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Find minimum area of an ellipse that can pack three unit circles such that all three touch the ellipse internally:

I took a point H as shown in the diagram and used the fact that the radius of the circle is 1, and that the circle touches the ellipse at point H. I am getting four equations for five unknowns, which means I can derive a relation between a and b of ellipse and use calculus to minimize the area. But those equations are tedious to solve and even after hours, I am not able to solve them.
Is there any easier way to solve this?
Here are the equations which I've got:

If we choose coordinates as in your sketch:
$$
A=(-2sintheta,0);quad B=(0,-2costheta);quad C=(2sintheta,0);
$$
then the tangent ellipse must be centered at $(0,0)$ and have $b=1+2costheta$ as the $y$ semi-axis. Hence its equation is:
$$
{x^2over a^2}+{y^2over(1+2costheta)^2}=1,
$$
where the unknown semi-axis $a$ must be determined so that the ellipse touches circle $C$. To find $a$ we can couple the equation of the ellipse to that of circle $C$:
$$
(x-2sintheta)^2+y^2=1
$$
and plug then $y^2=1-(x-2sintheta)^2$ into the ellipse equation.
The resulting quadratic equation in $x$ must have vanishing discriminant, which leads to:
$$
a={1+2costhetaoversqrt{costheta}}.
$$
Knowing both semi-axes $a$ and $b$ as a function of $theta$ you can then find by yourself the minimum value of the area, which occurs for $costheta=1/6$.

EDIT.

The above method to find $a$ works as long as tangency points have $yne0$, that is for $theta>theta_0$, with $theta_0approx69.65°$. For smaller values of $theta$ you get simply $a=1+2sintheta$, but you can check that those ellipses have larger area.

Let me try to help with the discriminant.

Start with $$frac{x^2}{a^2}+ frac{y^2}{(1+2cosθ)^2}=1$$

Put $$y^2=1−(x−2 sinθ)^2$$

So the equation becomes
$$frac{x^2}{a^2}+ frac{1−(x−2 sinθ)^2}{(1+2cosθ)^2}=1$$

Compare it with the standard quadratic equation $px^2 + qx + r = 0$. If the discriminant is zero, then the roots are equal and $q^2 = 4pr$

Here, $$p = frac{1}{a^2} - frac{1}{(1+2 cos θ)^2}$$
$$q = frac{4 sin θ}{(1+2 cos θ)^2}$$
$$r = frac{1- 4 sin^2 θ}{(1+2 cos θ)^2}-1$$

Now $q^2 = 4pr implies$

$$frac{16 sin^2 θ}{(1+2 cos θ)^4} = 4 left(frac{1}{a^2} - frac{1}{(1+2 cos θ)^2}right) left(frac{1- 4 sin^2 θ}{(1+2 cos θ)^2}-1right) (1)
$$

Next, observe that $$1- 4 sin^2 θ - (1+2 cos θ)^2 = -4(1+ cos θ)$$

Use this in (1) above:

$$frac{16 sin^2 θ}{(1+2 cos θ)^4} = 4 left(frac{1}{a^2} - frac{1}{(1+2 cos θ)^2}right) left(frac{-4(1+ cos θ)}{(1+2 cos θ)^2}right) (2)
$$

Next cancel $16$ from both LHS and RHS and write $$ sin^2 theta = (1 - cos theta)(1 + cos theta)$$ on the LHS of (2) to obtain

$$frac{(1 - cos theta)(1 + cos theta)}{(1+2 cos θ)^4} = left(frac{1}{a^2} - frac{1}{(1+2 cos θ)^2}right) left(frac{-(1+ cos θ)}{(1+2 cos θ)^2}right) (3) $$

Next cancel $(1 + cos theta)$ from both LHS and RHS of (3) and rearrange to obtain

$$ frac{1}{a^2} = frac{cos theta}{(1+2 cos theta)^2}$$