<s>[INST] 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! 
 [/INST] 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 $\sqrt{45}$.
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