Find the orthogonal trajectory of the equation $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1,\lambda$ being the parameter of the family.
In order to find the orthogonal trajectory, my algorithm is comprised of $4$ steps:
First, consider the given equation and identify, the parameter (if not specified), and let the given equation of the family of curves, whose orthogonal trajectory is to be calculated be $f(x,y,c)=0,$
Next, we differentiate the equation, $f$ with respect to $x$,
We proceed to eliminate the parameter between these two equations, one formed by differentiating with respect to $x$ and the other is the given one,
In the equation so obtained, after eliminating the parameter $c$ (here) we replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$,
Finally, we solve the differential equation to obtain at the equation of the orthogonal trajectory of the given curve.
Using this method/algorithm, I tried to solve the problem, by first differentiating the given equation( w.r.t x). I got $\frac{2x}{a^2+\lambda}+\frac{2y\frac{dy}{dx}}{b^2+\lambda}=0.$
After this, as $\lambda$ is given as the parameter, I tried eliminating $\lambda$ between $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1,$ and $\frac{2x}{a^2+\lambda}+\frac{2y\frac{dy}{dx}}{b^2+\lambda}=0.$
However, this is where the problem started. I couldn't really eliminate $\lambda$ as things got really messy. Bashing things randomnly, I found the eliminant to be something which seems unreasonable. I found, that as a hint, it was given that for the elimination of the parameter $\lambda$ value of $\lambda$ should be calculated and that comes out as, $\lambda=-\frac{b^2x+a^2y\frac{dy}{dx}}{x+y\frac{dy}{dx}}$.
After using this, everything went down smoothly. But I want to know how did they arrive at the value of $\lambda=-\frac{b^2x+a^2y\frac{dy}{dx}}{x+y\frac{dy}{dx}}$. Till now, I only heard that eliminating parameters between two equations might be challenging, but I never experienced it, up until now.
Any help regarding this, will be appreciated.
You made a careless mistake when differentiating with respect to $x$. The 1 on the RHS should be a 0.
From there, it is straightforward to make $\lambda$ the subject, EG By cross-multiplying, we get a linear expression in $\lambda$.
Note: In your (incorrect) version, cross-multiplying gives us a quadratic in $\lambda$, of which we can find solutions using the quadratic formula.