Shortest distance from a point

Question

Find the shortest distance from the point $(0,b)$ to the parabola $y=x^2+8$. Express your answer in terms of $b$.

(Comment: If $b \le \frac{33}{4}$ then the answer is just $|b|$, so assume that $b \gt \frac{33}{4}$.)

I know this is an optimization problem and so far I have my problem set up like this: $d/dx^2=(x-0)^2+((x^2+8)-b)^2$

Am I correct? What would be the next steps? is this a Lagrange equation?

Answer 1

Geometric Solution:

If $r$ is the shortest distance from $B(0,b)$ to the parabola $P:=\big\{(x,y)\in\mathbb{R}^2\,|\,y=x^2+8\big\}$, then the circle $\Gamma$ centered at $B$ with radius $r$ is tangent to this parabola. Suppose that $A(p,q)$ is a tangential point between $\Gamma$ and $P$. Then, $q=p^2+8$ and the line $\ell$ passing through $A$ tangent to $P$ is given by the equation $y-q=2p\cdot(x-p)$, since the slope of $P$ at $A$ is $2p$.

Now, $\ell$ is also tangent to $\Gamma$, whence $AB \perp \ell$. If $p=0$, then $q=8$, and so $$r=\sqrt{(0-0)^2+(b-8)^2}=|b-8|\,.$$ If $p\neq 0$, the slope of $AB$ is $\frac{q-b}{p-0}=\frac{q-b}{p}$. Since $AB\perp \ell$, $2p\cdot \left(\frac{q-b}{p}\right)=-1$. Hence, $q-b=-\frac{1}{2}$, or $p^2+8-b=-\frac{1}{2}$. This equation has a solution $p\in\mathbb{R}$ if and only if $b\geq \frac{17}{2}$, and the solutions are then $p=\pm\sqrt{b-\frac{17}{2}}$ for such $b$. In this case, $$r=\sqrt{(p-0)^2+(q-b)^2}=\sqrt{\left(b-\frac{17}{2}\right)+\left(-\frac{1}{2}\right)^2}=\sqrt{b-\frac{33}{4}}\,.$$

Now, if $b<\frac{17}{2}$, the only possible value of $(p,q)$ is $(p,q)=(0,8)$, whence $r=|b-8|$. For $b \geq \frac{17}{2}$, we can easily seen that $|b-8|=\sqrt{\left(b-\frac{17}{2}\right)^2+\left(b-\frac{33}{4}\right)}\geq\sqrt{b-\frac{33}{4}}$, so we must have $(p,q)=\left(\pm\sqrt{b-\frac{17}{2}},b-\frac{1}{2}\right)$ and $r=\sqrt{b-\frac{33}{4}}$.

P.S.: Either your result is wrong, or you have made a typo when you posted your question,

Answer 2

Hint

The square of the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is effectively given by $$D^2=(x_2-x_1)^2+(y_2-y_1)^2$$ In your case $x_1=0$, $y_1=b$, $x_2=?$, $y_2=x_2^2+8$. So $$D^2=x_2^2+(x_2^2+8-b)^2$$ Minimizing the distance is the same as minimizing the square of the distance. So, you have to consider $$\frac{dD^2}{dx_2}=0$$ and get $x_2$ from it.

When you have the solution, compute $D^2$ accordingly and take the square root.

You do not need Lagrange multipliers since there is no constraint in this problem.

If the equation given by the derivative has no root, a shortest distance still exists even if it does not correspond to a free minimum. So, you problem has different answers depending on the value of $b$.