Maximum/Minimum of Curvature - Ellipse 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. 
 A: 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 $\frac{28}{9}$. 
A: $ (x-1)^2 + ( y/3)^2 = 9$.
Major , minor axes are a, b
Sum of curvatures ( obtained by differentiation)
$$ \dfrac{b}{a^2} +\dfrac{a}{b^2},\, a= 1, b=3    $$
Plug in.
