ELL

370

ELL

Example .—What is the area of an ellipsis, the greater axisof which is 24, and the lesser 18 ?

576

.7854

2304

2880

4608

4032

452.3904 the area of the circumscribing circle.

Therefore 24 : 18 : : 452.3904

18

36191232

4523904

24 ) 8143.0272(339.2928 the ellipticarea.72

l i

223

216

222

216

192

192

But the proportion of 18 to 24, is as 4 to 3 : therefore theabove might have been considerably abridged ; we were, how-ever, desirous of working the operation at full length, aswould unavoidably happen in case of incommensurablenumbers, in order to compare it with the following ope-rations.

Now let us try whether we cannot find a more practicalrule for the area of an entire ellipsis, than that above.

Let p d 2 be the area of a circle circumscribing anellipsis, where

p r= .7854, and d — the diameter of the circle orgreater axis of the ellipsis,and c — the conjugate, or shorter axis jT) d 2 c

then we have d : c : : pd 2 : T -—— — pdc the area of the

entire ellipsis: we have therefore the following neat rule.

Multiply the two axes together, and the product by .7854,and this second product will be the area.

Suppose now, for the sake of comparison, that we take theformer example, viz., the greater axis 24, and the lesser 18.

432

.7854

17282 16034 56302 4

339.2928 area of the ellipsi 9, w

ft^

In the same manner, may the half, or the fl^er,found, viz., by multiplying the two dimensions to Dthe product by .7854. aX j g ,

Example .—In a semi-ellipsis upon the greater .ggjg S*the greater axis be 24, as above, and the lesser sthe area is required.

24

.7854

9

216

216

4 7124

7 854

157 08

of

tta®

169.6464 = the are " V° r6semi - ellipsis, which is half of the entire ashown. • a *is b

Example in a quadrant .—Let the greater se12, and the lesser semi-axis 9, the area is require •

12 .7854

9 108

108

6 283278 540

of thp pTT- • t. • 84.8232 the area of

°: r . e J h Pf s > being one quarter of the area of “{Lgis 13 'ellipsis, in the foregoing example. The area of a n e Ip ' inS 1a mean proportional between the area of the circu® 9and inscribing circle.

. is the area of the circumscribing cir

is the area of the inscribing circle : , re

.ZZf!- P c’; andtl^ f0

P t 2 “

the proposition is manifest.

To find the periphery of an ellipse

Let a = the greater semi-axis a c ;

e = the lesser semi-axis ; , „: 39 9 i

x — the distance c d from the centre, the ay = d h, the ordinate ;

z = E f, the arc : r . r6 A v

then will Ad * a — % and do = a + * thereX ® <* = « + « X a — x ~ a 2 — x 1 .

xiien by the property of the ellipsis,ac or cb : c G a : : a d x db ; dh 1a3 : c a ; ; a 3 — a?

consequently f ~ c -(fi _

at

role I

Now pt 2 : ptc :: ptc

that is: /

and therefore y = — (a 2 —