I know how to calculate the orthogonal trajectory of a given family of curves. And it is said in my text book that if the orthogonal trajectory of a curve is itself, then we say that the family of curves is self orthogonal. As an example, orthogonal trajectory of parabola $$y^2=4a\left(x+a\right)$$ Is the same function. But how this is possible? Orhogonal trajectory cut every member of given family normally. I cannot get an idea of self orthogonality. I tried to graph the above equation with suitable $a$ . But I cannot agree with the fact that the same function is its orthogonal trajectory. I try an algebraic approach. If $m$ is the slope of such function at a point $x$ , then $\frac{-1}{m}$ is also its slope at $x$ . This gives, $$m^2=-1\\
\implies {m=i}$$
thus, the slope I get is a complex value for such functions. But I cannot agree with it. Where is my mistake? Can anyone help me?
[Math] the logical idea of self orthogonality
ordinary differential equationsorthogonality
Best Answer
Note that if $m$ is the slope of a function in the given family, than $\frac{-1}{m}$ is the slope, at the same $x$, of another function of the same family ( this is your mistake, I suppose).
See the figure:
where this is illustrated for the values $a=1$ and $a=-1$ for the functions of your family ( the semiplane $y<0$ is symmetric). The two functions are: $$ a=1 \rightarrow y^2=4(x+1) \rightarrow y=2\sqrt{x+1} $$ $$ a=-1 \rightarrow y^2=4(1-x) \rightarrow y=2\sqrt{1-x} $$ that have orthogonal tangents at $x=0$.
For an analitic approach:
you can see that $m=\frac{a}{\sqrt{a(x+a)}}$ so the condition of othogonality for two curves of the family with parameters $a$ and $b$ is: $$ \frac{a}{\sqrt{a(x+a)}}=-\frac{\sqrt{b(x+b)}}{b} $$ squaring, with the condition $ab<0$, this becomes: $$ x^2+(a+b)x=0 $$ and this means that if the two curves have a common point at $x=0$, at this point they are orthogonal, or they can be orthogonal at $x=-(a+b)$.