This is not homework question. I am writing a research paper and studying the behaviors of complete algebraic curves and I came across this questions and I am interested in it.
A nodal function is well know. Obviously, $p=(1, \sqrt{2})$ is a point on this curve. Assume that $C = \mathcal{V} (f)$ is the variety of the nodal function. Is there a way to find a nonzero vector $v= \langle v_x, v_y \rangle$ such that $f(p + tv) = f(1+ tv_x, \sqrt{2} + tv_y)$ has a double zero at 0? Is it right to say that any nonzero point on the curve is $v_x$ and $v_y$?
Or what is an explicit way to compute $v= \langle v_x, v_y \rangle$?
When I checked online it suggest to me that "double zero" will mean both $f(p+tv)$ and its derivative with respect to $t$ are zero when $t$ is zero. To find the nonzero vector $v=\langle v_x, v_y\rangle$ what I did was to plug in $p+tv$ into the nodal function $y^2-x^2(x+1)$ but it seems complicated and I got stuck.
$$f(t+tv_x, \sqrt{2} + tv_y) = (\sqrt{2} + tv_y)^2 - (t+tv_x)^2(t+tv_x +1) = (tv_y)^2 + 2 tv_y \sqrt{2} - (tv_x)^3 -4 (tx)^2 - 5tv_x $$