Ellipses

Here's my attempt at an analytical answer:

I should have written: This is the value of k at the minimum area.  The minimum area itself is $k\pi$

I've just noticed something went wrong with the display of equation (3)! It should be $xp = \frac{a^2}{a^2-b^2}$

Have you graphed this with desmos?

I suggest you do so and attach a link to the graph.  It is always a good starting point.

Let's include a slightly more helpful version of your graph here.

I haven't done it yet and I always have to look up the 'rules' when I am dealing with ellipses

but it is pretty clear that the focii are (-1,0) and (1,0)

I'd take it from there.  I might do it later.

Here is a much better graph.   Unfortunately, the answer still alludes me. 

https://www.geogebra.org/classic/gcr4sdt4

And here is an ellipse formula page

It's right!!!! Here is the solution provided on the website:

We can assume that the ellipse is tangent to the circle $(x - 1)^2 + y^2 = 1.$ From this equation, $y^2 = 1 - (x - 1)^2.$ Substituting into the equation of the ellipse, we get $\frac{x^2}{a^2} + \frac{1 - (x - 1)^2}{b^2} = 1.$ This simplifies to $(a^2 - b^2) x^2 - 2a^2 x + a^2 b^2 = 0.$
By symmetry, the $x$-coordinates of both tangent points will be equal, so the discriminant of this quadratic will be 0: $(2a^2)^2 - 4(a^2 - b^2)(a^2 b^2) = 0.$
This simplifies to $a^4 b^2 = a^4 + a^2 b^4.$ We can divide both sides by $a^2$ to get $a^2 b^2 = a^2 + b^4.$ Then $a^2 = \frac{b^4}{b^2 - 1}.$
The area of the ellipse is $\pi ab$ Minimizing this is equivalent to minimizing $ab$, which in turn is equivalent to minimizing $a^2 b^2 = \frac{b^6}{b^2 - 1}.$
Let $t=b^2$, so $\frac{b^6}{b^2 - 1} = \frac{t^3}{t - 1}.$
Then let $u=t-1.$ Then $t=u+1,$ so $\frac{t^3}{t - 1} = \frac{(u + 1)^3}{u} = u^2 + 3u + 3 + \frac{1}{u}.$
By AM-GM,

 $\begin{align*} u^2 + 3u + \frac{1}{u} &= u^2 + \frac{u}{2} + \frac{u}{2} + \frac{u}{2} + \frac{u}{2} + \frac{u}{2} + \frac{u}{2} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} \\ &\ge 15 \sqrt[15]{u^2 \cdot \frac{u^6}{2^6} \cdot \frac{1}{8^8 u^8}} = \frac{15}{4}. \end{align*}$
Equality occurs when $u=\frac{1}{2}$ For this value of $u, t=\frac{3}{2}, b=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2},$ and $a=\frac{3\sqrt2}{2}.$ 

Hence, $k = ab = \boxed{\frac{3 \sqrt{3}}{2}}.$

Whew! that took some time to format lol