For super math whizzes only?

Question

A chord of length 1 is made to slide around the inside of an ellipse. Its midpoint traces out another convex curve inside the ellipse. Find the area between the two.

Okay, if you need to know, the ellipse's major diameter is 10 and minor diameter is 5.

Answer 1

Let x^2 / a^2 + y^2 /b^2 = 1

(x1, y1) = (X,Y) + (u,v)
(x2, y2) = (X,Y) - (u,v)

where
x1^2 / a^2 + y1^2 /b^2 = 1 (1)
x2^2 / a^2 + y2^2 /b^2 = 1 (2)
u^2 + v^2 = d^2. (3)


(1) - (2) :
Xu / a^2 + Yv / b^2 = 0
Let (X,Y) = R(a cos t, b sin t), then
u cos t + v sin t = 0, and combined with (3) we have

u = d sin t
v = -d cos t

(1) + (2):
R^2 (cos^2 t + sin^2 t ) + d^2 (sin^2 t / a^2 + cos^2 t / b^2) = 1, or
R^2 = 1 - d^2 (sin^2 t / a^2 + cos^2 t / b^2).

dR = - d^2/2R sin 2t (1/a^2 - 1/b^2) dt
dY = b(R d sin t + dR cos t) =
b (R cos t - d^2/2R sin 2t (1/a^2 - 1/b^2) cos t) dt

Area = integral of X dY =
ab int [ R^2 cos^2 t - d^2/2 sin 2t cos^2 t (1/a^2 - 1/b^2)] dt =

ab int [cos^2 t dt] -
abd^2 int [(sin^2 t / a^2 + cos^2 t / b^2) cos^2 t dt] -
abd^2/2 int [sin 2t cos^2 t (1/a^2 - 1/b^2)) dt ] =

********************* ***********************
S_ellipse( 1 - d^2/2 (1/a^2 + 1/b^2))
********************* ***********************

a = 5/2, b = 10/2, d = 1/2

S = 195/16 pi