394 
ANSWERS TO THE QUESTIONS PROPOSED 
1n Nos. II. and III. 
Question VI (No. IL.—)4z/wered by 
Mr. T. Hickman. 
Let T be the given point 
in the ide LF of the cone 
FLP, and TV the required 
fection, being an elipfis, 
PutLM=i, TM=MH=,, 
o='7354,and LN=1; then 
by menfuration, the area 
of the circle TMH=4/2a, 
Now TH=2;; andas LM 
(4): TH (ar) :: LN(x): F 
27x 
J} =CV=2NV, Then letting fall the perpen- 
dicular VQ=i—vx, it is evident that TQ== 
TM+NV=rF" and from the 47. I. of Euc. 

TQ: +VQ°=TV?, whence / ha ny EY ; 
2 
==T'V the tranverfe diameter of the clipf 
1S $ 
alfo_by Emerfon’s Conics alae £1 ei T. 
er Doctor Hutton’s Conics, cor. x. p. 6, 
/ GVxTH=OP the conjugate diameter== 
farx 
fz Xr aad from the property of the elipfis, 
‘we have TVX OP xa equal the area of the 
; . ae 27x = 
the elipfis, ee X21 X4/; gee] i 
—SS_—————————___.____ 
27k ra] 2 
whence a? x ign sacs pe] =16r43?, 
by Eran and he alg reduction xi2h x 
t= r 
appt =o an equation from 
whence LN, andwhatever elfe is required, may 
in any cafe be determined, 
Corollary 1. Put the expreffion found above 
for the area of the elipfis, into fluxions, and 
12 ee 2 
after proper reduction we have 17-44% ————: - 
72-42 
f2 . ‘ 
x-|-—=0, an equation from whence the greateft 
3 
and leaft elipfis in any given-cone may be found ; 
the {mallefi poflitive root fhowing the maximum, 
and the largeft pofitive root the mizimum; but 

eae 
r ? bp ? 
if 4 be lefs than ra/ avg or r X.02679, or 
3 
4f the vertical angle of the cone exceeds 176° 56! 
the cope will admit of no greateft or leaft elipfis. 
Scholium, the expreflion above given for the 
greateft and leaft elipfis though the true limits, 
does not always in reality fhaw the greateft and 
leaft fetions of which the cane is capable, if cut 
nearly parallel to the bafe or to the flant fide. 
Coroliary 2. Put again the expreffion for the 
acea of the elipfis equal to a given area 42, and 
Mathematical Correfpondence. 
[June 

A ee © 
after proper reduction, we have *21X ae 
; Ash3 
sie wie ———— ==0, an equation for 
4227? x 12h f2 
" estting an elipfis of a given area from any given 
cone. 
This quefiton was alfo anfwered by Mr. F, 
F——-r. 
SEE 
Question VII (No. If.—)Anfwered by Mr. 
0. G. Gregory. 

Let.AB in the annexed figure reprefent the 
pole, and D the place of the eye: then will a ray 
coming from the battom, A, of the pole, and 
ftriking upon the water’s furface at a, be re- 
flected into the direétion aD; and a ray 
from B the top of the pole, ftriking upon the 
water’s furface at 4, will be refie¢ied into the 
dire&tion 4D. It is a fundamertal law or 
principle in the doctrine of catoptrics, that the 
angle in which a ray falls upon any reflecting 
furface (called the angle of incidence) is equal 
to the angle in which it quits it, when it is re- 
fleéted from it (called the angle of refleGtion) : 
hence A2E=CaD, and BOE=D4C. From 
this law arifes another, which is, that rays, B4, 
Aa, &c. proceeding from various objects, wouid 
(if continued) converge to 2 paint as far below 
the reflecting furface as D, the point where the 
reflected rays meet, is above the faid furface: 
on thefe two principles the folytion chiefly de- 
pends. 
Here we have CD=—C d=8+s=13; BA 
=18; AH=3; BE==18-+8==26; CE=60. 
Alfo the triangle a Cd fimilar to zEA; and 
6Cd fimilar to SEB. 
Therefore, by fimilar triangles, asd CLAE 
(==21): CE (60) >: Cd (=13) : 375—Cae 
And, by the fame, as dC-++-BE (39): CE 
(=60) :: Cd{=13):20=C4. Confequently, 
371—20==17! feet=aa 4 the length of the imege.. 
The breadth of. the image at a will, by the 
the rules of perfpedtive, be to 6 inches (the 
diameter of the poles) as da to dA; or, as Cd 
to Cd+AE: hence, as 21: 13 :: 6:35 inches, 
breadth of the image at a. Again, the breadth 
at 6, will be to 6 inches, as db: dB; or, as Cd: 
Cd+EB Therefore, as 39: 13:3: 6:2 inches, 
breadth of the image at 6; which was te~ 
quired. 
\ 
N.B. 
