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 \(2\sqrt2\)

abcdefghijklmnopqrst Dec 3, 2021

#1**+1 **

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.

Melody Dec 3, 2021

#3**+1 **

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.

Melody Dec 4, 2021

#4**+6 **

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...

abcdefghijklmnopqrst
Dec 5, 2021

#6**+2 **

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

Melody Dec 6, 2021

#7**+5 **

Best Answer

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 \(xp = \frac{a^2}{a^2-b^2}\)

Alan Dec 6, 2021

#9**+7 **

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

abcdefghijklmnopqrst Dec 7, 2021

#12**+2 **

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

--. .-

GingerAle
Dec 7, 2021