To find the vertical tangent line of a curve, why must the numerator of the derivative be non-zero

Question

Let's say we are trying to find a point on a curve whose tangent line is vertical.

Example:

Consider $3y^2-2x^5=10$. Using implicit differentiation, it can be shown that

$\frac{\displaystyle dy}{\displaystyle dx}= \frac{\displaystyle 5x^4}{\displaystyle 3y}$

My textbook says in order to find a point whose tangent line is vertical, we need to solve the following system of equations:

$3y^2-2x^5=10$

$3y=0$

$5x^4 \neq 0$

which yields $x=-5^{1/5}$ and $y=0$.

Here's my problem:

Why does the numerator of the derivative(i.e $\space 5x^4$) need to be zero? In this case, if $3y=0$, then $5x^4$ isn't equal to zero but what if it were? Would the tangent line be vertical? Would it have a real value? I can't think of a curve whose derivative is $\frac{\displaystyle 0}{\displaystyle 0}$ at some point(and the curve is defined at that point).

Thanks in advance for any help!

Answer 1

An illustration with

$$3y^2-2x^5=a$$ for various $a$, including $0$.