Finding the shortest distance between two Parabolas Recently, a problem asked me to find the minimum distance between the parabolas $y=x^2$ and $y=-x^2-16x-65$.
I proceeded with the problem as thus. 
Let $P(a,a^2), Q(b, -b^2-16b-65), a-b=x$.
$\therefore PQ^2=x^2+(2a^2+2ax+16a+x^2+16x+65)^2$.
$PQ^2=x^2+(2(a+\frac{x+8}{2})^2+\frac{(x+8)^2+2}{2})^2 \ge (x^2+(\frac{(x+8)^2+2}{2})^2)(1+\frac{1}{4}) \times \frac{4}{5}$
Applying Cauchy gives us that 
$PQ^2 \ge (\frac{1}{4}x^2+3x+\frac{33}{2})^2 \times \frac{4}{5} \ge (\frac{15}{2})^2 \times \frac{4}{5}=75$
This implies that the answer is $\sqrt{75}$. 
However, it took me a long time to find the values for Cauchy, and the calculations proved tedious. 
What are other approaches to this problem?
EDIT: $(\frac{15}{2})^2 \times \frac{4}{5} \neq 75$, it`s $45$ actually! 
 A: $$y_1 = x_1^2, \qquad y_2 = -x_2^2 - 16x_2 -65$$
$$d = \sqrt{(x_2-x_1)^2 + (-x_2^2 - 16x_2 -65-x_1^2)^2}$$
Thus we want to minimize the following: (Note the change of variables, simply to make the notation easier to follow)
$$f(u,v) = (u-v)^2 + (-u^2 - 16u -65-v^2)^2$$
$$f_u(u,v) = 4u^3 + 96u^2 + u(4v^2 + 774) + 32v^2 - 2v + 2080$$
$$f_v(u,v) = 262 v + 4 u^2 v + 4 v^3 + u (-2 + 64 v)$$
Solving this, we get the real solution $u=-7, v = -1$ (it's quite tedious here, so I ended up using Wolfram Alpha). Thus our minimum distance is
$$\sqrt{((-7)-(-1))^2 + (-(-7)^2 - 16(-7) -65-(-1)^2)^2} = \color{red}{\sqrt{45}} \approx \color{red}{6.7082039325}$$   
Addendum 
While my answer differs from yours, this Desmos demonstration I created seems to numerically support my answer, as long as I interpreted the question correctly; feel free to mess around with it. I have attached a screenshot from the program showing the minimum distance value. Also note the green line in the image represent the normal line to the graphs at the corresponding points, and both points share a tangent line with a slope of $-2$: this is to be expected, and should hold for a number of problems of this type. Accordingly, this could provide another way to solve this problem with fewer steps and calculations than my solution. Nevertheless, I leave this work up to you!

A: You could define a new function as the distance between a point on one parabola and a point on the other. 
We know a generic point for the first parabola is given by $(x,x^2)$ and a generic point for the second parabola is given by $(y,-y^2-16y-65)$
We want to minimize
$$min{\sqrt{(y-x)^2+(-y^2-16y-65-x^2)^2}}$$
Note that this is the same as minimizing
$$min{(y-x)^2+(-y^2-16y-65-x^2)^2}$$
This is another approach.
A: Suppose the closest points are $(a,a^2)$ and $(b,-b^2-16b-65)$.
As @brevan-ellefsen noted, the slopes of these parabolas are equal at these points, so $2a=-2b-16$, or $b=-a-8$.
The closest points are therefore $(a,a^2)$ and $(-a-8,-(-a-8)^2-16(-a-8)-65)$ for some value of $a$. The squared distance between these points is $$s(a)=(2a+8)^2+(a^2+(a+8)^2-16(a+8)+65)^2=65 + 32 a + 8 a^2 + 4 a^4.$$ This is smallest when $s'(a)=16(a^3+a+2)=(a+1)(a^2-a+2)=0$, or when $a=-1$. When $a=-1$, $s(a)=65 -32 + 8 + 4=45$, so the minimum distance between the parabolas is $\sqrt{45}$.
