Problem Statement
Find the orthogonal trajectories of the family of parabolas $y^2 = 4ax$, where $a$ is an arbitrary parameter.
Verified Solution & Marking Scheme
Differentiate given family and eliminate parameter a
Given: $y^2 = 4ax \implies a = \frac{y^2}{4x}$.
Differentiating with respect to $x$:
$$2y \frac{dy}{dx} = 4a$$
Substitute $4a = \frac{y^2}{x}$:
$$2y \frac{dy}{dx} = \frac{y^2}{x} \implies \frac{dy}{dx} = \frac{y}{2x}$$
Replace dy/dx with -dx/dy for orthogonal family
For orthogonal trajectories, replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$:
$$-\frac{dx}{dy} = \frac{y}{2x} \implies 2x \, dx = -y \, dy$$
Integrate to find orthogonal trajectories
$$\int 2x \, dx = -\int y \, dy$$
$$x^2 = -\frac{y^2}{2} + C \implies 2x^2 + y^2 = 2C = K$$
This represents a family of concentric ellipses centered at the origin.