circles - Maximize the area of an ellipse inscribed in a semicircle.

An ellipse inscribed in semi-circle touches the circular arc at two distinct points and also touches the bounding diameter its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area, find its eccentricity.

I tried to approach this problem using coordinate geometry and tried to maximise the area of ellipse after construction of the area function using derivatives. The area function comes out to be $\displaystyle \frac{\pi a^2 \mathrm R}{\sqrt{\mathrm {R^2}+a^2}}$. Here $\mathrm R$ is the radius of semicircle and $a$ is semi-major axis of ellipse. But its derivative came out to be positive i.e. area will be maximised when $a$ is maximum. Here I am stuck since I can't find the maximum value of $a$ in terms of radius $\mathrm R$. Please help me with this.

Thanks!

Answer

If an ellipse with centre $(0,b>0)$ is tangent to the $x$-axis at the origin then its equation is given by $$\frac{x^2}{a^2}+\frac{(y-b)^2}{b^2} = 1$$ and if such ellipse is additionally tangent to the circle $x^2+y^2=1$, then the discriminant of the quadratic polynomial $\frac{1-y^2}{a^2}+\frac{(y-b)^2}{b^2}-1$ equals zero, hence $b^2=a^2-a^4$ and the area enclosed by the ellipse equals $\pi a b = \pi a^2\sqrt{1-a^2}$, which by the AM-GM (or Cauchy-Schwarz) inequality attains its maximum at $a=\sqrt{\frac{2}{3}}$, $b=\frac{\sqrt{2}}{3}$. It follows that the eccentricity of the solution (depicted below) is given by $$e = \sqrt{1-\left(\tfrac{b}{a}\right)^2}=\sqrt{\tfrac{2}{3}}.$$ $\hspace{1in}$

In particular the largest ellipse inscribed in a half-circle approximately covers the $77\%$ of the area of the half-circle.