To find the differential equation for a family of curves, the goal is to differentiate the general equation and then eliminate the arbitrary constant(s).

1. Differentiate the given equation for the family of parabolas:
The given equation is:
\[y = ax^2 \quad \cdots(1)\]
Differentiating both sides with respect to \(x\) gives the slope, \(y'\):
\[\frac{dy}{dx} = y' = 2ax \quad \cdots(2)\]
2. Eliminate the arbitrary constant 'a':
We can use the two equations to eliminate the constant \(a\). From equation (1), we can express \(a\) as:
\[a = \frac{y}{x^2}\]
Now, substitute this expression for \(a\) into equation (2):
\[y' = 2 \left( \frac{y}{x^2} \right) x\]
3. Simplify the resulting equation:
Simplifying the expression gives:
\[y' = \frac{2yx}{x^2}\]\[y' = \frac{2y}{x}\]
Rearranging this gives the correct differential equation for the family \(y=ax^2\):
\[xy' = 2y\]
4. Analysis of the Options:
The correctly derived differential equation, \(xy' = 2y\), does not match any of the provided options. There appears to be a significant typo in the question or the options. For instance, option (D) is \(y = y' \cdot x\), which is equivalent to \(xy' = y\).

It is highly probable that the question intended to ask for the differential equation for the family of **lines** passing through the origin, \(y = ax\). Let's quickly derive the DE for that family:
- Given: \(y = ax\)
- Differentiate: \(y' = a\)
- Substitute \(a=y'\) back into the original equation: \(y = (y')x\)
This derivation perfectly matches option (D). Therefore, we conclude the question mistakenly referred to parabolas instead of lines.

Result:
Assuming the question intended to describe the family of lines \(y=ax\), the corresponding differential equation is \(y = y' \cdot x\).