Minimise the area of an origin-centred ellipse with the constraint that it most enclose a particular circle

Question

If the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ is to enclose the circle $x^2+y^2 = 2y$, what values of $a$ and $b$ minimise the area of the ellipse?

Answer 1

Hint.

Calling

$$ f(x,y) = \frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0\\ g(x,y) = x^2+y^2-2y=0 $$

at the tangency points we have

$$ \nabla f(x,y) = \lambda \nabla g(x,y) $$

or

$$ \frac{2x}{a^2}=\lambda 2x\\ \frac{2y}{b^2}=\lambda(2y-2) $$

so we obtain

$$ \lambda = \frac{1}{a^2}\Rightarrow \frac{y}{b^2}=\frac{1}{a^2}(y-1)\Rightarrow y = \frac{b^2}{b^2-a^2} $$

etc.

NOTE

This can be used to solve the minimization problem. Calling $ S = \pi a b$ the area to be minimized we have

$$ L(a,b,x,y,\lambda,\mu,\xi) = \pi a b + \lambda f(x,y) + \mu g(x,y)+ \xi \left(y-\frac{b^2}{b^2-a^2}\right) $$

now considering instead the minimization of $S^2$ which in this case is equivalent to the minimization of $S$ we have

$$ L(a^2,b^2,x,y,\lambda,\mu,\xi) = \pi^2a^2b^2+\lambda\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)+\mu\left(x^2+y^2-2y\right)+\xi\left(y- \frac{b^2}{b^2-a^2}\right) $$

now making $\bar a = a^2, \bar b = b^2$

$$ L(\bar a,\bar b,x,y,\lambda,\mu,\xi) = \pi^2\bar a\bar b+\lambda\left(\frac{x^2}{\bar a}+\frac{y^2}{\bar b}-1\right)+\mu\left(x^2+y^2-2y\right)+\xi\left(y- \frac{\bar b}{\bar b-\bar a}\right) $$

The stationary conditions give

$$ \left\{ \begin{array}{rcl} -\frac{\lambda x^2}{\bar a^2}+\bar b-\frac{\bar b \xi }{(\bar b-\bar a)^2}=0 \\ -\frac{\lambda y^2}{\bar b^2}+\bar a+\left(\frac{\bar b}{(\bar b-\bar a)^2}-\frac{1}{\bar b-\bar a}\right) \xi =0 \\ \frac{2 \lambda x}{\bar a}+2 \mu x=0 \\ \xi +\frac{2 \lambda y}{\bar b}+\mu (2 y-2)=0 \\ \frac{x^2}{\bar a}+\frac{y^2}{\bar b}-1=0 \\ x^2+y^2-2 y=0 \\ y-\frac{\bar b}{\bar b-\bar a}=0 \\ \end{array} \right. $$

and solving those equations we have the feasible solutions

$$ \left[ \begin{array}{cccccccc} \bar a & \bar b & x & y & \lambda& \mu & \xi & S^2\\ \frac{3}{2} & \frac{9}{2} & -\frac{\sqrt{3}}{2} & \frac{3}{2} & \frac{27}{2} & -9 & 0 & \frac{27 \pi ^2}{4} \\ \frac{3}{2} & \frac{9}{2} & \frac{\sqrt{3}}{2} & \frac{3}{2} & \frac{27}{2} & -9 & 0 & \frac{27 \pi ^2}{4} \\ \end{array} \right] $$

but $\bar a = a^2 \to a = \sqrt{\frac 32}, \bar b = b^2\to b = \frac{3}{\sqrt{2}}$

Answer 2

If the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is tangent to the circle $x^2+(y-1)^2=1$ we have that the discriminant of $q(y) = a^2\left(1-\frac{y^2}{b^2}\right)+y^2-2y $ is zero, hence $a^4+b^2 = a^2 b^2$. Assuming $a,b>0$ the area enclosed by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi a b$, hence we want to solve the minimization problem $$ \min_{a^4+b^2=a^2 b^2}\pi a b = \pi \sqrt{\min_{a^4=b^2(a^2-1)}a^2 b^2}=\pi\sqrt{\min_{a>1} \frac{a^6}{a^2-1}}.$$ The minimum area is $\frac{3\pi}{2}\sqrt{3}$ and it is is achieved by $a=\sqrt{\frac{3}{2}}$, $b=\frac{3}{\sqrt{2}}$:

The origin and the tangency points are the vertices of an equilateral triangle inscribed in the given unit circle.