Find the curvature of the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ at the point $P=(a,0)$.
Let $x=t$ and $y=f(t)=\pm\frac{b}{a}\sqrt{a^2-t^2}$
I want to find the curvature, $\kappa$, of the ellipse at $P$. I already found that $$\kappa=\frac{|f''(t)|}{[1+(f'(t))^2]^{3/2}}$$ and then it should be $$\frac{ba^4}{[(bt)^2-(at)^2+a^4]^{3/2}}$$ But the first derivative is $$f'(t) = \mp\frac{b}{a}t\frac{1}{\sqrt{a^2-t^2}}$$ and the second derivative is $$f''(t)=\mp\frac{b}{a}\frac{1}{\sqrt{a^2-t^2}}\mp\frac{b}{a}t^2\frac{1}{(a^2-t^2)^{3/2}}$$
If I put $t=a$ into the equation, $0$ will be in the denominator! Why?
The problem here is that neither ${dy\over dx}$ nor ${d^2y\over dx^2}$ is defined at $t=\pm a$: the tangents to the ellipse are vertical there. The expression that you have for the curvature is then also undefined at these points. Sometimes you can get lucky and these “infinities” cancel, but this isn’t one of those cases. You also haven’t defined the values of $t$ for which you use the positive or negative square root in your parameterization. The expression $\pm\frac ba\sqrt{a^2-t^2}$ is not a well-defined function of $t$. Fixing that isn’t going to make the bigger problem of undefined derivatives go away, though.
To get an expression for the curvature that works everywhere, you need to choose a parameterization for which the derivatives are defined everywhere and don’t vanish simultaneously. You can then use the formula $$\kappa = {x'y''-x''y'\over(x'^2+y'^2)^{3/2}}.$$ Try the standard parameterization $(a\cos t,b\sin t)$.