Introduction:
In mathematics, orthogonal trajectories are the curves that intersect each other at right angles. When a family of curves is given, the orthogonal trajectories are the curves that are perpendicular to each member of the given family.

Given Curve:
The given curve is r=2a/(1+cosx).

Deriving the Equation:
To find the orthogonal trajectories of the given curve, we need to first derive the differential equation of the family of curves.
We can write the given curve in polar coordinates as
x = r cosθ
y = r sinθ
r = 2a/(1+cosx) = 2a/(1+cosθ)
Squaring both sides, we get
r^2 = 4a^2/(1+cosθ)^2
Substituting x and y in terms of r and θ, we get
x^2 + y^2 = r^2 = 4a^2/(1+cosθ)^2
Differentiating both sides with respect to θ, we get
2x(dx/dθ) + 2y(dy/dθ) = -8a^2sinθ/(1+cosθ)^3
Substituting y/x = dy/dx, we get
2(dx/dθ + y/x)dx/dθ = -8a^2sinθ/(1+cosθ)^3
Simplifying the equation, we get
dx/dθ + y/x = -4a^2sinθ/(x(1+cosθ)^3)

Deriving the Orthogonal Trajectories:
To find the orthogonal trajectories, we need to find the differential equation of the curves that intersect the given family of curves at right angles.
The slope of the curves in the given family is given by
dy/dx = (dy/dθ)/(dx/dθ) = (r sinθ)/(r cosθ) = tanθ
The slope of the orthogonal trajectories is given by
dy/dx = -1/tanθ = -cotθ
Therefore, the differential equation of the orthogonal trajectories is
dy/dx = -cotθ = -x/y
Substituting y/x = dy/dx, we get
y^2 = -x^2 + C
where C is the constant of integration.
Therefore, the orthogonal trajectories of the given curve r=2a/(1+cosx) are given by the equation y^2 = -x^2 + C.

Conclusion:
In summary, we derived the differential equation of the given family of curves and used it to find the differential equation of the orthogonal trajectories. We then solved the differential equation of the orthogonal trajectories to obtain the equation of the curves that intersect the given family of curves at right angles.