+483 An ellipse with equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
contains the circles \((x - 1)^2 + y^2 = 1\) and \((x + 1)^2 +y^2 = 1.\) Then the smallest possible area of the ellipse can be expressed in the form \(k\pi\). Find \(k\).
Any help is appreciated!
P.S. Answer is not \(2\sqrt2\)
+118687 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.
+118687 
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.
+483 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... 
+118687 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
+33661 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 \(xp = \frac{a^2}{a^2-b^2}\)
+483 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
+2511 You seem amazed that Alan derived the correct solution. He usually does. You’ll get used to it, if you stay around for awhile.
What really should amaze you is the massive and diverse amount of non-calculus math that Alan’s calculus solution replaces.
For relatively complex questions, the official solutions presented by AoPS professors often and usually consists of extended trigonometric functions, and/or finite and infinite arithmetic and geometric sequences, and/or elaborate generating functions, connected together using linear and non-linear algebra techniques.
Sometimes these methods are the only way to solve a problem. But if there is a calculus solution, learning these (laborious) methods will greatly enhance your understanding and appreciation of what the calculus is actually doing.
GA
--. .-
