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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.

Thomas Finley
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1 Answers1

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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.

Calvin Lin
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  • I am sorry, but that was a silly typo! ( I have fixed it, now.) But I would want to know how are they hitting at that value of $\lambda$. If you consider showing it, it would really be very helpful. – Thomas Finley Jun 19 '23 at 03:28
  • What do you mean by "how they are hitting?" If you refer to getting that expression, then like I said, cross-multiply and combine $\lambda$ terms. If you're stuck, explain what you've tried. Ideally, go through your writeup and fix the error (and it's consequences). – Calvin Lin Jun 19 '23 at 03:30
  • precisely, I wanna know how are they making $\lambda $ as the subject ? – Thomas Finley Jun 19 '23 at 03:32
  • @ThomasFinley Cross multiply and combine coefficients for $\lambda$. Do you know how to do that? – Calvin Lin Jun 19 '23 at 03:32
  • I am sorry, but no, not exactly... – Thomas Finley Jun 19 '23 at 03:33
  • @ThomasFinley Multiply the corrected equation throughout by $(a^2 + \lambda)(b^2 + \lambda)$. What do you get? – Calvin Lin Jun 19 '23 at 03:35
  • We get $(b^2+\lambda)x+(a^2+\lambda)y\frac{dy}{dx}=0$ ? – Thomas Finley Jun 19 '23 at 03:36
  • Great. Now combine the coefficients for $\lambda$ ... In particular, (from their answer) you want the coefficient of $(x + y \frac{dy}{dx} )$. Do you see that visually already? – Calvin Lin Jun 19 '23 at 03:36
  • I think you mean, $\lambda(x+y\frac{dy}{dx})=-(b^2x+a^2y\frac{dy}{dx})$ ? – Thomas Finley Jun 19 '23 at 03:38
  • Yup, and now make $\lambda $ the subject by ..... – Calvin Lin Jun 19 '23 at 03:38
  • Ok, then $\lambda=-\frac{(b^2x+a^2y\frac{dy}{dx})}{(x+y\frac{dy}{dx})}$ ? – Thomas Finley Jun 19 '23 at 03:40
  • Yup, that's it. – Calvin Lin Jun 19 '23 at 03:40
  • Thanks, so much but I have just one more thing to ask, that is, you know that after eliminating $\lambda$ I got the expression $(x-y\frac{dx}{dy})(x+y\frac{dy}{dx})=a^2-b^2$ . But when I substituted $-\frac{dx}{dy}$ with $\frac{dy}{dx}$ (as in the 2nd last step mentioned in OP) I got the same equation, i e $(x-y\frac{dx}{dy})(x+y\frac{dy}{dx})=a^2-b^2.$ Do I need to solve this? Or can I directly claim that the family of curves is self-orthogonal ? If it's the later case then how ? The thing is, I know this algorithm I stated just as an "algorithm". Sorry, if my question looks foolish. – Thomas Finley Jun 19 '23 at 03:46
  • @ThomasFinley I did not get that expression. I believe you're making careless arithmetic errors, or I might be incorrect. Can you write it up carefully? – Calvin Lin Jun 19 '23 at 04:08
  • You were correct. In the book, the expression evaluated is wrong. It should be: If we substitute the value of $\lambda=-\frac{(b^2x+a^2y\frac{dy}{dx})}{(x+y\frac{dy}{dx})}$ in the given equation i.e $\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$ we get, $(x+\frac{dy}{dx})(x\frac{dy}{dx}-y)=(a^2-b^2)\frac{dy}{dx}$. Now, we substitute $\frac{-dx}{dy}$ in place of $\frac{dy}{dx}$ to get, $(x-\frac{dx}{dy})(x\frac{dx}{dy}+y)=(a^2-b^2)\frac{dx}{dy},$ but am stuck on how to solve this differential equation? – Thomas Finley Jun 19 '23 at 05:18