# Conflicting results in finding orthogonal curves

I'm told to find the set of orthogonal curves to the curve of equation $$y=cx^2$$. Using implicit differentiation, $$\frac{dy}{dx}=2cx$$. Then, the desired curves obey the differential equation $$\frac{dy}{dx}=-\frac{1}{2cx}$$. Integrating, we find the curves $$y=-\frac{1}{2c}\log{|x|}+K$$, for some real constant $$K$$. The thing is in some solution I found, they rewrite $$2cx$$ as $$\frac{2y}{x}$$, because $$c=\frac{y}{x^2}$$, and in doing so they get a different differential equation with different curves $$\left(y=\pm \sqrt{K-\frac{1}{2}x^2}\right)$$. Which are clearly different curves. Plotting them on geogebra makes it seem like mine is wrong. What error did I make?

• @AmanKushwaha but shouldn't c be a free parameter? Commented Jan 29, 2022 at 16:49
• The procedure you are following to obtain the orthogonal trajectories of given family of parabolas requires you to eliminate $c$ after implicit differentiation with respect to $x$ which will give the differential equation of the family. That differential equation can not have a free parameter $c$. Solving $\frac{dy}{dx}=2cx$ will give you $y=cx^2 +c_1$ instead of $y=cx^2$ where as after eliminating $c$ you'd get $\frac{dy}{dx}=\frac{2y}{x}$ solving which will give you the family of parabolas Commented Jan 29, 2022 at 17:18

Let me expand on Aman Kushwaha's comment. The central issue here is exactly what problem you are solving. The way you stated it,

find the set of orthogonal curves to the curve of equation $$y=cx^2$$.

$$c$$ would be a fixed constant value, and you looking for a family of curves orthogonal to that single parabola. Finding all such curves is effectively impossible, because away from the parabola there is no requirement at all on the behavior of the orthogonal curves. They can twist and contort in uncountably infinitely many ways.

Your solution failed to find all such curves because you applied your differential equation $$\frac{dy}{dx} = \frac{-1}{2cx}$$ not just at the points $$(x, cx^2)$$ along this fixed parabola, but over the entire plane. As such you got the family of curves that are not just orthogonal to $$y=cx^2$$, but to every solution of $$\frac{dy}{dx} = 2cx$$. That is, you found the family of curves that are orthogonal to the entire family $$\{x \mapsto (x,cx^2 + c_1) \mid c_1 \in \Bbb R\}$$.

Evidently the problem you are intended to solve is

Find the family of curves orthogonal to $$y = cx^2$$ for all $$c \in \Bbb R$$.

Your equation $$\frac{dy}{dx} = \frac{-1}{2cx}$$ gives a curve that is orthogonal to one of those curves - the curve for the value of $$c$$ appearing in that equation. But it does not require the curve to be orthogonal to $$y = c'x^2$$ for any $$c' \ne c$$.

But when the differential equation is modified to remove $$c$$, we get an equation $$\frac{dy}{dx} = \frac{2y}x$$ whose solutions are exactly the family $$\{x \mapsto (x,cx^2) \mid c \in \Bbb R\}$$. And the orthogonal equation $$\frac{dy}{dx} = -\frac x{2y}$$ gives the set of curves that are orthogonal to all of the family, not just one.

• I should add that when I said the solutions to $\frac{dy}{dx} = \frac {2y}x$ are exactly the family $\{x \mapsto (x,cx^2) \mid c \in \Bbb R\}$, I was not just assuming it was true because that is what I was aiming for. Rather, I checked. It is very easy when modifying equations like this to introduce new solutions, or occasionally disallow an existing solution. Checking is essential. Commented Feb 5, 2022 at 0:21