Question:

Step 1: Obtain the differential equation of the given family of curves. First, differentiate the given equation $y = cx^2$ with respect to $x$ to find $\frac{dy}{dx}$. Then, eliminate the arbitrary constant $c$ using the original equation to express $\frac{dy}{dx}$ solely in terms of $x$ and $y$. We have the family of curves: $$y = cx^2 \quad (*)$$ Differentiating with respect to $x$: $$\frac{dy}{dx} = 2cx$$ From equation $(*)$, we can express $c$ as $c = \frac{y}{x^2}$. Substitute this expression for $c$ into the differential equation: $$\frac{dy}{dx} = 2 \left(\frac{y}{x^2}\right)x$$ $$\frac{dy}{dx} = \frac{2y}{x}$$ Step 2: Formulate the differential equation for the orthogonal trajectories. For orthogonal trajectories, the slope $\frac{dy}{dx}$ of the original family is replaced by its negative reciprocal, which is $-\frac{dx}{dy}$ (or $\frac{-1}{dy/dx}$). Replacing $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ in the differential equation from Step 1: $$-\frac{dx}{dy} = \frac{2y}{x}$$ Rearranging this to express $\frac{dy}{dx}$ for the orthogonal trajectories: $$\frac{dy}{dx} = -\frac{x}{2y}$$ Step 3: Solve the differential equation for the orthogonal trajectories. Separate the variables and integrate both sides of the differential equation obtained in Step 2. We have the differential equation: $$\frac{dy}{dx} = -\frac{x}{2y}$$ Separate the variables: $$2y \, dy = -x \, dx$$ Integrate both sides: $$\int 2y \, dy = \int -x \, dx$$ $$y^2 = -\frac{x^2}{2} + C$$ where $C$ is the constant of integration. Multiply the entire equation by 2 to clear the fraction and rearrange the terms: $$2y^2 = -x^2 + 2C$$ $$x^2 + 2y^2 = 2C$$ Let $k = 2C$, which is another arbitrary constant. Thus, the family of orthogonal trajectories is: $$x^2 + 2y^2 = k$$ The final answer is $\boxed{x^2+2y^2=k}$.