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        riddles >> putnam exam (pure math) >> Ellipse Area Triangle
        
(Message started by: Benoit_Mandelbrot on Feb 17^th, 2004, 8:48am)

Title: Ellipse Area Triangle
Post by Benoit_Mandelbrot on Feb 17^th, 2004, 8:48am

Find the equation of the line tangent to the ellipse:
b²*x² + a²*y² = a²*b²
in the first quadrant that forms with the coordinate axes the triangle of smallest possible area (a & b are positive constants).

Title: Re: Ellipse Area Triangle
Post by towr on Feb 17^th, 2004, 10:06am

interesting, I get a triangle area of ::[hide]a*b[/hide]::
And to answer the question, line ::[hide] y = b (sqrt(2) - x/a) [/hide]::

Title: Re: Ellipse Area Triangle
Post by John_Gaughan on Feb 17^th, 2004, 11:51am

Towr, how did you arrive at your answer? Given that this is the first quadrant, I was able to simplify to :: [hide] y == (b/a) sqrt (a² - x²), where x and y are positive and a > x. I imagine I need to take the derivative of this equation to get the slope of the line, but I don't remember offhand how to do it in this case. I am at work, my books are at home.[/hide] ::

Title: Re: Ellipse Area Triangle
Post by towr on Feb 17^th, 2004, 12:06pm

yep, that's how I started.. I have the luck of having my computer to do some of the 'hard' work, but you can just use the chain rule (I think that's what it's called) ::[hide]df(u)/dx = df(u)/du du/dx
and of course in this case f(u) = (b/a) sqrt(u), and u = (a² - x²) [/hide]::

Title: Re: Ellipse Area Triangle
Post by John_Gaughan on Feb 17^th, 2004, 1:17pm

I screwed something up I think... attached is the line I wound up with. It is in y=mx+b format.

Title: Re: Ellipse Area Triangle
Post by towr on Feb 17^th, 2004, 1:51pm

::[hide]
starting at y = f(x) = (b/a) sqrt (a² - x²),
the tangent line to the ellips at a point on the upper arc has direction
f'(x) = (b/a) * 1/2 / sqrt (a² - x²) * (-2x)
= -x (b/a) / sqrt (a² - x²)

the line we're looking for will cross through some point (x,y)_min, where the triangle is minimized
so we have
y - y_min = f'(x_min)(x-x_min)
y = f(x_min) + f'(x_min)(x-x_min )
y = b (a² - x_minx)/(a sqrt(a² - x_min²))

from this line we'll need y when x=0, and x when y=0 (the points where the line crosses the axis)
so y = ab/( sqrt(a² - x_min²))
and 0 = b (a² - x_minx)/(a sqrt(a² - x_min²))
=> x = a²/x_min

the size of the triangle (x*y/2) is minimized when we minimize
x*y = a³b/(x_minsqrt(a² - x_min²))
To do this we'll finally determine x_min, so take the derivative and find where it is 0; eventually we get x_min = sqrt(2)a/2 (among other solutions which don't fit our needs)

So filling this in in our description of the line, y = b (a² - x_minx)/(a sqrt(a² - x_min²))
we get y = b(sqrt(2) a - x)/a
[/hide]::