1. We are given

$x^2 = at^2 + 2bt + c$

where $a,b,c$ are constants.

We need to find how acceleration depends on $x$, i.e. determine $n$ if

$ext{acceleration} \propto x^{-n}.$

2. Differentiate the given equation with respect to time $t$:

$x^2 = at^2 + 2bt + c$

Differentiating,

$2x\frac{dx}{dt} = 2at + 2b$

$xv = at+b$

where $v = \dfrac{dx}{dt}$.

So,

$v = \frac{at+b}{x}.$

3. Differentiate again to get acceleration.

From

$xv = at+b$

differentiate with respect to $t$:

$\frac{d}{dt}(xv) = a$

Using product rule,

$v^2 + x\frac{dv}{dt} = a$

Since acceleration is

$\frac{dv}{dt} = \ddot x,$

we get

$x\ddot x = a - v^2$

So,

$\ddot x = \frac{a-v^2}{x}.$

4. Now find $v^2$ in terms of $x$.

From

$xv = at+b,$

square both sides:

$x^2v^2 = (at+b)^2 = a^2t^2 + 2abt + b^2.$

But from the given relation,

$x^2 = at^2 + 2bt + c.$

Multiply this by $a$:

$ax^2 = a^2t^2 + 2abt + ac.$

Thus,

$(at+b)^2 = ax^2 + b^2 - ac.$

Hence,

$x^2v^2 = ax^2 + b^2 - ac$

$v^2 = a + \frac{b^2-ac}{x^2}.$

5. Substitute into acceleration formula:

$\ddot x = \frac{a-v^2}{x}$

$\ddot x = \frac{a-\left(a + \dfrac{b^2-ac}{x^2}\right)}{x}$

$\ddot x = \frac{-\dfrac{b^2-ac}{x^2}}{x}$

$\ddot x = -\frac{b^2-ac}{x^3} = \frac{ac-b^2}{x^3}.$

So acceleration varies as

$\ddot x \propto x^{-3}.$

Therefore,

$n=3.$

6. Comparison with stored answer:

Stored correct answer = $3$.

Our derived answer also gives $n=3$.

So the stored answer is correct.