which determines the abscissa of the point required, and substituting in equation (1) we get the corresponding ordinate, which will determine the actual position of the point. There are two solutions giving points on different sides of the axis of y.

Then Hy being

PT and

88. Let S, H, (Fig. 51,) be the foci of the ellipse APa, and PT a tangent at any point P. 'Cy, SP joined, Cy is parallel to SP.

For, producing SP, Hy to meet in h, since PT (by property of the ellipse) bisects the angle hPH, we easily prove that yh = yH. Hence, since SC CH, we have CHy similar to SHh.

= SH, or Cy is parallel SP.

:. LyCa

Similarly, Sy' being drawn PT, and Cy' joined, it may be shewn that Cy' is parallel to PH.

89.

Let C, A (Fig. 52,) be the circle of curvature to the parabola AP,P,, &c. at its vertex A, and suppose a series of circles whose centres are C2, C,, &c., to be described each touching the one preceding it and the two branches of the parabola, then r1, 72, 73, &c., denoting the radii of the circles, required to prove that r¡,„, 73, &c., I'ng are in the proportion of 1, 3, 5 ..... 2n ·

1

1.

Joining C, P,, C, P1, &c., and making AN,= x, P2N. = y1⁄2, &c. we have r ̧ = C ̧a = aN ̧ + C‚N, = 1, − 2r, + 1⁄2

卫

2

By the same process with respect to the next circle, we get

Now it is well known that the diameter of curvature 2r1 at the vertex of a parabola = latus rectum p.

which give the relations required, the negative values merely indicating those circles which touch both the preceding one, and the branches internally, and evidently coincide in toto with it. If x=(y) be the equation of the curve, then we easily find

whence it appears possible to find a series of circles such as are represented (r, however being any given O touching the branches, which somewhat generalizes the problem,) in the figure, only for constant, i. e., for the common parabola

such curves as give

only.

ydy
dx

90.

This enunciation ought to have specified that the radius vector in arriving at SP had just completed a revolution. Let SP = g, and the corresponding angle = 6, then the equation to the spiral (which is that of Archimedes), is

91. This enunciation is very ambiguously expressed. By the "acute angle" is meant that of the angles which is the most acute, and the tangent is considered "the base."

Let R, r, be the radii of the respective circles, C, c their circumferences, L the straight line C, and I the one to be found; then from similar ▲ whose sides are L, C; l, c, we have

L: Cl: c

But LC, :. l = c.

92.

Q. E. D.

Let R the radius of the circle, r that of the circular base of the cone, and s = side of the cone; then R' = area of the circle, 2 × r = circumference of the base.

Hence the surface of the cone =

#R2 = πr x s

.. Rrs, or

r: R: Rs.

rxs, and by the question

Q. E. D.

93.

Let R be the radius of the sphere. Then the solidity of the cylinder whose base R' and altitude 2R, is 2R3, and that of the cone with same base and altitude as the cylinder, is

94.

Let PL (Fig. 53) = CD = a', CP = b', CK = KL

= KM = KN, then PM a the axis-major of the ellipse, and

m2 + n2 = (MK + PK)2+ (MK — PK)2

= 2MK2 + 2PK2 = 2KL2 + ?PK2

= PL2 + PC2 = a'2 + b22

= a2 + b2 by property of the ellipse.

Again, (mn) = CL

CP2 + PL2+2PL x PE

.. m2 + n2 + 2mn = a22 + b'2 + 2a'b'′ sin. PCE.

Hence mn = a'b' sin. ≤ C

ab by property of the ellipse.

:. m2 + n2 = a2 + b2

mn = ab

}

whence m = a }
n = b j

95.

Given the figure of an ellipse, to find its centre and foci practically.

In the ellipse draw any two parallel lines GH, gh, (Fig. 53) bisect them in I, i, and Ii being joined, let it be produced to meet the curve in P,p. Then, since every diameter bisects its ordinates and reciprocally, Pp is a diameter, of which the middle point C is the required centre of the ellipse.

Again, through C draw Dd parallel HG, and we get Dd a diameter conjugate to Pp. Produce EP, (which is 1 CD,) to L, making PL CD, join CL and through its middle point K and P draw MKN, making MK = KN = CK or KL. Then, by the preceding problem, PM a the axis-major, and PN = b the

=

PROPERTIES OF CURVES.

axis-minor of the ellipse. Hence with C as centre and radii = a, b describing circular arcs, the points A, B, where they will meet the ellipse, will determine the position of its axes AC, BC; and since the distance of the focus from B AC, if with centre B and radius AC we describe arcs cutting AC in S, S', the foci S, S' of the ellipse will also be determined.

96.

=

Let Ioi, Qoq (Fig. 54), intersecting in o, be parallel to any two conjugate diameters BCD, ACM; then lo xoi: Qox oq:: BC: AC2.

For drawing Com through o and C and mn parallel to BD, we have

Iv2: mn2:: AC- Cv : AC2

And mn2 ov2 :: Cn2: Cv2

.. Iv2: ov2 :: AC2. Cn2 - Cva.Cn2 : AC.Cv2.

.. Iv2 – Ov2 : Iv2 :: ACa (Cn2 – Cv2): Cn (AC2 –

Cn2

}

Cn2.Cv

}

Or Io x oi : mo x om' :: BC2: Cm2 Similarly moom': Qo × og :: Cm2: AC2

.. Io x oi: Qo x og :: BC: AC2

BC, AC being any conjugate diameters whatever, which is more general, although solved with the same facility as the problem proposed.

97. Let the cycloid acb (Fig. 55), equal and similar toACB, have its vertex c in the base AB, and its base ab parallel to AB, and let them intersect in P; they cut at right angles.

For drawing PEn axes CD, cd, and intersecting the generating circles in E, e, and the axes in N, n, and joining EC, ed, we have the tangents at P to the curves AC, ac, parallel to CE, ce respectively. But it is easily shewn that ed is parallel to EC, and dec is a right angle; .. the tangents at P are at right angles, and consequently the curves themselves.