Find Shortest Distance from z2+3x-xy=9 to Origin

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Homework Statement


a) what is the shortest distance from the surface z2+3x-xy=9 to the origin?

Homework Equations


I know that when you take the gradient of an equation the gradient is perp. to the vector and gives the direction of largest rate of inc.


The Attempt at a Solution


so I took the gradient of f(x,y,z) and i got
fx=3-y
fy=-x
fz=2z
I tried looking at these and thinking of it as a parameterized line and i got an answer of (3,0,0) but I don't think that it is right..
now I'm still a little confused on what i should do next, can someone help point me in the right direction?
 
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The gradient won't help you find distance to the origin. I would recommend Lagrange multipliers here.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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