Question

find the shortest distance between the origin and the surface x ^ 2y-z ^ 2 + 9 = 0

Answer 1

Let P(x,y,z) be the point on the surface $x^{2}y-z^2+9=0$ . Then the distance from the origin is $d = \sqrt{x^2+y^2+z^2}$ . We have to minimize the value of $d$ subject to the constraint $x^{2}y-z^2+9=0$.

Let $f(x,y,z) = x^{2}+y^{2}+z^{2}$ and let $g(x,y,z)=x^{2}y-z^{2}+9$.

From the given constraint, $z^{2}=x^{2}y+9$. Using this in f(x,y,z) it becomes $x^{2}+y^{2}+x^{2}y+9$, which is a function in two variables.

Let $F(x,y)=x^{2}+y^{2}+x^{2}y+9.$ Now we have to solve,

$\dfrac{\partial F}{\partial x} = 0; \dfrac{\partial F}{\partial y} = 0\\$

which gives,

$2x+2xy = 0~~~~(1)\\ 2y+x^{2}=0~~~~~~~(2)$

Solving equation (1), we get either $x=0~~\text{or}~~y=-1.$

When $x = 0, y = 0$ and When $y = -1, x = \pm \sqrt{2}$ , from equation (2).

Now from the constraint equation, $z^{2} = 9 \Rightarrow z = \pm 3$, when x = 0 and y=0.

Also, $z^{2} = -2 + 9 = 7 \Rightarrow z = \pm \sqrt{7}$, when $y = -1, x = \pm \sqrt{2}$.

Therefore the critical points are, $(0,0,\pm3), (\pm \sqrt{2},-1,\pm\sqrt{7})$.

Now,

$f(0,0,\pm3) = 9\\ f(\pm\sqrt2,-1,\pm\sqrt7)=10$

Hence the minimum distance is $d = 3$ which occurs at the points $(0,0,\pm3)$.