The Area of an Ellipse
Kyla Basnett and Sabrina Allen
October 3, 2011
Archimedes showed using the double reductio ad absurdum that the area
of an ellipse, A, with major axis a and minor axis b is, A = πab. His proof is
= ab . (see poster 1).
based off the ratio PP Q
R
Today’s geometry gives a method of proving this ratio. First, start with
2
2
the equation of an ellipse: xa2 + yb2 = 1. With manipulation of this equation
we obtain:
y2
b2
2
2
= 1 − xa2
2 2
y = b2 − b ax2
2 2
y 2 a2 = b√
a − b2 x 2
ya = b√ a2 − x2
y = ab a2 − x2
√
By image 1, we see that y = P Q and a2 − x2 = P R by the Pythagorean
Theorem. Thus, by plugging in, we obtain the ratio that Archimedes used.
P Q = ab P R ⇒
PQ
PR
= ab .
Note: We’re not sure how Archimedes would have proved this without
declaration of the equation of an ellipse.
1
AE = πab
Consider image 2 and image 3. Let E be the ellipse with major axis a and
minor axis b. Let C be the circle with radius a, circumscribed√about E, called
its auxiliary circle. Finally, let C 00 be the circle with radius ab. Achimedes
proved the area of an ellipse by showing that a(E) = a(C 00 ).
1
1.1
Case 1
Assume a(E) < a(C 00 ). Archimedes showed by exhaustion that a(P 00 ) converges to a(C 00 ), so replace a(C 00 ) with a(P 00 ) to get a(E) < a(P 00 ). Let P 0
be a regular polygon, similar to P 00 , inscribed in C 0 . Thus, we have the ratio
2
a(P 00 )
= ar 2 = aab2 = ab . Let P be the polygon inscribed in the ellipse, such that
a(P 0 )
the vertices are the intersections between E and the perpendiculars from the
vertices of P 0 to the horizontal axis of E. Now, looking at the polygons P
and P 0 , consider the corresponding pairs of triangles like Qrs and QRS and
the corresponding pairs of trapezoids like klmn and KLM N . Based on how
Q)
kn
rs
lm
= KN
= RS
= 2(P
= PP Q
= ab .
they were constructed, we get the ratios LM
2(P R)
R
a(klmn)
a(Qrs)
It then follows that a(KLM
= a(QRS)
= ab , since the sides are proportional
N)
to the areas. Sum all the areas of the trapezoids and triangles to obtain a(P 0 )
a(P )
a(P 00 )
b
a
and a(P ) Thus is follows that a(P
0 ) = a . Earlier we obtained a(P 0 ) = b , thus
00
a(P )
a(P )
00
we have a(P
0 ) = a(P 0 ) ⇒ a(P ) = a(P ). Plugging this in to the inequality
we assumed, we get a(E) < a(P ), which is the contradiction desired because
P was defined to be the polygon inscribed in E, and by the Archimedean
Postulate a(P ) must be less than a(E).
1.2
Case 2
Assume a(E) > a(C 00 ). By Archimedean Postulate a(P ) < a(E). Start with
the polygon P like the one in Case 1 inscribed in E such that a(P ) > a(C 00 ).
a(P )
a(P 00 )
b
By the same computations as in Case 1, we conclude that a(P
0 ) = a = a(P 0 ) .
Again this gives a(P ) = a(P 00 ), and plugging this in to the inequality we assumed gives a(P 00 ) > a(C 00 ). This is the contradiction desired because P 00 is
defined to be the polygon inscribed in C 00 and by the Archimedean Postulate
a(P 00 ) must be less than a(C 00 ).
This completes the double reductio ad absurdum proof that a(E) =
00
00
2
a(C
√ ). 2Archimedes already proved the area of a circle, so a(C ) = πr =
π( ab) = πab. Thus a(E) = πab.
In essence, Archimedes has shown that the area of an ellipse is b/a times
the area πa2 of its auxiliary circle. This corresponds to the observation that
the circle is transformed into the ellipse by shrinking the vertical dimension
by the factor b/a.
2
Note: In this proof Archimedes used the fact that an ellipse can be exhausted using polygons. This is true because if you translate kl to KL in
image 2, the edge of the ellipse will lie between KL and the edge of the circle.
Thus, since we know that the circle is exhausted by the polygons, the ellipse
is as well since it is being squeezed between P 0 and C 0 . This is very similar
to the squeeze theorem.
3
Image 1
Image 2
Image 3