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!
On
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
$$ a=\frac{\sqrt{6}}{2}\quad b=\frac{3\sqrt{2}}{2}$$
The objective is simple:
minimize $ab$
The contraint is tricker.
By inspection we can guess
$b>2, b>a, a>1$
If the circle is inside the ellipse, the two curves do not intersect, and this system has:
$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\\ x^2 +y^2 = 2y$
has no real solutions.
So, we try to find solutions, apply the quadratic formula, and then set the discriminant $\le 0.$
$x^2 + \frac {a^2}{b^2}y^2 = a^2\\ x^2 +y^2 = 2y\\ (1-\frac {a^2}{b^2})y^2 - 2y + a^2 = 0\\ y = 1 \pm \sqrt{1 - a^2(1-\frac {a^2}{b^2})}$
$1 - a^2(1-\frac {a^2}{b^2})\le 0$
And, the minimal ellipse will be when the two curves are tangent.
$a^4 + b^2 - a^2b^2 = 0$
$f(a,b,\lambda) = ab + \lambda(a^4 + b^2 - a^2b^2)\\ \nabla f = 0$
$b + 4a^3\lambda + 2ab^2\lambda= 0\\ a + 2b\lambda + 2a^2b\lambda=0\\ b^2 = 2a^4$
$a^4 +2a^4 - 2a^6 = 0\\ a^4(3 -2a^2)= 0\\ a^2= \frac 32\\ b^2 = \frac {9}{2}$
$ab = \frac {3\sqrt 3}{2}$
$\pi ab = \frac {3\sqrt 3 \pi}{2}$