Area of Ellipse Problem – Duke Math Meet 2009: Individual Round Problem 7 Solution

Let \(R_I , R_{II} , R_{III} , R_{IV} \) be areas of the elliptical region \(\frac{(x-10)^2}{10} + \frac {(y-31)^2}{31} \le 2009 \) that lie in the first, second, third, and fourth quadrants, respectively. Find \(R_I – R_{II} + R_{III} – R_{IV} \)

Discussion:

Special Note: The answer to this problem is given as 1240. This is a wrong answer. It approximates the region \(R_I – R_{II} + R_{III} – R_{IV} \) as a rectangle. However we provide a solution using symmetry using that assumption and that works fine. Computing area of ellipse is a tricky business.

First we draw an approximate picture of the ellipse. The centre is at (10, 31). Here is a figure of it.

To find \(R_{I} – R_{II} \) reflect region \(R_{II} \) about y axis. Look at the figure. The shaded region is \(R_{I} – R_{II} \). It’s width along the line through centre (y=31) is 20 by symmetry (as the centre is 10 unit away from x axis so \(R_I \) is 20 unit ‘thicker’ than \(R_{II} \) . You may convince yourself about this by solving for x setting y = 31 ).

Now to find \(R_{III} – R_{IV} \) we reflect \(R_{III} \) about y axis again. The strip (shaded in blue) is negative of \(R_{III} – R_{IV} \)

Note that the blue region ‘begins’ 31 unit ‘below’ the minor axis of the ellipse. So if we go 31 unit ‘above’ the minor axis and take the portion of the red strip, by symmetry it will be equal to the blue strip. We have shaded it in black and red.

So if we want to find \(R_{I} – R_{II} – { – (R_{III} -R_{IV}) } \) we remove the black strip from the red strip and get the final region whose area is \(R_I – R_{II} + R_{III} – R_{IV} \).

Here is where we apply approximation. Width of the strip is 20 and it’s height is (31+31) = 62. Hence if we approximate the area as a rectangle, then answer is 1240 (62*20).

But note that the strip is not ACTUALLY a rectangle. So this is only an approximate answer.