Let $f(x) = c(x-h)^2 + d$ where $c,h,d$ are constants. The condition gives that if $f(a) = f(b)$, $a+b = 2h$ and it stays consistent for some constant $h$. As $f(a) = f(b) (a \neq b)$ implies $f(a^2-6b-1) = f(b^2 + 8)$, we can split this into two cases. Case 1: $a^2-6b-1 = b^2 + 8$. Since $a+b = 2h$, we can let $b = 2h-a$. Plugging in, we realize that $2a(2h+3) = (2h+3)^2$ for any arbitrary $a$. Therefore, $h$ must be $-\frac{2}{3}$. Plugging $h$ into our definition of $f(x)$, the answer cancels out extra constants $c$ and $d$, so we get $13$. Case 2: $a^2-6b-1 + b^2 + 8 = 2h$. Doing the same, we plug in $b = 2h-a$. Writing our equation based on $a$ where $h$ is a constant, we realize that this can't hold for any arbitrary value of $a$ so there is a contradiction.

frazer wrote: Let $f(x) = c(x-h)^2 + d$ where $c,h,d$ are constants. The condition gives that if $f(a) = f(b)$, $a+b = 2h$ and it stays consistent for some constant $h$. As $f(a) = f(b) (a \neq b)$ implies $f(a^2-6b-1) = f(b^2 + 8)$, we can split this into two cases. Case 1: $a^2-6b-1 = b^2 + 8$. Since $a+b = 2h$, we can let $b = 2h-a$. Plugging in, we realize that $2a(2h+3) = (2h+3)^2$ for any arbitrary $a$. Therefore, $h$ must be $-\frac{2}{3}$. Plugging $h$ into our definition of $f(x)$, the answer cancels out extra constants $c$ and $d$, so we get $13$. Case 2: $a^2-6b-1 + b^2 + 8 = 2h$. Doing the same, we plug in $b = 2h-a$. Writing our equation based on $a$ where $h$ is a constant, we realize that this can't hold for any arbitrary value of $a$ so there is a contradiction. $h$ is supposed to be $-\frac{3}{2}$.

In general $f(x)$ can be written $k(x-p)^2+q$ where $k \neq 0$. The property can be summarized to be \[ a+b = 2p \implies a^2-6b-1=b^2+8 \vee a^2+b^2-6b+7=2p \]which can be rewritten as \[ \forall a ((2 p+3) (2 a-2 p-3)(-4 a p+2 a (a+3)+4 p^2-14 p+7)=0) \]Therefore we have $p = -\frac{3}{2}$ which implies $\frac{f(8)-f(2)}{f(2)-f(1)}=13$.