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
