Minimal area ellipse containing a given circumference 
Out of all the ellipses
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
that contain the following circumference
$$x^2 + y^2 = 2y$$
determine the one that has the minimal area.

So far I've done this (in the picture), but now I'm kinda lost on using Lagrange multipliers!


We want to minimize $f(a,b)=\pi ab$ subject to the constraint that the circle $x^2+y^2=2y$ lies inside the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Letting $x^2=2y-y^2$ the two figures will intersect for any $a,\,b$ satisfying
$$ \frac{2y-y^2}{a^2} + \frac{y^2}{b^2}=0 $$
The discriminant of this quadratic is
$$ 4b^4+4a^2b^2(a^2-b^2) $$
Since we wish for no intersection or at most two points of tangency we require that
$$ 4b^4+4a^2b^2(a^2-b^2) \le0$$
which simplifies to the requirement
$$ g(a,b)=b^2+a^4-a^2b^2\le0 \tag{1}$$
One may then apply 
$$\nabla f(a,b)=\lambda\nabla g(a,b)$$
together with $(1)$ to obtain the result that
$$ a=\frac{\sqrt{6}}{2}\quad b=\frac{3\sqrt{2}}{2}$$

