I understand that the standard form of a hyperbola is $$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2} = 1$$ and I also know that via algebra we can make this general form: $$a_0x^2+a_1y^2+a_2x+a_3y+a_4=0$$
In my experience, which is not a lot for hyperbolas, I've only encountered hyperbolas where $|a_0| = b^2$ and $|a_1|=a^2$.
i.e. $$9x^2 - 16y^2 = 144 \rightarrow \frac{x^2}{4^2} - \frac{y^2}{3^2} = 1$$
Is this true for all hyperbolas? My teacher has told me that it is not, but it seems intuitive because of the algebra we do to get the standard form. I just wanted to make sure I'm not making a dumb mistake.
This may not always be true, depending on interpretation. Consider the hyperbola $$36x^2-16y^2-576=0$$ $$\Rightarrow 9x^2-4y^2-144=0$$ While both of these equations are the same hyperbola, the second equation is simplified and so I assume that is the desired form. If we now look back to our first equation, we find that the standard form is $$\frac{x^2}{16}-\frac{y^2}{36}=1$$ We now notice that $|a_0|\neq b^2$ and $|a_1|\neq a^2$. This is a result of having factored our equation.
In general, it is always possible to write a hyperbola in standard form in general form with the property given above. We can show this simply by expanding the standard form: $$(b^2)x^2+(-a^2)y^2+(2hb^2)x+(-2ka^2)y+(h^2-k^2-a^2b^2)=0.$$
However, whether this equation is simplified depends heavily on the prime factorizations of $a$, $b$, $h$, and $k$. This means that even though you can always write an equation that fits your description, it may not always have co-prime integer coefficients which is usually desirable.