How to find tangents to curves at points with undefined derivatives

Question

I will explain my question with the help of an example. We need to find the tangent at origin to the curve $$x^3 + y^3 =3axy$$

The derivative at origin is $0/0$ or indeterminate, found after implicit differentiation. But the tangents exist (via Wolfram Alpha) and they are $x=y=0$.

1. If the derivative at the origin does not exist, how are we getting the tangents? At least $y=0$ has a determinate slope (0).
2. Also how should I find tangents to more general curves at points where the derivative doesn't exist? Is there a general method using differentiation?

3. My professor told me that as $x,y\to0$, $x^3 + y^3\ll3axy$ and hence the zeroes of the function will be approximately where the zeroes of $3axy$ are. Now I couldn't understand the next line that he said:

   Near the origin the curve will look like the solutions to $3axy$.

   What does he mean by this? Of course the solutions to $3axy=0$ are $x=0$ and $y=0$, which are the tangents, but the curve isn't like that.

Can anyone please explain me this? And is there a general method to find tangents at points where the derivative doesn't exist?

Answer 1

For algebraic curves, you can use the notion of tangent cone.

Consider the curve defined by $P(x, y) = 0$, where $P(x, y)$ is a polynomial. Write $$P(x, y) = P_m(x, y) + P_{m+1}(x, y) + \dotsb + P_{m+k}(x, y)$$ where each $P_i(x, y)$ is a polynomial of degree $i$, and $P_m(x, y) \neq 0$, i.e. $P_m(x, y)$ is the homogeneous component of $P(x, y)$ of the lowest degree. Then, the equation $P_m(x, y) = 0$ defines the tangent cone to the curve at the origin, and the line of equation $a x + b y = 0$ is tangent to the curve at the origin if and only if $a x + b y$ divides $P_m(x, y)$.

In your case, since $P(x, y) = 3axy - x^3 - y^3 = P_2(x, y) + P_3(x, y)$, the tangent cone is given by $3axy = 0$, and so the tangent lines have equations $x = 0$ and $y = 0$.