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**lllll** Find the point on the surface $\displaystyle y^2=9+xz$ that are closest to the origin.

so far I have

$\displaystyle d= \sqrt{x^2+y^2+z^2} = \sqrt{x^2+9+xz+z^2}$

$\displaystyle d^2=x^2+9+xz+z^2 $

$\displaystyle F_x = 2x+z \ \mbox{and} \ F_z = 2z+x$

$\displaystyle F_x = 2x+z = 0\ \mbox{and} \ F_z = 2z+x = 0$

$\displaystyle x = 0\ \mbox{and} \ z = 0$

$\displaystyle F_{xx} = 2, \ F_{zz} = 2, \ F_{xz} = F_{zx} = 1$

$\displaystyle D=(2)(2) -1^1 = 3 > 0 \ mbox{and} \ F_{xx} = 2 > 0 \therefore \mbox{minimum}$

and the shortest distance would be $\displaystyle \sqrt{0^2+9+(0)(0)+0^2} = 3$

is this correct?