Find the differential equation of the family of ellipses (x2)/(a2) + (y2)/(4) = 1

The equation is:

$$\frac{x^2}{a^2} + \frac{y^2}{4} = 1$$

Here, $a$ is an arbitrary constant.

Since there is one arbitrary constant ($a$), we differentiate the equation once with respect to $x$.

$$\frac{d}{dx}\left(\frac{x^2}{a^2}\right) + \frac{d}{dx}\left(\frac{y^2}{4}\right) = \frac{d}{dx}(1)$$ $$\frac{1}{a^2}(2x) + \frac{1}{4}(2y)\frac{dy}{dx} = 0$$ $$\frac{2x}{a^2} + \frac{y}{2}\frac{dy}{dx} = 0$$

$$\frac{2x}{a^2} = -\frac{y}{2}\frac{dy}{dx}$$ $$\frac{1}{a^2} = -\frac{y}{2x} \cdot \frac{1}{2}\frac{dy}{dx}$$ $$\frac{1}{a^2} = -\frac{y}{4x}\frac{dy}{dx}$$

Substitute $\frac{1}{a^2}$ into $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$:

$$x^2 \left(-\frac{y}{4x}\frac{dy}{dx}\right) + \frac{y^2}{4} = 1$$ $$-\frac{xy}{4}\frac{dy}{dx} + \frac{y^2}{4} = 1$$

Multiply the entire equation by 4 to eliminate the denominators:

$$-xy\frac{dy}{dx} + y^2 = 4$$

Rearrange the terms to isolate $xy\frac{dy}{dx}$:

$$xy\frac{dy}{dx} = y^2 - 4$$

Comparing this result with the given options:

1. $xy\frac{dy}{dx} = 4 - y^2$
2. $xy\frac{dy}{dx} = 4 - x^2$
3. $xy\frac{dy}{dx} = x^2 - 4$
4. $xy\frac{dy}{dx} = y^2 - 4$

The derived differential equation matches option 4.

The final answer is $\text{4. } xy\frac{dy}{dx} = y^2 - 4$.