# How does equating to zero the coefficients of lowest degree terms in x and y give the tangent at origin to a polynomial curve?

My course book on differential calculus says, "... the equation of the tangent at the origin can be written down by equating to zero the lowest degree terms in x and y." I don't see how this is true.

Why is this true ?

• '' the lowest degree terms'' of what? Jul 12, 2017 at 15:11
• Sorry. I've edited the question now. Jul 12, 2017 at 15:34
• A term to search for is "tangent cone". Jul 12, 2017 at 16:10

Let the curve be $$f(x)=g(y)$$

Assuming that $$f(\cdot), g(\cdot)$$ are polynomials, the respective constant terms must be zero such that the curve passes through the origin.

Differentiating w.r.t $$x$$ gives $$f'(x)+g'(y)\frac {dy}{dx}=0\\\frac {dy}{dx}=-\frac {f'(x)}{g'(y)}$$ Hence slope at origin is $$-\dfrac{f'(0)}{g'(0)}$$ and tangent at origin is $$y=-\frac {f'(0)}{g'(0)}x\\ \color{red}{f'(0)\ x+g'(0)\ y=0}$$

NB:

If $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1 x+a_0$$, then $$f'(0)\ x = a_1 x$$ which is the lowest term in $$x$$ in $$f(x)$$. Similarly for $$g(y)$$.

"...by equating to zero the sum of the lowest degree terms in $$x$$ and $$y$$."
Note that, assuming $$g'(0)\neq 0$$, if $$f(x)$$ does not have an $$x$$ term then the tangent at the origin must have a slope of zero i.e. the $$x$$-axis is the tangent at the origin.