The distance of point from paraboloid

Question

I need to find the distance from the point $(0,0,0)$ to the paraboloid $c=ax^2+by^2-z$. I have tried it with Lagrange multipliers and I have wierd outcomes.. I need help in this one.

Answer 1

If $c<0$, but $a$ and $b$ are positive, then the nearest point is $(0,0,-c)$ and the distance is $-c$.
If $a,b,c$ are all positive, then take the narrower of the two parabolas $z=ax^2-c$ or $z=by^2-c$. That is the first if $a>b$, otherwise the second.
$D^2 = x^2+z^2=x^2+(ax^2-c)^2$, so to minimize $D$ - or rather, minimize $D^2$, differentiate the quartic function of $x$.

Answer 2

You have to give relative values of a,b and c to consider all cases. The minimum distance is simply c for certain values of a and b. This is because the origin lies on the axis of an oblate/prolate paraboloid.

Answer 3

We want to minimize the distance of the point $(x,y,z)$ from the origin $(0,0,0)$ subject to the constraint $ax^2 + by^2 - z - c=0$. Our objective function becomes $f = \sqrt{x^2+y^2+z^2}$. In fact, since the square root is a monotonic function (preserves order), we can choose to minimize the simpler objective function $f = x^2 + y^2 + z^2$. Setting up the Lagrange function we get $\Lambda = x^{2} + y^{2} + z^{2}- \lambda \left(a x^{2} + b y^{2} - c - z\right)$. We need to solve $\nabla \Lambda = \textbf{0}$: $$\left[\begin{matrix}- 2 a \lambda x + 2 x\\- 2 b \lambda y + 2 y\\\lambda + 2 z\\- a x^{2} - b y^{2} + c + z\end{matrix}\right] = \textbf{0}$$ So our solutions are: $(x,y,z,\lambda) = (0, 0, -c, 2c), \left(0,-\frac{\sqrt{4bc-2}}{2b},-\frac{1}{2 b},\frac{1}{b}\right), \left( 0, \frac{\sqrt{4 b c - 2}}{2 b}, - \frac{1}{2 b}, \frac{1}{b}\right)$, $\left( - \frac{\sqrt{4 a c - 2}}{2 a}, 0, - \frac{1}{2 a}, \frac{1}{a}\right),\left( \frac{\sqrt{4 a c - 2}}{2 a}, 0, - \frac{1}{2 a}, \frac{1}{a}\right)$. Characterizing which of these are minima is a bit of a pain since you need to do the 2nd derivative test for constrained optimization.

The first two points are a minimum if $b < a$. The third and fourth points are minima if $a < b$. The fifth point is a minima if $2ac<1$ and $2bc<1$. Depending on what your choice of $a,b$ and $c$, are, you can find all the minima. Then you just need to compute the objective function for those points and find which point makes $f$ the smallest.