1. The problem asks to find and graph the orthogonal trajectories of the family of curves given by $$x^2 - y^2 = c_1$$. 2. The given family of curves satisfies the implicit equation $$x^2 - y^2 = c_1$$, where $$c_1$$ is a constant. 3. Differentiate implicitly with respect to $$x$$: $$2x - 2y \frac{dy}{dx} = 0$$ which gives $$\frac{dy}{dx} = \frac{x}{y}$$. 4. Orthogonal trajectories have slopes that are negative reciprocals of the original family, so $$\frac{dy}{dx}_{\perp} = -\frac{y}{x}$$. 5. Rewrite the differential equation for the orthogonal trajectories: $$\frac{dy}{dx} = -\frac{y}{x}$$ which can be rearranged as $$\frac{dy}{y} = -\frac{dx}{x}$$. 6. Integrate both sides: $$\int \frac{dy}{y} = -\int \frac{dx}{x}$$ which yields $$\ln|y| = -\ln|x| + \ln|c_2|$$ where $$c_2$$ is a constant of integration. 7. Simplify the logarithmic expression: $$\ln|y| + \ln|x| = \ln|c_2|$$ which implies $$\ln|xy| = \ln|c_2|$$ and therefore $$xy = c_2$$. 8. The orthogonal trajectories are the family of rectangular hyperbolas given by $$xy = c_2$$. 9. To graph the orthogonal trajectories similar to the circles and lines in problem 2, plot the curves $$xy = c_2$$ for various values of $$c_2$$. These hyperbolas intersect the original family $$x^2 - y^2 = c_1$$ orthogonally.