y = - 00, are placed between the asymptotes BD, BM and bd, bm.

5. Since the subtangent of the curve = n times the abscissa, and the subtangent =

ydx

(r and y being the co-ordinates), we

dy have

ndy

dir
y
And lya = la + const.

= lx + Ic

=l.cz :: y = c. The curve is, therefore, parabolic. The locus of the vertex of a A of given base, when the sum of its angles at the base are also given, is evidently a circular aro described

upon

the base, so as to contain an angle = the supplement of the given sum of the L at the base.

For since the sum of the Z of a A=2 right 2, the angle at the vertex = the supplement of the 2 at the base, and is :: given.

6. Again, let BC, the given base of the A (Fig. 6) = a, and let P be any point in the required locus. Make C the origin of abscissæ, or PM = y and CM = Also put the angle PBC = 0. Then by the question PCB = 20, and we have

y = x. tan. 24 = (a — x) tan. 8
2r, tan. 6

(a – r). tan. 0, whence
1 - tan.20

..

And
:: y = (a – x) tan. 6 = ✓ (a-x). (a– 31)

=Nai - fat+ 3x2) which indicates a conic section. Let y = 0

y=

2ax'+x'?, demonstrates that the locus is an hyper

a

bola whose semi-axes are a and

3 ✓ 3 The branch QAq corresponds to the positive values of x'. From

2a = 0 to = AB we have y imaginary; but when s' is taken

3

2a

negatively > my again becomes real, and describes the branch Q' Bg'.

(7). Let the given area be denoted by a, the given Zat A (Fig. 7) by A, and the given ratio of BF : FC by n : 1.

Draw FG, Fg parallel to AB, AC respectively, and CN, FM I AB. Then GF: AB :: FC : BF :: 1:n + 1

AB
.. GF =

n+1
Also CN : FM :: CB : FB :: n +1:n

nCN
:. FM = = also Fg. sin. A

n+1

AB nCN

2an Hence GFx Fg = X

n+1 (n+1)sin. A (n+1)'sin. A Or making GF = Ag = x

and gf = y, we have

2an

= m2 a constant quantity, which is the (n+1)'sin.A equation to an hyperbola whose asymptotes are AD, AE.

To find its principal diameters, bisect the ZA by AV, make
I= y, or yo = m', and .. y = m; and we have Ah = hV –VH

sin. AhV
sin. A

A - HA. Hence AV =

X Ah =

-.m = 2 cos. sin. AV

A sin.

2

2

2

COS.

✓ zan
2
x van

A

an cot. (1+1) sin A

nti

2 (n+1). sin. which is the semi-axis major. The semi-axis minor :: (= VK)

A - AV. tan.

2

A ✓

an tan.? cot. n+1

nti

2

Х

2

2

2

an tan.

8. Let PM = y (see Fig. p. 54 in the Problem)

AM = x, and AB = a.
Then y’ = AZ = x + ZM' = x + x. (a –x)

= ar, the equation to a parabola whose Latus-Rectum - a the diameter of the circle.

2

Let r=a. Then y = a, and the area required =

a.

3

9. Let ABCD (Fig. 8) be any position of the generating square, QM the axis of the solid and PQp the circular section 1 to AD and BC. Let also P'Qp' be a section inclined to AD, BC at any given 2 (c), and passing through the axis QM; and put QM = x P’M = y, and the radius of PQp = a.

a

and a;

sin. a

a

which is the equation to an ellipse whose axes are the origin of abscissæ being at the extremity of the axis-minor, because is necessarily > a. sin, a

x2 + 10. To trace the curve whose equation is ao (t – b) = x y', we first reduce it to a + 63 - 2 bx = x y'; then we have

and putting r = 0, we get y = to which are represented by AB, Ab (fig. 9). Again, making a =

a' + b

= AC, y = ,

26 and we have a cusp of the first species at C. No positive value of x

a' + ba can exceed

26

.

Let x =

Then y= + 0, and Ac is an asymptote to the branches FE, fe. Also Bb is an asymptote to each of the branches.

11. To prove that every section of a conoid is a conic section,

We will here exhibit a method of ascertaining the nature of the section made by a plane with any solid of revolution whatever.

Let DAd (fig. 10) be the solid formed by the revolution of DAQ about AQ as an axis, and let BP bp be a circle described by any point B. The plane of this circle will evidently be I plane DAD, and their intersection Bb will be 1 axis AQ.

Again, let CP op be the section made by a plane, passing any how through the solid, and intersecting the circle by the line Pp; which latter also cuts Bb in M. Join CM. Then since the points C, M are both in the planes CAd, CPc, CM being produced will meet the curve in c, and cut the axis AQ in R.

Now if the circle BPb move parallel to itself, the intersection

= BNP
= BN

Pp will also move parallel to itself, and we, therefore, put the con-
stant angle BMP = a.
Also let ZRMN = B, CR=b, AR= a, cM= x, and PM = y.

By the property of the circle, we have
Pm = BN - Nm = BN? - (NM + Mm)

(RM . cos. ß + y cos. a)

- b) cos. ß + y cos. a}”. But Pm = yo sin.”ce, and supposing y'a = f (x') to be equation of the generating curve, we have

BN? = f. (AN) = f(a + RN) = f.(a + RM. sin. B) f. {a + (x – b) sin. B}.

:. by substitution y? sin.° c = f.(a + 5—6. sin. B) – (x — . cos. ß + y cos. a)"; whence by reduction we get ya – 2 cos. a . cos. B. (x b) y = f. (at a

6. sin. B) b) cos.' B, which is the general equation of the section of a solid of revolution.

Now, since in the conic sections y' = f (') is always of two dimensions, the equation of the section of a conoid is likewise of two dimensions, and :. the section itself is a conic section Q.E.D.

Again, let y'? pr', or the conoid be a paraboloid. Also let vcw, the cutting plane, be parallel to the axis AQ or B = 90°. Then a = 90°, a= 60, and b = on, and PM becomes I Cc. Hence substituting in the general equation, we have

yo = p.r. The section is .. a parabola, similar to, because of equal parameter with, the generating one Q E.D.

If in other applications it should be necessary to ascertain the angle at which PM is inclined to Cc, in order to transform the Co-ordinates of the section to rectangular or otherwise, the following process will serve that

purpose. Take MA' = MB = MC = 1, and with M, as centre, describe the arcs A'B', B'C', C'A' forming a spherical A A'B'C'.

Then since the angle A' measures the inclination of the planes CPc, BPb, A'B' measures the 2 and B'C' measures B, and are