Is $x^2=4ay$ a function while $y^2=4ax$ is not? [closed]

I just want to know if $x$ is always the independent variable.

A function has three components:

• Domain $(X)$
• Codomain $(Y)$
• A rule that assigns each element of the domain, to exactly one element of the codomain $(f)$

This is typically written as $f : X\to Y.$

The symbol that stands for an arbitrary input, or the representing element of the domain, is called the independent variable. And similarly the dependent variable. It is a tradition to write $x$ for the independent variable and $y$ for the dependent variable.

Now, first, let me assume $a\neq 0.$ Otherwise you wouldn't have meaningful functions, but two equations. Also, I will assume that your domain and codomain are the set of real numbers. In your first example $$y=\dfrac{x^2}{4a}.$$ Using this formula for each real numbered value of $x,$ you can compute a value for $y.$ Hence, by definition, we have a function.

On the other hand, the second formula is $$y^2=4ax,$$ which produces a quadratic equation for $y$ as soon as we assign a value for $x.$ For example, lets set $x=a,$ then $y^2=4a^2$ which has two solutions $y=\pm 2a.$ This isn't unique; hence doesn't agree with the definition of a function.

Not only that, suppose $a$ is positive, then for $x=-1,$ we get $y^2=-4a$ which is impossible. This means that there is no value assigned to $-1.$ One can restrict the domain and codomain to non-negative real values to remedy these two problems. Then even the second formula defines an honest function.