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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 $(x - 1)^2 + y^2 = 1.$ 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 

By the law of exponents, if $3^a=2,$ and $3^b=5,$ then $3^{a+2b}=2 \cdot 5 \cdot 5 = 50.$ Now we know that $50 \cdot 3 = 150,$ so $3^{a+2b+1}=150,$ so in terms of $a$ and $b,$ $\boxed{x=a+2b+1}$

Hello Guest,

Here is my attempt:

There are 16 people who own cats, and 5 people who own cats and dogs, so there are 11 people who only own cats. Now there are 9 people who own dogs or both, so the answer would be $11 + 9 = \boxed{20}$ people own a cat or a dog

First of all, thank you! 

But how do you know the foci must be at (-1,0) and (1,0)? Why does this minimize the area... 

Yes, I have. Here is the graph:

wouldnt k=1 work?

Thank you!!!! 

Use the Wolfram-Alpha Theorem to get x=-1/6

can you tell me how you did this?

Let us call a position where you can win if you play correctly a winning position, and any other position a losing one. We can see that having 1-7 left is a winning position, and 8 is a losing position, etc. Notice that whatever amount $x$ that Morty takes on a given turn, Rick can always take $8-x$ french fries, and thus 8 will be taken in total. This means that if Rick can turn the number of french fries into a multiple of 8, Morty will be in a losing position, and so Rick will win by just taking $8-x$ french fries each turn, lowering the number by 8 each time until it reaches 0. In order to turn 60 into a multiple of 8, you need to take away 4, so the answer should be $\boxed{b) \space 4}$