Statement-1: The slope of the tangent at any point P on a parabola, whose axis is the axis of x and vertex is at the origin, is inversely proportional to the ordinate of the point P.
Statement-2: The system of parabolas $$y^2 = 4ax$$ satisfies a differential equation of degree 1 and order 1.

First, consider Statement-1: The slope of the tangent at any point P on a parabola with its axis along the x-axis and vertex at the origin is inversely proportional to the ordinate (y-coordinate) of point P.

The standard equation of such a parabola is $$ y^2 = 4ax $$, where $$ a $$ is a constant. To find the slope of the tangent at any point $$ (x, y) $$ on this parabola, differentiate the equation implicitly with respect to $$ x $$.

Differentiate both sides of $$ y^2 = 4ax $$:

$$ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) $$

$$ 2y \frac{dy}{dx} = 4a $$

Solve for $$ \frac{dy}{dx} $$:

$$ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} $$

So, the slope $$ m = \frac{2a}{y} $$. Since $$ a $$ is a constant, $$ m $$ is inversely proportional to $$ y $$, the ordinate. Therefore, Statement-1 is true.

Now, consider Statement-2: The system of parabolas $$ y^2 = 4ax $$ satisfies a differential equation of degree 1 and order 1.

The equation $$ y^2 = 4ax $$ represents a family of parabolas with parameter $$ a $$. To form a differential equation, eliminate the arbitrary constant $$ a $$. Differentiate both sides of $$ y^2 = 4ax $$ with respect to $$ x $$:

$$ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) $$

$$ 2y \frac{dy}{dx} = 4a $$

Solve for $$ a $$:

$$ a = \frac{2y \frac{dy}{dx}}{4} = \frac{y \frac{dy}{dx}}{2} $$

Substitute this expression for $$ a $$ back into the original equation $$ y^2 = 4ax $$:

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

Simplify:

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

Assuming $$ y \neq 0 $$ (valid for points not on the x-axis), divide both sides by $$ y $$:

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

Rearrange to standard form:

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

The highest derivative is $$ \frac{dy}{dx} $$, which is of order 1. The exponent of this derivative is 1, so the degree is 1. Thus, the differential equation is of order 1 and degree 1. Therefore, Statement-2 is true.

Now, check if Statement-2 correctly explains Statement-1. The differential equation $$ 2x \frac{dy}{dx} - y = 0 $$ gives $$ \frac{dy}{dx} = \frac{y}{2x} $$. Statement-1 requires the slope to be inversely proportional to $$ y $$, but this expression depends on both $$ x $$ and $$ y $$. To relate it to Statement-1, substitute $$ x = \frac{y^2}{4a} $$ from the parabola equation:

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

This shows the slope is inversely proportional to $$ y $$, but the substitution relies on the specific parabola equation, not solely on the differential equation from Statement-2. The differential equation describes the family and relates slope to both $$ x $$ and $$ y $$, but without the original equation, it does not directly show inverse proportionality to $$ y $$ alone. Thus, Statement-2 is true but does not directly explain Statement-1.

Hence, the correct answer is Option B.