# vertices of a hyperbola the silliest question ever

I'm given that the center of the hyperbola is $(2,1)$ and $a=3$ and asked to find the vertices. Since vertices are on the same line with the axis of symmetry I thought the coordinates should be $(2,1 \pm3)$ which is $(2,4)$ and $(2,-2)$ but for some reason the system gives me wrong answer, am i doing this wrong? pls help

The equation of the standard form is $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1.$

• Are you using a certain equation for the hyperbola? Presumably either $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ or the one where $(y - k)^2$ comes first, but the order matters here, and it's not quite clear how you should be using $a$ (whether its square is with the $x$ term or the $y$ term) from what you've given. – pjs36 May 25 '15 at 3:25
• it's the one you mentioned, sorry – john May 25 '15 at 3:28

In the hyperbola $$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$ the vertices happen where the second term disappears. That is, where $y-k=0$ (file for later: $y=k$) and therefore: $$\frac{(x-h)^2}{a^2}=1$$ This amounts to requiring $$|x-h|=a\text{,}$$ so either $x=a-h$ or $x=a+h$.
All this is to say that you should do what you did to the $x$-coordinate instead of to the $y$-coordinate.