Orthogonal Trajectories

Orthogonal trajectories are curves that intersect another set of curves at right angles. In order to find the orthogonal trajectories of a given curve, we need to follow a few steps:

Step 1: Find the differential equation of the given curve.

Step 2: Solve the differential equation to find the general solution.

Step 3: Find the slope of the given curve at any point.

Step 4: Find the negative reciprocal of the slope obtained in step 3. This will give us the slope of the orthogonal trajectories.

Step 5: Solve the differential equation of the orthogonal trajectories using the slope obtained in step 4.

Step 6: Find the general solution of the orthogonal trajectories.

Step 7: If necessary, apply initial conditions to find the particular solution.

Now let's apply these steps to find the orthogonal trajectories of the given curve (1-x)^2y^2-2ax=0.

Step 1: Find the differential equation of the given curve.

The given curve equation can be rearranged as follows:
y^2 = \(\frac{2ax}{(1-x)^2}\)

Taking the derivative of both sides with respect to x, we get:
2yy' = \(\frac{2a(1-x)^2 - 4ax(1-x)}{(1-x)^4}\)

Simplifying this further, we have:
yy' = \(\frac{a(1-x) - 2ax}{(1-x)^3}\)
yy' = \(\frac{a(1-3x)}{(1-x)^3}\) -- (Equation 1)

Step 2: Solve the differential equation to find the general solution.

We can rewrite Equation 1 as:
y' = \(\frac{a(1-3x)}{y(1-x)^3}\) -- (Equation 2)

Separating variables, we get:
\(\frac{dy}{dx}\) = \(\frac{a(1-3x)}{y(1-x)^3}\)

Cross multiplying and rearranging, we have:
y(1-x)^3 dy = a(1-3x) dx

Integrating both sides, we get:
\(\int y(1-x)^3 dy\) = a \(\int (1-3x) dx\)

Simplifying the integrals, we have:
\(\frac{y}{4}(1-x)^4\) = a \(\left(\frac{x^2}{2} - \frac{3x^3}{3}\right)\)

Simplifying further, we get:
y(1-x)^4 = 2a(x^2 - x^3) -- (Equation 3)

Step 3: Find the slope of the given curve at any point.

Taking the derivative of the given curve equation with respect to x, we get:
\(\frac{d}{dx}\)(1-x)^2y^2 - 2ax = 0

Applying the product rule and simplifying, we have:
2(1-x)y^2 - 2(1-x)^2yy' -