391. Ex. 1. To find the direction of the tangent at any point of a circle.

B

D

In any isosceles triangle AOB, either base angle OBA is less than a right angle by half the vertical angle; but as the points A and B approach to coincidence, the vertical angle may be supposed less than any assignable angle, therefore the angle OBA which the tangent makes with the radius is ultimately a equal to a right angle. We shall frequently have occasion to use the principle here proved, viz. that two indefinitely near lines of equal length

are at right angles to the line joining their extremities.

Ex. 2. The circumferences of two circles are to each other as their radii.

If polygons of the same number of sides be inscribed in the circles, it is evident, by similar triangles, that the bases ab, AB, are to each other as the radii of the circles, and, therefore, that the whole perimeters of the polygons are to each other in the same ratio; and since this will be true, no matter how the number of sides of the polygon be increased, the circumferences are to each other in the same ratio.

Ex. 3. The area of any circle is equal to the radius multiplied by the semi-circumference.

For the area of any triangle OAB is equal to half its base multiplied by the perpendicular on it from the centre; hence the area of any inscribed regular polygon is equal to half the sum of its sides multiplied by the perpendicular on any side from the centre; but the more the number of sides is increased, the more nearly will the perimeter of the polygon approach to equality with that of the circle, and the more nearly will the perpendicular on any side approach to equality with the radius, and the difference between them can be made less than any assignable quantity; hence ultimately the area of the circle is equal to the radius multiplied by the semi-circumference; or = r2.

392. Ex. 1. To determine the direction of the tangent at any point on an ellipse.

T

Let P and P' be two indefinitely near points on the curve, then FP+ PF' = FP' + P'F'; or, taking FR=FP, FR'=F'P', we have P'R=PR'; but in the triangles PRP', PR'P', we have also the base PP' common, and (by Ex. 1, Art. 391) the angles PRP' PRP right; hence the angle

T

F'

N

PP'R=P'PR'. Now TPF is ultimately equal to PPF, since their difference PFP' may be supposed less than any given angle; hence TPF= T'PF", or the focal radii make equal angles with the tangent.

Ex. 2. To determine the direction of the tangent at any point on a hyperbola.

We have

F'P'-F'P=FP' – FP,

or, as before,

P'R=P'R'.

Hence the angle

PP'R=PP'R',

RP

F

or, the tangent is the internal bisector of the angle FPF". Ex. 3. To determine the direction of the tangent at any voint of a parabola.

We have FP=PN, and FP' = P'N'; hence P'R=P'S, or the angle N'P'P=FP'P. The tangent, therefore, bisects the angle FPN.

393. Ex. 1. To find the area of the parabolic sector FVP.

=

Since PS PR, and PN= FP, we have the triangle FPR half the parallelogram PSNN'. Now if we take a number of points P'P", &c. between V and P, it is evident that the closer we take them, the more nearly will the sum of all the parallelograms PSN'N, &c. approach

N'

N

D

F

SP

R

to equality with the area DVPN, and the sum of all the triangles PFR, &c. to the sector VFP; hence ultimately the sector PFV is half the area DVPN, and therefore one-third of the quadrilateral DFPN.

Ex. 2. To find the area of the segment of a parabola cut off

by any right line.

T

R

RN

MM'

Draw the diameter bisecting it, then the parallelogram PR is equal to PM', since they are the complements of parallelograms about the diagonal; but since TM is bisected at V', the parallelogram PN' is half PR'; if, therefore, we take a number of points P, P, P, &c., it follows that the sum of all the parallelograms PM' is double the sum of all the parallelograms PN', and therefore ultimately that the space V'PM is double V'PN; hence the area of the parabolic segment V'PM is to that of the parallelogram V'NPM in the ratio 2; 3,

394. Ex. 1. The area of an ellipse is equal to the area of a circle whose radius is a geometric mean between the semi-axes of the ellipse.

For if the ellipse and the circle on the transverse axis be divided by any number of lines

Dd

B

2

5

C m m 172"

parallel to the axis minor, then since mb: md:: m′b′ : m'd' :: b: a, the quadrilateral mbb'm' is to mdd'm' in the same ratio, and the sum of all the one set of quad- A rilaterals, that is, the polygon Bbb'b" A inscribed in the ellipse is to the corresponding polygon Ddd'd'A inscribed in the circle, in the same ratio. Now this will be true whatever be the number of the sides of the polygon; if we suppose them, therefore, increased indefinitely, we learn that the area of the ellipse is to the area of the circle as b to a; but the area of the circle being = Tra3, the area of the ellipse =πab.

D'

COR. It can be proved, in like manner, that if any two figures be such that the ordinate of one is in a constant ratio to the corresponding ordinate of the other, the areas of the figures are in the same ratio.

Ex. 2. Every diameter of a conic bisects the area enclosed by the curve.

For if we suppose a number of ordinates drawn to this diameter, since the diameter bisects them all, it also bisects the trapezium formed by joining the extremities of any two adjacent ordinates, and by supposing the number of these trapezia increased without limit, we see that the diameter bisects the area.

395. Ex. 1. The area of the sector of a hyperbola made by joining any two points of it to the centre, is equal to the area of the segment made by drawing parallels from them to the asymptotes.

For since the triangle PKC=QLC, the area PQC=PQKL.
Ex. 2. Any two segments PQLK, RSNM, are equal, if
PK: QL:: RM: SN.

For

PK: QL:: CL: CK,

but (Art. 197)

CL=MT', CK=NT;

we have, therefore,

RM: SN:: MT': NT,

CKL

R

M TNT

and therefore QR is parallel to PS. We can now easily prove that the sectors PCQ, RCS are equal, since the diameter bisecting PS, QR will bisect both the hyperbolic area PQRS, and also the triangles PCS, QCR.

If we suppose the points Q, R to coincide, we see that we can bisect any area PKNS by drawing an ordinate QL, a geometric mean between the ordinates at its extremities.

Again, if a number of ordinates be taken, forming a continued geometric progression, the area between any two is constant.

396. The tangent to the interior of two similar, similarly placed, and concentric conics cuts off a constant area from the exterior conic.

For we proved (Art. 236, Ex. 4) that this tangent is always bisected at the point of contact; now if we draw any two tangents, the angle AQA' will be equal to BQB' and the nearer we suppose the point Q to P, the more nearly will the sides AQ, A'Q approach to equality with the sides BQ, B'Q; if, therefore, the two

P P

A'

B

B

tangents be taken indefinitely near, the triangle AQA' will be equal to BQB, and the space AVB will be equal to A'VB'; since, therefore, this space remains constant as we pass from any tangent to the consecutive tangent, it will be constant whatever tangent we draw.

COR. It can be proved, in like manner, that if a tangent to one curve always cuts off a constant area from another, it will be bisected at the point of contact; and, conversely, that if it be always bisected it cuts off a constant area.

Hence we can draw through a given point a line to cut off from a given conic the minimum area. If it were required to cut off a given area, it would be only necessary to draw a tangent through the point to some similar and concentric conic, and the greater the given area, the greater will be the distance between the two conics. The area will, therefore, evidently be least when this last conic passes through the given point; and since the tangent at the point must be bisected, the line through a given point which cuts off the minimum area is bisected at that point.

In like manner, the chord drawn through a given point which cuts off the minimum or maximum area from any curve is bisected at that point. In like manner can be proved the following two theorems, due to the late Professor Mac Cullagh.

Ex. 1. If a tangent AB to one curve cut off a constant arc from another, it is divided at the point of contact, so that AP: PB inversely as the tangents to the outer curve at A and B.

Ex. 2. If the tangent AB be of a constant length, and if the perpendicular let fall on AB from the intersection of the tangents at A and B meet AB in M, then AP will = MB.

397. To find the radius of curvature at any point on an ellipse. The centre of the circle circumscribing any triangle is the intersection of perpendiculars erected at the middle points of the sides of that triangle; it follows, therefore, that the centre of the circle passing through three consecutive points on the curve is the intersection of two consecutive normals to the curve.

Now, given any two triangles FPF', FP'F', and PN, P'N, the two bisectors of their vertical angles, it is easily proved by elementary geometry, that twice the angle PNP'= PFP'+ PF'P'. (See figure, Art. 392, Ex. 1).