Maximizing Area In An Ellipse

Auxiliary circle is a circle concentric with the ellipse and having a radius equal to the semi-major axis of the ellipse.

A point on the ellipse,
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
can be given with a parameter, $\theta$ , as $P(\theta) = (a\cos \theta , b\sin \theta)$ .

A complimentary point to P, namely P' can be defined on the auxiliary circle with the same parameter , $\theta$ .
It can be given as $P'(\theta) = (a\cos \theta , a\sin \theta)$ .

Using the fact that the ratio of area of the triangle $ABC$ to that of the triangle $A'B'C'$ equals $\frac{b}{a}$ , we just need to maximize the area of the triangle on the auxiliary circle to maximize the area of the triangle in ellipse since they bear a constant ratio. (Proof of the above statement is in the solution of the problem linked with the question.)

Area of triangle in terms of the circumradius and the angles is given as,
$Area = 2R^{2} \sin A' sin B' sin C'$ , where $R$ is the circumradius.
Here, $R = a$ . Therefore, to maximize the area we need to maximize the product $\sin A' sin B' sin C'$ ,

We know that the maximum value of the product is $\frac{3\sqrt{3}}{8}$ .
(It can be proved by a combination of Jensen's Inequality and AM $\geq$ GM .)

Hence the maximum area is
$Area_{max} = \frac{b}{a} \times \frac{3\sqrt{3}}{4}a^{2}$
$= 3ab\frac{\sqrt{3}}{4}$

Plug in the given values to get the required answer.

We can get the given ellipse by apply the scale transform $scale(\sqrt{3}, 1)$ on the circle with equation $\frac{x^2}{4}+\frac{y^2}{4}=1$ . Since the this scale transform will increase the area of all triangle $ABC$ in the ellipse by $\sqrt{3}$ , we just need to find the area of the triangle in the circle mentioned above with maximum area (the equilateral triangle) and multiply it by $\sqrt{3}$ . The circle mentioned above has the radius of 2 and so, the area of the equilateral triangle inside it is $\frac{3R^2\sqrt{3}}{4}=3\sqrt{3}(R=2)$ . Finally, we get the maximum area of the triangle in the ellipse is $3\sqrt{3}*\sqrt{3}=9$

the max. SA of the triangle in the ellipse/ the max SA of that in a circle = semiminor/semimajor=2/sqrt(12)

max SA of triangle in a circle is when its equilateral triangle, whose side length is x

x can be calculated from the cosine law

SA of equilateral triangle= sqrt(3)*x^2 / 4

A diagram here