Find the orthogonal trajectories of the family of curves $y^2=4ax$.

Solution

1. Given the family of curves $y^2=4ax$, differentiate with respect to $x$ to get $2y\frac{dy}{dx}=4a$.
2. Substitute $a=\frac{y}{2}\frac{dy}{dx}$ into the original equation to obtain $y^2=2xy\frac{dy}{dx}$, which simplifies to the differential equation $\frac{dy}{dx}=\frac{y}{2x}$.
3. The orthogonal trajectories satisfy the negative reciprocal differential equation $\frac{dy}{dx}=-\frac{2x}{y}$.
4. Separate the variables to get $ydy=-2xdx$ and integrate both sides to obtain $\frac{y^2}{2}=-x^2+C$.
5. Rearrange the integrated equation to the form $2x^2+y^2=C$.
6. Hence the answer is (A).