Find the sum of the maximum and minimum of the curvature of the ellipse: $9(x-1)^2 + y^2 = 9$.
Hint (Use the parametrization $x(t) = 1 + \cos(t)$)
Tried to use parametrization like that, but then get stuck trying to find the curvature function and max/minimizing it.
OK. So you used that, and got $$ 9(\cos^2 t) + y^2 = 9. $$ Then you divided by 9 to get $$ \cos^2 t + (\frac{y}{3})^2 = 1. $$ From this, you figured out that $y/3 = \sin(t)$, because $\cos^2t + \sin^2 t = 1$. Now you've got an explicit parameterization of your curve, namely
$$ x(t) = 1 + \cos(t)\\ y(t) = 3 \sin(t) $$
You have a formula for curvature that looks something like $$ \kappa = \frac{x'(t) y''(t) - y'(t) x''(t)}{\sqrt{(x'(t)^2 + y'(t)^2}^{3}} $$ (I might have the sign wrong), which becomes $$ \kappa(t) = \frac{3\sin^2(t) + 3\cos^2(t)}{\sqrt{\sin^2(t) + 9 \cos^2(t)}^{3}}\\ = \frac{3}{\sqrt{\sin^2(t) + 9 \cos^2(t)}^{3}} $$
Now what's the problem with finding the max and min curvature? The max happens when the denominator is small; the min happens when it's large. Those values are $$ \kappa_{max} = \frac{3}{\sqrt{1}^{3}} = 3 \\ \kappa_{min} = \frac{3}{\sqrt{9}^3} = 1/9. $$ The product is $1/3$.
And the sum is $\frac{28}{9}$.