I have to find orthogonal trajectory of $xy = 2$ hyperbola.
Differentiating wrt x:
$$x\dfrac{dy}{dx}+y = 0$$
$$\dfrac{dy}{dx} = \dfrac{-y}{x} \ \ \ \ \ … (1) \ \ \ \ \ \ \\=\dfrac{-2}{x^2}\ \ \ \ \ …(2)$$
Now we replace $\dfrac{dy}{dx}$ with $\dfrac{-dx}{dy}$ and solve.
The problem is that eqn (1) gives wrong curve but eqn (2) gives correct curve.
Edit:
From 1 and 2 here we get two different answers. Are they both correct?
Best Answer
I'm not sure what you mean by "give the correct ans". So I'll solve both equations:
(1) If $\frac{dy}{dx} = -\frac{y}{x}$, then the orthogonal trajectory satisfies $\frac{dy}{dx} = \frac{x}{y}$. So $$ x \,dx = y\, dy \implies \frac{1}{2} x^2 = \frac{1}{2} y^2 + K $$ for some constant $K$. Renaming the constant $k=2K$, we see the orthogonal trajectories have equation $x^2-y^2 = k$.
(2) If $\frac{dy}{dx} = -\frac{2}{x^2}$, then the orthogonal curve has equation $\frac{dy}{dx} = \frac{x^2}{2}$. Then $y = \frac{x^3}{6}+C$.
As long as $xy=2$ and $y = x^{3}/6 +C$ intersect at some point $(a,b)$, then their tangents are orthogonal. But we usually talk about orthogonal families of curves. For any point $(a,b)$ in the plane, there is a curve of the form $xy=c$ through it, and a curve of the form $x^2-y^2 = k$ through it, and those two curves have orthogonal tangents at $(a,b)$. This is the stronger result, and what I would call the expected answer.