# Math Help - shortest distance question

1. ## shortest distance question

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

so far I have

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

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

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

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

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

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

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

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

is this correct?

2. ## Hmm

Originally Posted by lllll
Find the point on the surface $y^2=9+xz$ that are closest to the origin.

so far I have

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

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

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

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

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

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

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

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

is this correct?
At a precursory glance...it looks great!