Find shortest distance from the parabola $y=x^2-9$ to the origin. 
Find shortest distance from the parabola $y=x^2-9$ to the origin.

First, I find minima of $\sqrt{x^2+(x^2-9)^2}$, so use derivative and ...
Is have an easier way?
 A: Another approach: if $P=(x,y)$ is the point of minimum on the parabola, then line $OP$ must be perpendicular to the tangent at $P$. Hence:
$$2x=-{x\over y},\quad\text{that is}\quad y=-{1\over2}.$$
A: Here is an easier way: the shortest distance $r$ is taken at the minimum of 
$r^2 = y^2 + x^2 = y^2 + y + 9$ This gets minimal at $y = -0.5$ (take the derivative or write $r^2 = (y+ 0.5)^2 + 8.75 \ge 8.75 $),  so $r = \sqrt{y^2 + y + 9} \ge  \sqrt{8.75} \simeq 2.9580$.
A: $$x^2+(x^2-9)^2=(x^2-9)^2+x^2-9+\frac{1}{4}+\frac{35}{4}=\left(x^2-9+\frac{1}{2}\right)^2+\frac{35}{4}\geq\frac{35}{4}.$$
The equality occurs for $x^2=\frac{17}{2},$ which says that $\frac{\sqrt{35}}{2}$ is a minimal value.
A: The direct way is the easier way.
$$ d^2 = x^2+ (x^2-9)^2 = x^4 -17x^2+81 $$
Differentiating $d^2$ (since derivative point is same as with $d$) and removing common facor $2x$
$$ 2 x^2-17 =0,\, x_{min}=\sqrt{17/2}, y_{min}= -1/2, d_{min}=\sqrt{x_{min}^2+y_{min}^2} = \sqrt{35/4}$$
One more differentiation can verify minimum here.
A: The distance $d$ is minimized when $d^2$ is minimized. We have $d^2=(x^2-9)^2+x^2. $ Substituting $x^2=z$ we have $d^2=(z-9)^2+z=z^2-17z+81.......$  I will assume you can find the minimum value of a quadratic in $z$.
A: firstly, we have:
$$f(x)=x^2-9=(x+3)(x-3)$$
from this we can see that $f(-x)=f(x)$. At $x=3$ and $x=-3$ the distance from the origin is $3$ so $l_{min}\le3$ and this can be our upper limit. 
If you want the exact answer differentiation is the best way to do it. Note:
$$l=\sqrt{x^2+(x^2-9)^2}$$
$$l^2=x^2+(x^2-9)^2$$
$$2l\frac{dl}{dx}=2x+4x(x^2-9)$$
It is a bit of a shortcut for the differentiation
