Form the differential equation representing the family of curves given by $(x-a)^{2}+2 y^{2}=a^{2},$ where $a$ is an arbitrary constant.

(D) Given equation: $(x-a)^{2}+2 y^{2}=a^{2}$
Expanding the equation: $x^{2}-2ax+a^{2}+2y^{2}=a^{2}$
Simplifying: $x^{2}-2ax+2y^{2}=0$
$2ax = x^{2}+2y^{2}$
$a = \frac{x^{2}+2y^{2}}{2x}$
Now,differentiate the equation $x^{2}-2ax+2y^{2}=0$ with respect to $x$:
$2x - 2a + 4y \frac{dy}{dx} = 0$
$x - a + 2y \frac{dy}{dx} = 0$
Substitute the value of $a$:
$x - \frac{x^{2}+2y^{2}}{2x} + 2y \frac{dy}{dx} = 0$
Multiply by $2x$:
$2x^{2} - (x^{2}+2y^{2}) + 4xy \frac{dy}{dx} = 0$
$2x^{2} - x^{2} - 2y^{2} + 4xy \frac{dy}{dx} = 0$
$x^{2} - 2y^{2} + 4xy \frac{dy}{dx} = 0$
$4xy \frac{dy}{dx} = 2y^{2} - x^{2}$
$\frac{dy}{dx} = \frac{2y^{2}-x^{2}}{4xy}$