Vo. IV.) Anfwers to Queftions propofed. | 565 
Now the abfolute velocity at G is as 4/ AF, and that in the diretion of GH as a/ VE, and 
this proportion will be exprefled by that of AG to AF ; whence we have a/y : ANE ieee iy 
Jy? Cyan ys 
3 i 
ory: VE::¢?: y*, then VE=2, and CV=y—=—= 3 but y?==c?——x’, there- 
cz". b? c3 

x? : —_—_—_—— 
fore, CV = => =} the parameter; and, by a common property, EG=/VEx4CV— 
| 2xy? be? —c2x 
a LE a 
74 amy? y? , : 
J =. but HR =b—x, then —. +5—x =DR, then, by 
c ? 
Axy?—c2x 1 hc? 
— m4 e 
2 
conics, we fhall have 4CV : DR-LEG:: HR: BF, that is, sey : > b—x 
cz 
c2 
: a—y, therefore, Abxy?—2bc2 xb c2-2 1 b2;2—4 ax? 3 but y2=Fmx?, and y==4/c?—x", thems 
by fubftitution, xmmm x3 Labs? — hx /2— 43 (where m=6'4, and f=19'2). And this 
fquared and refolved, gives x=1-7376 feet, then y==1'802, and x—-c=="69504, the natural 
fine of the required; = 44° 2’ nearly. . 
This queftion was alfo anfwered by Mr. ohn Collins, Mr. T. Hickman, and Virgo, 

Question XXXIII (No. XX).—Aifwered by Mr. R. Simpfon, of Bath. 
Maxe AB equal to the given bafe, alfo BD perpendicular to it, fo 
that AB be to BD in the given ratio. Join AD, which produce 
indefinitely, and make the angle DBC equal to the angle A; fo fhall 
ABC be the triangle fought. AL SB 
For, draw CE parallel to DB; then the angle BCE = DBC = DAB by the conftruion ; 
therefore their complements are alfo equal; that is, the complement of the angle A is equal the 
fupplement of the angle B. Alfo, the triangles AEC, ABD, BCE, are equiangular 5 and 
therefore AC : BC :: AE: CE :: AB: BD, that is, in the given ratio by conftruction. 
This Queftion was alfo anfwered by Mr. Fames Afkton, Mr. D. Booth, Mr. F. Collins, Mr. F. H. 
Mr. T. Hickman, and Virgo. 
Question XXXIV (No, XXI)—4nfwered by the Rev. L. Evans, of Froxfield, Wilts. 
Cy 

_Havinc given the ratio of the bafe to one of the fides of an ifofceles 
‘triangle, as 1 tor, and the area of its greateft infcribed ellipfis =a, the 
dimenfions of both will be found thus: Eg 
Suppofe, firft, the annexed figure to reprefent a triangle and ellipfis 
fimilar to thofe required ; in which AB = 1, AC=BC =r, and confe- 
_ I " 
quently CF =,4/r?— —, are given quantities. Put AF =4, CF =f, 
and DF =x, Then, by fimilar triangles, CF: AF::CD:DE=s x Att # BR 
px 

po ee 
en ; and by the property of the ellipfis 24/ AF x DE = 2by/ the conjugate diametes 
pm 
£ 
Nd it : j Si. NEE: 2 ; 
x3 ==a maximum ; the fluxion of*which made= 0, gives == —/, or the vertical axis ane 
3 

ef the eilipfes ; alfo, by menfuration, - 78 54 X 2bx4/ = its ayea, or maximnm ; or x? =m 
the triangle’s perpendicular. 
Next, let the above figure now reprefent the real triangle and ellipfis in queftion, the area of 
the latter being =a ; alfo, let = AB, the bafe j then 34/7? — whe CF the perpendicular, and 
na 2 4 
2 2 I = 3 E ' 
Seals — z == DF the vertical axis; alfo, by fimilar triangles, CF: AF :: Cr vie By 
6 
Le I e . 
and 24/AF x baggie » the horizontal axis ; confequently + 7854 x af ik =m/r A Bp 
4 
2.747 
@ and hence z = 4/ » where x is =: 7854. Hence the remaining dimenfions 
47171 Fame 732 
y Ollow,. bes 
This Queftion was alfo anfwered by Mr. Fames Aiton, Mr. Fohn Collins, and Mr. T. Eickmani 
% 
QuEsTion 
