# Find the distance from the point P(0,0,1) to the surface z=x^2+2y^2

Find the distance from the point P(0,0,1) to the surface z=x^2+2y^2
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Jovanni Salinas
Find the distance from the point P(0,0,1) to the surface $z={x}^{2}+2{y}^{2}$
Let $d\left(x,y,z\right)=\sqrt{\left(x-0{\right)}^{2}+\left(y-0{\right)}^{2}+\left(z-1{\right)}^{2}}$ be the distance from point P.
We need to find the shortest distance wr.t constant

d is always positive , so , for simplicity we can work on ${d}^{2}={x}^{2}+{y}^{2}+\left(z-1{\right)}^{2}$.
Subject to $z-{x}^{2}-2{y}^{2}=0$
Lagrangian function
$L\left(x,u,z,\lambda \right)=d\left(x,y,z\right)+\lambda g\left(x,y,z\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+{y}^{2}+\left(z-1{\right)}^{2}+\lambda \left(z-{x}^{2}-2{y}^{2}\right)$
where $\lambda$ is lagrange multiplier
By the lagrangian method we need to solve

when
$x=y=z=0\phantom{\rule{0ex}{0ex}}⇒d\left(x,y,z\right)=1\phantom{\rule{0ex}{0ex}}x=±\frac{1}{\sqrt{2}},y=0,z=\frac{1}{2}\phantom{\rule{0ex}{0ex}}d\left(x,y,z\right)=\sqrt{\frac{1}{2}+0+\left(\frac{1}{2}-1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{2}+\frac{1}{4}}=\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}x=0,{y}^{2}=\frac{3}{8},z=\frac{3}{4}\phantom{\rule{0ex}{0ex}}d\left(x,y,z\right)=\sqrt{0+\frac{3}{8}+\left(\frac{3}{4}-1{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{3}{8}+\frac{1}{16}}=\frac{\sqrt{7}}{4}$
shortest distance is from (0,0,0)