A Lagrange Multiplier Problem : How to deal with this case when $b< 8$

Question

I was trying to solve the following problem of several variables calculus given in my class.I am stuck in a particular case of the problem.Please help me to solve the problem.Thnx in advance.

Find the shortest distance from $(0,b)$ to the parabola $x^2-16y=0$ using Langrange multiplier method.

What I done so far:

To minimise $f(x,y)=x^2+(y-b)^2$ subject to $g(x,y)=0$ where $g(x,y)=x^2-16y$ I formed $F(x,y)= x^2+(y-b)^2+\lambda (x^2-16y)$ and thus solving for stationary points using $$F_x=2x(\lambda+1)=0,F_y=2(y-b)-16\lambda=0,x^2-16y=0$$ I got $(0,0)$ is a staionary point and when $\lambda=1$ staionary points are $(\pm4\sqrt{b-8},b-8)$ provided $b\ge8$

In this case ($b\ge8$) minimum value of $f$ can easily checked by putting those points in $f(x,y)$

My problem is that what will happen if $b< 8$ ?How should I deal with this case? I am totally stuck here. Please help me. Thnx again.

Answer 1

Compute the various distances, You will find that when $b\gt 8$ there are two nearest points, symmetric about the $y$-axis. When $b\le 8$, the nearest point is the origin.

Answer 2

You have to be careful when solving the system $$2x(\lambda+1)=0,\quad y-b=8\lambda,\quad x^2-16y=0\ .\tag{1}$$ According to the first equation $x=0$ or $\lambda=-1$.

From $x=0$ we get $y=0$ and then $\lambda=-{b\over8}$. It follows that $(0,0)$ is a conditionally stationary point of $f$ in any case.

When $\lambda=-1$ we get $y=b-8$ and then $x=\pm 4\sqrt{b-8}$. It follows that when $b>8$ we get two more conditionally stationary points, namely $(\pm4\sqrt{b-8},b-8)$.

Therefore we have a candidate list consisting of one point when $b\leq8$ (which then gives the global minimum), and of three points when $b>8$. Which of these three points corresponds to the global minimum of $f$ on the parabola $x^2-16y=0$ has to be determined by a comparison of values; it does not come out of solving the equations $(1)$.