Vertices on the y axis don't exist though. Sub in x = 0 and you get y^2 = -b^2 which unless y = b = 0, is impossible, and b=0 can't be the case either.I don't think so- what about vertices on y axis? You have to deal with a quadratic for those
I'm guessing, either I'm wrong in doing the negative, or even if you can technically do a negative they won't care about it. Like, I won't pretend to know all the definitions in conics, so you could totally be right. And in that case doing the negative is wrong.
I did N>=0, so sin(theta)>=h/r cos(theta)
As written in the question, a can be negative.