abcdefghijklmnopqrst

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 #9
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It's right!!!! Here is the solution provided on the website:

 

We can assume that the ellipse is tangent to the circle (x1)2+y2=1. From this equation, y2=1(x1)2. Substituting into the equation of the ellipse, we get x2a2+1(x1)2b2=1. This simplifies to (a2b2)x22a2x+a2b2=0.
By symmetry, the x-coordinates of both tangent points will be equal, so the discriminant of this quadratic will be 0: (2a2)24(a2b2)(a2b2)=0.
This simplifies to a4b2=a4+a2b4. We can divide both sides by a2 to get a2b2=a2+b4. Then a2=b4b21.
The area of the ellipse is πab Minimizing this is equivalent to minimizing ab, which in turn is equivalent to minimizing a2b2=b6b21.
Let t=b2, so b6b21=t3t1.
Then let u=t1. Then t=u+1, so t3t1=(u+1)3u=u2+3u+3+1u.
By AM-GM,

 u2+3u+1u=u2+u2+u2+u2+u2+u2+u2+18u+18u+18u+18u+18u+18u+18u+18u1515u2u626188u8=154.
Equality occurs when u=12 For this value of u,t=32,b=32=62, and a=322. 

Hence, k=ab=332.

 

Whew! that took some time to format lol 

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