Answer
$P(\pm \sqrt{5}, 0,0)$
Work Step by Step
The distance from a piven point $P(x, y, x)$ to the origin is given by
\[
\sqrt{x^{2}+y^{2}+z^{2}}=d
\]
Since all point on the given surface satisfy $x^{2}=5+y=$, we have that
\[
\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{5+y z+y^{2}+z^{2}}=d
\]
Using the first derivative test:
\[
\begin{array}{l}
d_{y}=0 \Rightarrow \frac{1}{2 d} \frac{\partial}{\partial y}\left(5+y z+y^{2}+z^{2}\right)=0 \Rightarrow z+2 y=0 \\
d_{z}=0 \Rightarrow \frac{1}{2 d} \frac{\partial}{\partial z}\left(5+y z+y^{2}+z^{2}\right)=0 \Rightarrow y+2 z=0
\end{array}
\]
From the first of the last two equations, we have $-2 y=z$. Using this in the second equation gives:
\[
0=y+2(-2 y) \Rightarrow 0=y
\]
Hence $z=-2 y=-2(0)=0$ and $x^{2}=5+y z=5+0(0)=5 \Rightarrow x=\pm \sqrt{5}$
Then $P(\pm 5,0,0)$ are the only critical points.
Using the second derivative test at this point we have:
\[
\begin{aligned}
f_{\mathrm{vv}}(0,0) &=\frac{1}{4 d^{3}}\left(20+3(0)^{2}\right)=\frac{20}{4 d^{3}} \\
f_{x z}(0,0) &=\frac{1}{4 d^{3}}\left(20+3(0)^{2}\right)=\frac{20}{4 d^{3}} \\
f_{z v}(0,0) &=\frac{20(20)}{\left(-4 d^{3}\right)^{2}}-\frac{10^{2}}{\left(4 d^{3}\right)^{2}}=300 /\left(4 d^{3}\right)^{2} \frac{1}{4 d^{3}}(10-3(0)(0))=\frac{10}{4 d^{3}} \\
D &=f_{v y} f_{z z}-f_{z y}^{2}=\frac{20(20)}{\left(4 d^{3}\right)^{2}}-\frac{10^{2}}{\left(4 d^{3}\right)^{2}}=300 /\left(4 d^{3}\right)^{2}>0
\end{aligned}
\]
At $P(\pm \sqrt{5}, 0,0),$ the distance from the surface to the origin is a minimum.
