Minimizing area of ellipse 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:

 A: If we choose coordinates as in your sketch:
$$
A=(-2\sin\theta,0);\quad B=(0,-2\cos\theta);\quad C=(2\sin\theta,0);
$$
then the tangent ellipse must be centered at $(0,0)$ and have $b=1+2\cos\theta$ as the $y$ semi-axis. Hence its equation is:
$$
{x^2\over a^2}+{y^2\over(1+2\cos\theta)^2}=1,
$$
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 $C$:
$$
(x-2\sin\theta)^2+y^2=1
$$
and plug then $y^2=1-(x-2\sin\theta)^2$ into the ellipse equation. 
The resulting quadratic equation in $x$ must have vanishing discriminant, which leads to:
$$
a={1+2\cos\theta\over\sqrt{\cos\theta}}.
$$
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 $\cos\theta=1/6$.
EDIT.
The above method to find $a$ works as long as tangency points have $y\ne0$, that is for $\theta>\theta_0$, with $\theta_0\approx69.65°$. For smaller values of $\theta$ you get simply $a=1+2\sin\theta$, but you can check that those ellipses have larger area.
A: Let me try to help with the discriminant.
Start with $$\frac{x^2}{a^2}+ \frac{y^2}{(1+2\cosθ)^2}=1$$ 
Put $$y^2=1−(x−2 \sinθ)^2$$
So the equation becomes
$$\frac{x^2}{a^2}+ \frac{1−(x−2 \sinθ)^2}{(1+2\cosθ)^2}=1$$
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 $q^2 = 4pr \implies$
$$\frac{16 \sin^2 θ}{(1+2 \cos θ)^4} = 4 \left(\frac{1}{a^2} - \frac{1}{(1+2 \cos θ)^2}\right) \left(\frac{1- 4 \sin^2 θ}{(1+2 \cos θ)^2}-1\right)  (1)
$$ 
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 
$$ \frac{1}{a^2} = \frac{\cos \theta}{(1+2 \cos \theta)^2}$$
