sqrt{          \left(\frac{10+2\sqrt2+1}{4}\right) }\\
d=\sqrt{          \left(\frac{11+2\sqrt2}{4}\right) }\\
smallest\;\; distance=\frac{\sqrt{  11+2\sqrt2  }}{2}\\
smallest\;\; distance\approx 1.86\;units