The differential equation for the family of curves $x^2-y^2-2 a y=0$, where $a$ is an arbitrary constant, is

$\left(x^2+y^2\right) \frac{d y}{d x}=2 x y$

The correct answer is Option (1) → $\left(x^2+y^2\right) \frac{d y}{d x}=2 x y$

Given: $x^2 - y^2 - 2ay = 0$

Differentiate w.r.t. $x$:

$2x - 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0$

$2x = 2(y + a) \frac{dy}{dx}$

From original equation:

$x^2 - y^2 = 2ay \Rightarrow a = \frac{x^2 - y^2}{2y}$

Substitute:

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

$2x = \frac{x^2 + y^2}{y} \frac{dy}{dx}$

$2xy = (x^2 + y^2) \frac{dy}{dx}$

Hence, Option (1) is correct.