Can the Second Plane Autonomous System Have a Unique Critical Point?

[TheMercury79]

TL;DR

Can we still show a unique critical point if we change one variable?

The plane autonomous system $$x' = -ax + xy\\y' = 1 - by - x^2$$ has a unique critical point when $ab>1$

Since if we set x' = 0, then either x=0 or y=a

If y=a, then for y' = 0: $1 - ab - x^2= 0$ , but if $ab>1$ this leads to $x^2<0$ which has no solutions and thus $y\ne a$

The only option left is to let x=0 so that $y' = 1-by = 0$, so $y=1/b$

Therefore $(0 , 1/b)$ is a unique critical point of the given plane autonomous system.

This is pretty straightforward, but consider instead a similar system with a slight difference: $$x' = -ax + xy\\y' = 1 - bx - x^2$$
How can you do the same process as above and show that this system has a unique critical point when $ab>1$?

We had the second one on a test the other day and I couldn't get anywhere with it. I thought I recognized the question because the first
system is from the text-book and I suspect that the teacher intended to use the first system but mispelled y for x.

Though, I am reluctant to say anything to the teacher because I don't want to look silly, he could have switched the y and x intentionally
and it's just me that can't show what the unique critical point is. I tried and tried but couldn't show anything.

So, is it me or the teacher? I want to be sure before I point this out to him. Can it be shown that the second system has a unique critical point?

 

[TheMercury79]

After further inspection I still come up with nothing. I found that if y=a, then $bx+x^2=1$, but since this is a quadratic, it yields two possible values for x in terms of b, so there can't be a unique point for the second system. And how to use it with ab>1 is a mystery to me since it seems ab>1 doesn't even enter the picture for the second system.
Also there was a follow up question asking us to show that this unique point is always stable for b>0.
He must have mispelled the question