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the same radical axis as the circles

j.2^y-2 â€” 4. an^ x^ + y^ + 2x + 4:y = &.

12. Find the equations to the straight lines joining the origin to

the points of intersection of

x^ + y^-4:x-2y = 4: and x'- + y^-2x~'iy -4^ = 0.

13. The polars of a point P with respect to two fixed circles meet

in the point Q. Prove that the circle on PQ as diameter passes

through two fixed points, and cuts both the given circles at right

angles.

14. Prove that the two circles, which pass through the two points

(0, a) and (0, - a) and touch the straight line y = mx + c, will cut ortho-

gonally if c2 = a2 (2 + m^).

15. Find the locus of the centre of the circle which cuts two given

circles orthogonally.

16. If two circles cut orthogonally, prove that the polar of any

point P on the first circle with respect to the second passes through

the other end of the diameter of the first circle which goes through P.

Hence, (by considering the orthogonal circle of three circles as

the locus of a point such that its polars with respect to the circles

meet in a point) prove that the orthogonal circle of three circles,

given by the general equation is

\x+9i> y+fi, 9i^+fiy+ci

\x + go, y + f^, g^x + f^y + c^ =0.

\x + gz, y + fs, g-i^+UJ+H

166 COORDINATE GEOMETEY.

188. Coaxal Circles. Def. A system of circles

is said to be coaxal when they have a common radical axis,

i.e. when the radical axis of each pair of circles of the

system is the same.

Tojind the equation of a system of coaxal circles.

Since, by Art. 183, the radical axis of any pair of the

circles is perpendicular to the line joining their centres, it

follows that the centres of all the circles of a coaxal system

must lie on a straight line which is perpendicular to the

radical axis.

Take the line of centres as the axis of x and the radical

axis as the axis of y (Figs. I. and II., Art. 190), so that

is the origin.

The equation to any circle with its centre on the axis

of Â£c is

x^ + y'^ â€” 2gx + c-0 (1).

Any point on the radical axis is (0, y^).

The square on the tangent from it to the circle (1) is,

by Art. 168, y^' + c.

Since this quantity is to be the same for all circles of

the system it follows that c is the same for all such circles ;

the different circles are therefore obtained by giving dif-

ferent values to g in the equation (1).

The intersections of (1) with the radical axis are then

obtained by putting a^ = in equation (1), and we have

If c be negative, we have two real points of intersection

as in Fig. I. of Art. 190. In such cases the circles are said

to be of the Intersecting Species.

If c be positive, we have two imaginary points of in-

tersection as in Fig. II.

'&â€¢

189. Limiting points of a coaxal system.

The equation (1) of the previous article which gives any

circle of the system may be written in the form

(x-gY + 2/2 ^ / - c = [Jf - cf.

COAXAL CIRCLES.

167

It therefore represents a circle whose centre is the point

(^, 0) and whose radius is J g^ â€” c.

This radius vanishes, i.e. the circle becomes a point-

circle, when g^ â€” G, i.e. when g â€” Â±Jc.

Hence at the particular points (+ Jc, 0) we have point-

circles which belong to the system. These point-circles are

called the Limiting Points of the system.

If G be negative, these points are imaginary.

But it was shown in the last article that when c is

negative the circles intersect in real points as in Fig. I.,

Art. 190.

If c be positive, the limiting points L^ and L^ (Fig. II.) are

real, and in this case the circles intersect in imaginary points.

The limiting points are therefore real or imaginary

according as the circles of the system intersect in imaginary

or real points.

190. Orthogonal circles of a coaxal system.

Let T be any point on the common radical axis

of a system of coaxal circles, and let TR be the tangent

from it to any circle of the system.

Then a circle, whose centre is T and whose radius is TR^

will cut each circle of the coaxal system orthogonally.

168

COORDINATE GEOMETRY.

[For the radius TR of this circle is at right angles to

the radius O^R, and so for its intersection with any other

circle of the system.]

Fig. II.

Hence the limiting points (being point- c^rcZes of the

system) are on this orthogonal circle.

The limiting points are therefore the intersections with

the line of centres of any circle whose centre is on the

common radical axis and whose radius is the tangent from

it to any of the circles of the system.

Since, in Eig. I,, the limiting points are imaginary these

orthogonal circles do not meet the line of centres in real

points.

In Fig. II. they jDass through the limiting points Z^

and 7^2 .

These orthogonal circles (since they all pass through two

points, real or imaginary) are therefore a coaxal system.

Also if the original circles, as in Fig. I., intersect in

real points, the orthogonal circles intersect in imaginary

points; in Fig. II. the original circles intersect in imaginary

points, and the orthogonal circles in real points.

We therefore have the following theorem :

A set of coaxal circles can be cut orthogonally hy another

set of coaxal circles, the centres of each set lying on the

radical axis of the other set ; also one set is of the limiting-

point sjyecies and the other set of the other species.

ORTHOGONAL CIRCLES. 169

191. Without reference to the limiting points of the original

system, it may be easily found whether or not the orthogonal circles

meet the original line of centres.

For the circle, whose centre is T and whose radius is TR, meets

or does not meet the line 0-fi,2 according as TR^ is > or < TO^,

i.e. according as TO^^-O^R^ is > TO^,

L e. according as TO^ +00^^- 0-fi^ is > TO^,

i.e. according as 00-^ is < O^R,

i.e. according as the radical axis is without, or within, each of the

circles of the original system.

192. In the next article the above results will be

proved analytically.

To find the equation to any circle y)hich cuts two ghien

circles orthogonally.

Take the radical axis of the two circles as the axis of y^

so that their equations may be written in the form

a? ^y^ â€” 2gx + c = (1),

and a? 4- y'^ â€” 2g^x +c â€” (2),

the quantity c being the same for each.

Let the equation to any circle which cuts them or-

thogonally be

{x-Ay + {y-Bf = E^ ....(3).

The equation (1) can be written in the form

{x-gf + f-^[J^::rcy (4).

The circles (3) and (4) cut orthogonally if the square of

the distance between their centres is equal to the sum of

the squares of their radii,

i.e. if {A - gf ^E^-^m^. [V/^]^

i.e. if A^-^B'-1Ag = R^-c (5).

Similarly, (3) will cut (2) orthogonally if

A - vB''-'lAg^ = E'-c (6).

Subtracting (6) from (5), we have A (g - g^) ^^^ 0.

Hence ^ = 0, and i?- = B^ + c.

170 COORDINATE GEOMETRY.

Substituting these values in (3), the equation to the

required orthogonal circle is

ar^ + 2/'-2%-c = (7),

where B is any quantity whatever.

Whatever be the value of B the equation (7) represents

a circle whose centre is on the axis of y and which passes

through the points (+ Jc, 0).

But the latter points are the limiting points of the

coaxal system to which the two circles belong. [Art. 189.]

Hence any pair of circles belonging to a coaxal system

is cut at right angles by any circle of another coaxal

system ; also the centres of the circles of the latter system

lie on the common radical axis of the original system, and

all the circles of the latter system pass through the limiting

points (real or imaginary) of the first system.

Also the centre of the circle (7) is the point (0, B) and

its radius is JB"^ + c.

. The square of the tangent drawn from (0, B) to the

circle {I) = B"" + c (by Art. 168).

Hence the radius of any circle of the second system is

equal to the length of the tangent drawn from its centre to

any circle of the first system.

193. The equation to the system of circles which cut

a given coaxal system orthogonally may also be obtained

by using the result of Art. 182.

For any circle of the coaxal system is, by Art. 188,

given by

a? + y'^- Igx + c = (1),

where c is the same for all circles.

Any point on the radical axis is (0, y).

The square on the tangent drawn from it to (1) is

therefore y"^ + c.

The equation to any circle cutting (1) orthogonally is

therefore

^^ + (2/ - 2/T = y"" + ^'

i.e. x^ + y'^â€” lyy â€”c = 0.

ORTHOGONAL CIRCLES. 171

Whatever be the value of y this circle passes through

the points (+ ^]c, 0), i.e. through the limiting points of the

system of circles given by (1).

194. We can now deduce an easy construction for the

circle that cuts any three circles orthogonally.

Consider the three circles in the figure of Art. 186.

By Art. 192 any circle cutting A and B orthogonally

has its centre on their common radical axis, i.e. on the

straight line OD.

Similarly any circle cutting B and C orthogonally has

its centre on the radical axis OE.

Any circle cutting all three circles orthogonally must

therefore have its centre at the intersection of OD and OE.,

i.e. at the radical centre 0. Also its radius must be the

length of the tangent dravi^n from the radical centre to

any one of the three circles.

Ex. Find the equation to the circle lohich cuts orthogonally each

of the three circles

x^ + 'ij^ + 2x + ny+ 4 = (1),

x'^ + 2/ + lx+ 6y + ll = (2),

x^ + if- x + 22y+ 3 = (3).

The radical axis of (1) and (2) is

5x-lly + 7 = 0.

The radical axis of (2) and (3) is

8x-l&y + 8 = 0.

These two straight Hnes meet in the point (3, 2) which is therefore

the radical centre.

The square of the length of the tangent from the point (3, 2) to

each of the given circles =57.

The required equation is therefore {x - 3)^ +{y - 2)^ = 57,

i. e. x^ + ?/2 â€” 6a; - 4?/ - 44 = 0.

195. Ex. Find the locus of a point which moves so that the length

of the tangent draion from it to one given circle is X times the length of

the tangent from it to another given circle.

As in Art. 188 take as axes of x and y the line joining the centres

of the two circles and the radical axis. The equations to the two

circles are therefore

x'^ + y''-2g^x + c = Q (1),

and a;2 + ^2_2^^a; + c = (2).

172 COORDINATE GEOMETRY.

Let (h, k) be a point such that the length of the tangent from it to

(1) is always X times the length of the tangent from it to (2).

Then Ji^ + k^- 2gji + c = \^ [/i^ + A;^ - ^.g^h + c].

Hence {h, h) always lies on the circle

x'^ + y^-<ix^-^^ + cr=0 (3).

A"' â€” 1

This circle is clearly a circle of the coaxal system to which (1) and

(2) belong.

Again, the centre of (1) is the point (f/^, 0), the centre of (2) is

(^2, 0), whilst the centre of (3) is ( ^toZ\^ ' ^) '

Hence, if these three centres be called O^ , 0.^ , and O3 , we have

"\ 2 1

and O^Os = "^^ _ { ^ ~ ^3 = -^YZi (^2 " ^i)>

so that O1O3 : 0^0^ : : X^ : 1.

The required locus is therefore a circle coaxal with the two given

circles and whose centre divides externally, in the ratio X" : 1, the line

joining the centres of the two given circles.

EXAMPLES. XXIV.

1. Prove that a common tangent to two circles of a coaxal

system subtends a right angle at either limiting point of the system.

2. Prove that the polar of a limiting point of a coaxal system

with respect to any circle of the system is the same for all circles of

the system.

3. Prove that the polars of any point with respect to a system of

coaxal circles all pass through a fixed point, and that the two points

are equidistant from the radical axis and subtend a right angle at a

limiting point of the system. If the first point be one limiting point

of the system prove that the second point is the other limiting point.

4. A fixed circle is cut by a series of circles all of which pass

through two given points ; prove that the straight line joining the

intersections of the fixed circle with any circle of the system always

passes through a fixed point.

5. Prove that tangents drawn from any point of a fixed circle of

a coaxal system to two other fixed circles of the system are in a

constant ratio.

[EXS. XXIV.] COAXAL CIRCLES. EXAMPLES. 173

6. Prove that a system of coaxal circles inverts with respect to

either limiting point into a system of concentric circles and find the

position of the common centre.

7. A straight line is drawn touching one of a system of coaxal

circles in P and catting another in Q and R. Shew that PQ and PR

subtend equal or supplementary angles at one of the limiting points

of the system.

8. Find the Ipcus of the point of contact of parallel tangents

which are drawn to each of a series of coaxal circles.

9. Prove that the circle of similitude of the two circles

x^- + y'^-2kx + b = and x- + if-2k'x + 5 = Q

{L e. the locus of the points at which the two circles subtend the same

angle) is the coaxal circle

10. From the preceding question shew that the centres of simili-

tude {i.e. the points in which the common tangents to two circles

meet the line of centres) divide the line joining the centres internally

and externally in the ratio of the radii.

11. If x + y sj -l = tan{u + v /sf -1), where x, y, u, and v are all

real, prove that the curves tt=: constant give a family of coaxal circles

passing through the points (0, Â±1), and that the curves ?; = constant

give a system of circles cutting the first system orthogonally.

12. Find the equation to the circle which cuts orthogonally each

of the circles

x^-\-y'^ + '2.gx + c = 0, x'^-\-y^ + 2g'x-{-c = 0,

and x^ + y'^ + 2hx + 1ky + a=.0.

13. Find the equation to the circle cutting orthogonally the

three circles

x^ + y'^â€”a?', {x-cf-\-y'^=a'^, and x^ + {y -l))'" = a'^.

14. Find the equation to the circle cutting orthogonally the

three circles

and a;2 + 2/2 + 7^-9?/ + 29 = 0.

15. Shew that the equation to the circle cutting orthogonally the

circles

{x-aY + {y-hf = h\ {^x - bf+{y-af=d^

and {x-a-h-cY + y^^ah + c^,

is a;2 + 2/^-2a:(a + 6)-'?/(a + 6) + a2 + 3a6 + 62 = o.

CONIC SECTIONS.

CHAPTER X.

THE PARABOLA.

196. Conic Section. Def. The locus of a point

P, which moves so that its distance from a fixed point is

always in a constant ratio to its perpendicular distance

from a fixed straight line, is called a Conic Section.

The fixed point is called the Focus and is usually

denoted by S.

The constant ratio is called the Eccentricity and is

denoted by e.

The fixed straight line is called the Directrix.

The straight line passing through the Focus and per-

pendicular to the Directrix is called the Axis.

When the eccentricity e is equal to unity, the Conic

Section is called a Parabola.

When e is less than unity, it is called an SUipse.

When e is greater than unity, it is called a Hyper-

bola.

[The name Conic Section is derived from the fact that

these curves were first obtained by cutting a cone in

various ways.]

THE PARABOLA.

175

197. To find the equation to a Farahola.

Let S be the fixed point and ZM the directrix. We

require therefore the locus

of a point F which moves

so that its distance from S

is always equal to PM, its

perpendicular distance from

ZM.

Draw 8Z perpendicular

to the directrix and bisect

8Z in the point A ; produce

ZA to X,

The point A is clearly a

point on the curve and is

called the Vertex of the

Parabola.

Take A as origin, AX as the axis of x, and J.F,

perpendicular to it, as the axis of y.

Let the distance ZA^ or A8^ be called Â«, and let P be

any point on the curve whose coordinates are x and y.

Join /S'P, and draw PN and PM perpendicular respec-

tively to the axis and directrix.

We have then aSP^ = Pif ^

^.e. {x - of + 2/2 r= ZN'' = {a + xf,

y2 = 4ax (1).

This being the relation which exists between the co-

ordinates of any point P on the parabola is, by Art. 42, the

equation to the parabola.

Cor. The equation (1) is equivalent to the geometrical

proposition

PN^ = 4:AS.AIi.

198. The equation of the preceding article is the

simplest possible equation to the parabola. Throughout

this chapter this standard form of the equation is assumed

unless the contrary is stated.

176 COORDINATE GEOMETRY.

If instead of AX and A Y we take the axis and the

directrix ^M as the axes of coordinates, the equation

would be

(x â€” 2a)- + y^ = x",

i.e. y'^ = ia{x â€” a) (1).

Similarly, if the axis SX and a perpendicular line SL

be taken as the axes of coordinates, the equation is

aj2 + 2/2 = (a? + 2a)2,

i.e. y^ = 4:a(x + a) (2).

These two equations may be deduced from the equation

of the previous article by transforming the origin, firstly to

the point (- a, 0) and secondly to the point (a, 0).

199. The equation to the parabola referred to any focus and

directrix may be easily obtained. Thus the equation to the parabola,

whose focus is the point' (2, 3) and whose directrix is the straight

line X - 4y + S â€” 0, is

i. e. 17 [x^ + y^ - ix - 62/ + 13] :== [x^ + 16i/ + 9 - 8xy + Gx - 24.y] ,

i.e. 16x'^ + y^ + 8xy-7ix-78y + 212 = 0.

200. To trace the curve

2/2= iax (1).

If X be negative, the corresponding values of y are

imaginary (since the square root of a negative quantity is

unreal) ; hence there is no part of the curve to the left of

the point A.

If y be zero, so also is x, so that the axis of x meets

the curve at the point A only.

If X be zero, so also is y, so that the axis of y meets

the curve at the point A only.

For every positive value of x we see from (1), by taking

the square root, that y has two equal and opposite values.

Hence corresponding to any point P on the curve there

is another point P' on the other side of the axis which is

obtained by producing PN to P' so that PN and NP' are

THE PARABOLA. 177

equal in magnitude. The line PP' is called a double

ordinate.

As X increases in magnitude, so do the corresponding

values of y ; finally, when x becomes infinitely great, y

becomes infinitely great also.

By taking a large number of values of x and the

corresponding values of y it will be found that the curve is

as in the figure of Art. 197.

The two branches never meet but are of infinite length.

201. The quantity y"^ â€” 4aa;' is negative^ zero, or positive

according as the point {x\ y') is within, upon, or without the

parabola.

Let Q be the point {x , y') and let it be within the

curve, i.e. be between the curve and the axis AX. Draw

the ordinate QN and let it meet the curve in P.

Then (by Art. 197), PN^ - la . a;'.

Hence y"^, i.e. QN^, is < PiV^, and hence is < iax .

.'. y'^ â€” 4:ax is negative.

Similarly, if Q be without the curve, then y"-^, i.e. QJV%

is > PN^j and hence is > 4iax'.

Hence the proposition.

202. Latus Rectum. Def. The latus rectum of

any conic is the double ordinate LSL' drawn through the

focus S.

In the case of the parabola we have SL = distance of L

from the directrix â€” SZ= 2a.

Hence the latus rectum = la.

"When the latus rectum is given it follows that the

equation to the parabola is completely known in its

standard form, and the size and shape of the curve

determined.

The quantity la is also often called the principal

parameter of the curve.

Focal Distance of any point. The focal distance

of any point P is the distance 8P.

This focal distance = PM = ZN= ZA+AN'=a + x.

L. 12

178 COORDINATE GEOMETEY.

Ex. Find the vertex, axis, focus, and latus rectum of the parabola

4?/2 + 12ar-202/ + 67 = 0.

The equation can be written

y^-5y=-Sx - ^^,

i.e. {y - ^f=-Sx-s^- + ^^=-3{x + i).

Transform this equation to the point (-|, f) and it becomes

y^= -Sx, which represents a parabola, whose axis is the axis of x

and whose concavity is turned towards the negative end of this axis.

Also its latus rectum is 3.

Eef erred to the original axes the vertex is the point i-^, f ), the

axis is 2/ = f, and the focus is the point (-| -|, f), Â«-e. ( -V-i f)-

EXAMPLES. XXV.

Find the equation to the parabola with

1. focus (3, -4) and directrix Gcc- 7?/ + 5 = 0.

X XI

2. focus (a, &) and directrix - + f = 1.

^ ah

Find the vertex, axis, latus rectum, and focus of the parabolas

3. y^ = 4:X + ^y. 4. x'^ + 2y = 8x-7.

5. x^-2ax + 2ay = 0. 6. 2/^=4y-4a:.

7. Draw the curves

(1) y'2=-4:ax, (2) x'^=4:ay, and (3) x-z=-4:ay.

8 Find the value of p when the parabola y'^ = 4px goes through

the point (i) (3, - 2), and (ii) (9, - 12).

9. For what point of the parabola y^ = 18x is the ordinate equal

to three times the abscissa ?

10. Prove that the equation to the parabola, whose vertex and focus

are on the axis of x at distances a and a' f . om the origin respectively,

is y^ = 4:{a'-a){x-a).

11. In the parabola y^=Qx, find (1) the equation to the chord

through the vertex and the negative end of the latus rectum, and

(2) the equation to any chord through the point on the curve whose

abscissa is 24.

12. Prove that the equation y^ + 2Ax + 2By + C = represents a

parabola, whose axis is parallel to the axis of x, and find its vertex and

the equation to its latus rectum.

13. Prove that the locus of the middle points of all chords of

the parabola ?/2 = 4aa; which are drawn through the vertex is the

parabola y'^ = 2ax.

[EXS. XXV.] THE PARABOLA. EXAMPLES. 179

14. Prove that the locus of the centre of a circle, which intercepts

a chord of given length 2a on the axis of x and passes through a given

point on the axis of y distant 6 from the origin, is the curve

a;2-2'Â«/& + &2 = a2.

Trace this parabola.

15. PQ is a double ordinate of a parabola. Find the locus of its

point of trisection.

16. Prove that the locus of a point, which moves so that its

distance from a fixed line is equal to the length of the tangent drawn

from it to a given circle, is a parabola. Find the position of the

focus and directrix.

17. If a circle be drawn so as always to touch a given straight

line and also a given circle, prove that the locus of its centre is

a parabola.

18. The vertex ^ of a parabola is joined to any point P on the

curve and PQ is drawn at right angles to AP to meet the axis in Q.

Prove that the projection of PQ on the axis is always equal to the

latus rectum.

19. If on a given base triangles be described such that the sum of

the tangents of the base angles is constant, prove that the locus of

the vertices is a parabola.

20. A double ordinate of the curve y^=^px is of length 8p ; prove

that the lines from the vertex to its two ends are at right angles.

21. Two parabolas have a common axis and concavities in oppo-

site directions ; if any line parallel to the common axis meet the

parabolas in P and P', prove that the locus of the middle point of PP'

is another parabola, provided that the latera recta of the given para-

bolas are unequal.

22. A parabola is drawn to pass through A and P, the ends of

a diameter of a given circle of radius a, and to have as directrix a

tangent to a concentric circle of radius h ; the axes being AB and

a perpendicular diameter, prove that the locus of the focus of the

parabola IS - + ,^^=1,

203. To find the points of intersection of any straight

line with the parabola

2/^ = 4acc (1).

The equation to any straight line is

y = 7nx + c .(2).

The coordinates of the points common to the straight

line and the parabola satisfy both equations (1) and (2),

and are therefore found by solving them.

12â€”2

180 COORDINATE GEOMETRY.

Substituting the value of y from (2) in (1), we have

{mx + cf â€” 4:ax,

i.e. m^oc^ + 2x (mc - 2a) + G^ - (3).

This is a quadratic equation for x and therefore has two

roots, real, coincident, or imaginary.

The straight line therefore meets the parabola in two

points, real, coincident, or imaginary.

The roots of (3) are real or imaginary according as

{2(mc-2a)f-47^iV

is positive or negative, i.e. according as â€” aTnc + a^ is

positive or negative, i.e. according as mc is ^ ia.

204. To find the length of the chord intercepted by the parabola on

the straight line

y â€” mx + c (1).

If (o^i , y-^ and {x^, y^) he the common points of intersection, then,

as in Art. 154, we have, from equation (3) of the last article,

{^x^ â€” x^j = (ajj + x^) â€” ^x^x^

_4(mc-2a)^ 4:C^ _l&a{a-mc)

~ m* m^ wi'* '

and yj^-y^ = m{x^-x^).

Hence the required length = \/(l/i - 2/2)^ + (^1 ~ ^"2)^

= Jl + m'-^ (x^ - ajg) = ^2 Jl + mP J a (a - mc).

205. To find the equation to the tangent at any point

(^'3 y) 9f i^^^ parabola y^ = 4:ax.

The definition of the tangent is given in Art. 149.

Let P be the point [x, y) and Q a point (x", y") on the

parabola.

The equation to the line PQ is

2/-y' = f5f'(^-Â»'') (!)â€¢

Since P and Q both lie on the curve, we have

j.2^y-2 â€” 4. an^ x^ + y^ + 2x + 4:y = &.

12. Find the equations to the straight lines joining the origin to

the points of intersection of

x^ + y^-4:x-2y = 4: and x'- + y^-2x~'iy -4^ = 0.

13. The polars of a point P with respect to two fixed circles meet

in the point Q. Prove that the circle on PQ as diameter passes

through two fixed points, and cuts both the given circles at right

angles.

14. Prove that the two circles, which pass through the two points

(0, a) and (0, - a) and touch the straight line y = mx + c, will cut ortho-

gonally if c2 = a2 (2 + m^).

15. Find the locus of the centre of the circle which cuts two given

circles orthogonally.

16. If two circles cut orthogonally, prove that the polar of any

point P on the first circle with respect to the second passes through

the other end of the diameter of the first circle which goes through P.

Hence, (by considering the orthogonal circle of three circles as

the locus of a point such that its polars with respect to the circles

meet in a point) prove that the orthogonal circle of three circles,

given by the general equation is

\x+9i> y+fi, 9i^+fiy+ci

\x + go, y + f^, g^x + f^y + c^ =0.

\x + gz, y + fs, g-i^+UJ+H

166 COORDINATE GEOMETEY.

188. Coaxal Circles. Def. A system of circles

is said to be coaxal when they have a common radical axis,

i.e. when the radical axis of each pair of circles of the

system is the same.

Tojind the equation of a system of coaxal circles.

Since, by Art. 183, the radical axis of any pair of the

circles is perpendicular to the line joining their centres, it

follows that the centres of all the circles of a coaxal system

must lie on a straight line which is perpendicular to the

radical axis.

Take the line of centres as the axis of x and the radical

axis as the axis of y (Figs. I. and II., Art. 190), so that

is the origin.

The equation to any circle with its centre on the axis

of Â£c is

x^ + y'^ â€” 2gx + c-0 (1).

Any point on the radical axis is (0, y^).

The square on the tangent from it to the circle (1) is,

by Art. 168, y^' + c.

Since this quantity is to be the same for all circles of

the system it follows that c is the same for all such circles ;

the different circles are therefore obtained by giving dif-

ferent values to g in the equation (1).

The intersections of (1) with the radical axis are then

obtained by putting a^ = in equation (1), and we have

If c be negative, we have two real points of intersection

as in Fig. I. of Art. 190. In such cases the circles are said

to be of the Intersecting Species.

If c be positive, we have two imaginary points of in-

tersection as in Fig. II.

'&â€¢

189. Limiting points of a coaxal system.

The equation (1) of the previous article which gives any

circle of the system may be written in the form

(x-gY + 2/2 ^ / - c = [Jf - cf.

COAXAL CIRCLES.

167

It therefore represents a circle whose centre is the point

(^, 0) and whose radius is J g^ â€” c.

This radius vanishes, i.e. the circle becomes a point-

circle, when g^ â€” G, i.e. when g â€” Â±Jc.

Hence at the particular points (+ Jc, 0) we have point-

circles which belong to the system. These point-circles are

called the Limiting Points of the system.

If G be negative, these points are imaginary.

But it was shown in the last article that when c is

negative the circles intersect in real points as in Fig. I.,

Art. 190.

If c be positive, the limiting points L^ and L^ (Fig. II.) are

real, and in this case the circles intersect in imaginary points.

The limiting points are therefore real or imaginary

according as the circles of the system intersect in imaginary

or real points.

190. Orthogonal circles of a coaxal system.

Let T be any point on the common radical axis

of a system of coaxal circles, and let TR be the tangent

from it to any circle of the system.

Then a circle, whose centre is T and whose radius is TR^

will cut each circle of the coaxal system orthogonally.

168

COORDINATE GEOMETRY.

[For the radius TR of this circle is at right angles to

the radius O^R, and so for its intersection with any other

circle of the system.]

Fig. II.

Hence the limiting points (being point- c^rcZes of the

system) are on this orthogonal circle.

The limiting points are therefore the intersections with

the line of centres of any circle whose centre is on the

common radical axis and whose radius is the tangent from

it to any of the circles of the system.

Since, in Eig. I,, the limiting points are imaginary these

orthogonal circles do not meet the line of centres in real

points.

In Fig. II. they jDass through the limiting points Z^

and 7^2 .

These orthogonal circles (since they all pass through two

points, real or imaginary) are therefore a coaxal system.

Also if the original circles, as in Fig. I., intersect in

real points, the orthogonal circles intersect in imaginary

points; in Fig. II. the original circles intersect in imaginary

points, and the orthogonal circles in real points.

We therefore have the following theorem :

A set of coaxal circles can be cut orthogonally hy another

set of coaxal circles, the centres of each set lying on the

radical axis of the other set ; also one set is of the limiting-

point sjyecies and the other set of the other species.

ORTHOGONAL CIRCLES. 169

191. Without reference to the limiting points of the original

system, it may be easily found whether or not the orthogonal circles

meet the original line of centres.

For the circle, whose centre is T and whose radius is TR, meets

or does not meet the line 0-fi,2 according as TR^ is > or < TO^,

i.e. according as TO^^-O^R^ is > TO^,

L e. according as TO^ +00^^- 0-fi^ is > TO^,

i.e. according as 00-^ is < O^R,

i.e. according as the radical axis is without, or within, each of the

circles of the original system.

192. In the next article the above results will be

proved analytically.

To find the equation to any circle y)hich cuts two ghien

circles orthogonally.

Take the radical axis of the two circles as the axis of y^

so that their equations may be written in the form

a? ^y^ â€” 2gx + c = (1),

and a? 4- y'^ â€” 2g^x +c â€” (2),

the quantity c being the same for each.

Let the equation to any circle which cuts them or-

thogonally be

{x-Ay + {y-Bf = E^ ....(3).

The equation (1) can be written in the form

{x-gf + f-^[J^::rcy (4).

The circles (3) and (4) cut orthogonally if the square of

the distance between their centres is equal to the sum of

the squares of their radii,

i.e. if {A - gf ^E^-^m^. [V/^]^

i.e. if A^-^B'-1Ag = R^-c (5).

Similarly, (3) will cut (2) orthogonally if

A - vB''-'lAg^ = E'-c (6).

Subtracting (6) from (5), we have A (g - g^) ^^^ 0.

Hence ^ = 0, and i?- = B^ + c.

170 COORDINATE GEOMETRY.

Substituting these values in (3), the equation to the

required orthogonal circle is

ar^ + 2/'-2%-c = (7),

where B is any quantity whatever.

Whatever be the value of B the equation (7) represents

a circle whose centre is on the axis of y and which passes

through the points (+ Jc, 0).

But the latter points are the limiting points of the

coaxal system to which the two circles belong. [Art. 189.]

Hence any pair of circles belonging to a coaxal system

is cut at right angles by any circle of another coaxal

system ; also the centres of the circles of the latter system

lie on the common radical axis of the original system, and

all the circles of the latter system pass through the limiting

points (real or imaginary) of the first system.

Also the centre of the circle (7) is the point (0, B) and

its radius is JB"^ + c.

. The square of the tangent drawn from (0, B) to the

circle {I) = B"" + c (by Art. 168).

Hence the radius of any circle of the second system is

equal to the length of the tangent drawn from its centre to

any circle of the first system.

193. The equation to the system of circles which cut

a given coaxal system orthogonally may also be obtained

by using the result of Art. 182.

For any circle of the coaxal system is, by Art. 188,

given by

a? + y'^- Igx + c = (1),

where c is the same for all circles.

Any point on the radical axis is (0, y).

The square on the tangent drawn from it to (1) is

therefore y"^ + c.

The equation to any circle cutting (1) orthogonally is

therefore

^^ + (2/ - 2/T = y"" + ^'

i.e. x^ + y'^â€” lyy â€”c = 0.

ORTHOGONAL CIRCLES. 171

Whatever be the value of y this circle passes through

the points (+ ^]c, 0), i.e. through the limiting points of the

system of circles given by (1).

194. We can now deduce an easy construction for the

circle that cuts any three circles orthogonally.

Consider the three circles in the figure of Art. 186.

By Art. 192 any circle cutting A and B orthogonally

has its centre on their common radical axis, i.e. on the

straight line OD.

Similarly any circle cutting B and C orthogonally has

its centre on the radical axis OE.

Any circle cutting all three circles orthogonally must

therefore have its centre at the intersection of OD and OE.,

i.e. at the radical centre 0. Also its radius must be the

length of the tangent dravi^n from the radical centre to

any one of the three circles.

Ex. Find the equation to the circle lohich cuts orthogonally each

of the three circles

x^ + 'ij^ + 2x + ny+ 4 = (1),

x'^ + 2/ + lx+ 6y + ll = (2),

x^ + if- x + 22y+ 3 = (3).

The radical axis of (1) and (2) is

5x-lly + 7 = 0.

The radical axis of (2) and (3) is

8x-l&y + 8 = 0.

These two straight Hnes meet in the point (3, 2) which is therefore

the radical centre.

The square of the length of the tangent from the point (3, 2) to

each of the given circles =57.

The required equation is therefore {x - 3)^ +{y - 2)^ = 57,

i. e. x^ + ?/2 â€” 6a; - 4?/ - 44 = 0.

195. Ex. Find the locus of a point which moves so that the length

of the tangent draion from it to one given circle is X times the length of

the tangent from it to another given circle.

As in Art. 188 take as axes of x and y the line joining the centres

of the two circles and the radical axis. The equations to the two

circles are therefore

x'^ + y''-2g^x + c = Q (1),

and a;2 + ^2_2^^a; + c = (2).

172 COORDINATE GEOMETRY.

Let (h, k) be a point such that the length of the tangent from it to

(1) is always X times the length of the tangent from it to (2).

Then Ji^ + k^- 2gji + c = \^ [/i^ + A;^ - ^.g^h + c].

Hence {h, h) always lies on the circle

x'^ + y^-<ix^-^^ + cr=0 (3).

A"' â€” 1

This circle is clearly a circle of the coaxal system to which (1) and

(2) belong.

Again, the centre of (1) is the point (f/^, 0), the centre of (2) is

(^2, 0), whilst the centre of (3) is ( ^toZ\^ ' ^) '

Hence, if these three centres be called O^ , 0.^ , and O3 , we have

"\ 2 1

and O^Os = "^^ _ { ^ ~ ^3 = -^YZi (^2 " ^i)>

so that O1O3 : 0^0^ : : X^ : 1.

The required locus is therefore a circle coaxal with the two given

circles and whose centre divides externally, in the ratio X" : 1, the line

joining the centres of the two given circles.

EXAMPLES. XXIV.

1. Prove that a common tangent to two circles of a coaxal

system subtends a right angle at either limiting point of the system.

2. Prove that the polar of a limiting point of a coaxal system

with respect to any circle of the system is the same for all circles of

the system.

3. Prove that the polars of any point with respect to a system of

coaxal circles all pass through a fixed point, and that the two points

are equidistant from the radical axis and subtend a right angle at a

limiting point of the system. If the first point be one limiting point

of the system prove that the second point is the other limiting point.

4. A fixed circle is cut by a series of circles all of which pass

through two given points ; prove that the straight line joining the

intersections of the fixed circle with any circle of the system always

passes through a fixed point.

5. Prove that tangents drawn from any point of a fixed circle of

a coaxal system to two other fixed circles of the system are in a

constant ratio.

[EXS. XXIV.] COAXAL CIRCLES. EXAMPLES. 173

6. Prove that a system of coaxal circles inverts with respect to

either limiting point into a system of concentric circles and find the

position of the common centre.

7. A straight line is drawn touching one of a system of coaxal

circles in P and catting another in Q and R. Shew that PQ and PR

subtend equal or supplementary angles at one of the limiting points

of the system.

8. Find the Ipcus of the point of contact of parallel tangents

which are drawn to each of a series of coaxal circles.

9. Prove that the circle of similitude of the two circles

x^- + y'^-2kx + b = and x- + if-2k'x + 5 = Q

{L e. the locus of the points at which the two circles subtend the same

angle) is the coaxal circle

10. From the preceding question shew that the centres of simili-

tude {i.e. the points in which the common tangents to two circles

meet the line of centres) divide the line joining the centres internally

and externally in the ratio of the radii.

11. If x + y sj -l = tan{u + v /sf -1), where x, y, u, and v are all

real, prove that the curves tt=: constant give a family of coaxal circles

passing through the points (0, Â±1), and that the curves ?; = constant

give a system of circles cutting the first system orthogonally.

12. Find the equation to the circle which cuts orthogonally each

of the circles

x^-\-y'^ + '2.gx + c = 0, x'^-\-y^ + 2g'x-{-c = 0,

and x^ + y'^ + 2hx + 1ky + a=.0.

13. Find the equation to the circle cutting orthogonally the

three circles

x^ + y'^â€”a?', {x-cf-\-y'^=a'^, and x^ + {y -l))'" = a'^.

14. Find the equation to the circle cutting orthogonally the

three circles

and a;2 + 2/2 + 7^-9?/ + 29 = 0.

15. Shew that the equation to the circle cutting orthogonally the

circles

{x-aY + {y-hf = h\ {^x - bf+{y-af=d^

and {x-a-h-cY + y^^ah + c^,

is a;2 + 2/^-2a:(a + 6)-'?/(a + 6) + a2 + 3a6 + 62 = o.

CONIC SECTIONS.

CHAPTER X.

THE PARABOLA.

196. Conic Section. Def. The locus of a point

P, which moves so that its distance from a fixed point is

always in a constant ratio to its perpendicular distance

from a fixed straight line, is called a Conic Section.

The fixed point is called the Focus and is usually

denoted by S.

The constant ratio is called the Eccentricity and is

denoted by e.

The fixed straight line is called the Directrix.

The straight line passing through the Focus and per-

pendicular to the Directrix is called the Axis.

When the eccentricity e is equal to unity, the Conic

Section is called a Parabola.

When e is less than unity, it is called an SUipse.

When e is greater than unity, it is called a Hyper-

bola.

[The name Conic Section is derived from the fact that

these curves were first obtained by cutting a cone in

various ways.]

THE PARABOLA.

175

197. To find the equation to a Farahola.

Let S be the fixed point and ZM the directrix. We

require therefore the locus

of a point F which moves

so that its distance from S

is always equal to PM, its

perpendicular distance from

ZM.

Draw 8Z perpendicular

to the directrix and bisect

8Z in the point A ; produce

ZA to X,

The point A is clearly a

point on the curve and is

called the Vertex of the

Parabola.

Take A as origin, AX as the axis of x, and J.F,

perpendicular to it, as the axis of y.

Let the distance ZA^ or A8^ be called Â«, and let P be

any point on the curve whose coordinates are x and y.

Join /S'P, and draw PN and PM perpendicular respec-

tively to the axis and directrix.

We have then aSP^ = Pif ^

^.e. {x - of + 2/2 r= ZN'' = {a + xf,

y2 = 4ax (1).

This being the relation which exists between the co-

ordinates of any point P on the parabola is, by Art. 42, the

equation to the parabola.

Cor. The equation (1) is equivalent to the geometrical

proposition

PN^ = 4:AS.AIi.

198. The equation of the preceding article is the

simplest possible equation to the parabola. Throughout

this chapter this standard form of the equation is assumed

unless the contrary is stated.

176 COORDINATE GEOMETRY.

If instead of AX and A Y we take the axis and the

directrix ^M as the axes of coordinates, the equation

would be

(x â€” 2a)- + y^ = x",

i.e. y'^ = ia{x â€” a) (1).

Similarly, if the axis SX and a perpendicular line SL

be taken as the axes of coordinates, the equation is

aj2 + 2/2 = (a? + 2a)2,

i.e. y^ = 4:a(x + a) (2).

These two equations may be deduced from the equation

of the previous article by transforming the origin, firstly to

the point (- a, 0) and secondly to the point (a, 0).

199. The equation to the parabola referred to any focus and

directrix may be easily obtained. Thus the equation to the parabola,

whose focus is the point' (2, 3) and whose directrix is the straight

line X - 4y + S â€” 0, is

i. e. 17 [x^ + y^ - ix - 62/ + 13] :== [x^ + 16i/ + 9 - 8xy + Gx - 24.y] ,

i.e. 16x'^ + y^ + 8xy-7ix-78y + 212 = 0.

200. To trace the curve

2/2= iax (1).

If X be negative, the corresponding values of y are

imaginary (since the square root of a negative quantity is

unreal) ; hence there is no part of the curve to the left of

the point A.

If y be zero, so also is x, so that the axis of x meets

the curve at the point A only.

If X be zero, so also is y, so that the axis of y meets

the curve at the point A only.

For every positive value of x we see from (1), by taking

the square root, that y has two equal and opposite values.

Hence corresponding to any point P on the curve there

is another point P' on the other side of the axis which is

obtained by producing PN to P' so that PN and NP' are

THE PARABOLA. 177

equal in magnitude. The line PP' is called a double

ordinate.

As X increases in magnitude, so do the corresponding

values of y ; finally, when x becomes infinitely great, y

becomes infinitely great also.

By taking a large number of values of x and the

corresponding values of y it will be found that the curve is

as in the figure of Art. 197.

The two branches never meet but are of infinite length.

201. The quantity y"^ â€” 4aa;' is negative^ zero, or positive

according as the point {x\ y') is within, upon, or without the

parabola.

Let Q be the point {x , y') and let it be within the

curve, i.e. be between the curve and the axis AX. Draw

the ordinate QN and let it meet the curve in P.

Then (by Art. 197), PN^ - la . a;'.

Hence y"^, i.e. QN^, is < PiV^, and hence is < iax .

.'. y'^ â€” 4:ax is negative.

Similarly, if Q be without the curve, then y"-^, i.e. QJV%

is > PN^j and hence is > 4iax'.

Hence the proposition.

202. Latus Rectum. Def. The latus rectum of

any conic is the double ordinate LSL' drawn through the

focus S.

In the case of the parabola we have SL = distance of L

from the directrix â€” SZ= 2a.

Hence the latus rectum = la.

"When the latus rectum is given it follows that the

equation to the parabola is completely known in its

standard form, and the size and shape of the curve

determined.

The quantity la is also often called the principal

parameter of the curve.

Focal Distance of any point. The focal distance

of any point P is the distance 8P.

This focal distance = PM = ZN= ZA+AN'=a + x.

L. 12

178 COORDINATE GEOMETEY.

Ex. Find the vertex, axis, focus, and latus rectum of the parabola

4?/2 + 12ar-202/ + 67 = 0.

The equation can be written

y^-5y=-Sx - ^^,

i.e. {y - ^f=-Sx-s^- + ^^=-3{x + i).

Transform this equation to the point (-|, f) and it becomes

y^= -Sx, which represents a parabola, whose axis is the axis of x

and whose concavity is turned towards the negative end of this axis.

Also its latus rectum is 3.

Eef erred to the original axes the vertex is the point i-^, f ), the

axis is 2/ = f, and the focus is the point (-| -|, f), Â«-e. ( -V-i f)-

EXAMPLES. XXV.

Find the equation to the parabola with

1. focus (3, -4) and directrix Gcc- 7?/ + 5 = 0.

X XI

2. focus (a, &) and directrix - + f = 1.

^ ah

Find the vertex, axis, latus rectum, and focus of the parabolas

3. y^ = 4:X + ^y. 4. x'^ + 2y = 8x-7.

5. x^-2ax + 2ay = 0. 6. 2/^=4y-4a:.

7. Draw the curves

(1) y'2=-4:ax, (2) x'^=4:ay, and (3) x-z=-4:ay.

8 Find the value of p when the parabola y'^ = 4px goes through

the point (i) (3, - 2), and (ii) (9, - 12).

9. For what point of the parabola y^ = 18x is the ordinate equal

to three times the abscissa ?

10. Prove that the equation to the parabola, whose vertex and focus

are on the axis of x at distances a and a' f . om the origin respectively,

is y^ = 4:{a'-a){x-a).

11. In the parabola y^=Qx, find (1) the equation to the chord

through the vertex and the negative end of the latus rectum, and

(2) the equation to any chord through the point on the curve whose

abscissa is 24.

12. Prove that the equation y^ + 2Ax + 2By + C = represents a

parabola, whose axis is parallel to the axis of x, and find its vertex and

the equation to its latus rectum.

13. Prove that the locus of the middle points of all chords of

the parabola ?/2 = 4aa; which are drawn through the vertex is the

parabola y'^ = 2ax.

[EXS. XXV.] THE PARABOLA. EXAMPLES. 179

14. Prove that the locus of the centre of a circle, which intercepts

a chord of given length 2a on the axis of x and passes through a given

point on the axis of y distant 6 from the origin, is the curve

a;2-2'Â«/& + &2 = a2.

Trace this parabola.

15. PQ is a double ordinate of a parabola. Find the locus of its

point of trisection.

16. Prove that the locus of a point, which moves so that its

distance from a fixed line is equal to the length of the tangent drawn

from it to a given circle, is a parabola. Find the position of the

focus and directrix.

17. If a circle be drawn so as always to touch a given straight

line and also a given circle, prove that the locus of its centre is

a parabola.

18. The vertex ^ of a parabola is joined to any point P on the

curve and PQ is drawn at right angles to AP to meet the axis in Q.

Prove that the projection of PQ on the axis is always equal to the

latus rectum.

19. If on a given base triangles be described such that the sum of

the tangents of the base angles is constant, prove that the locus of

the vertices is a parabola.

20. A double ordinate of the curve y^=^px is of length 8p ; prove

that the lines from the vertex to its two ends are at right angles.

21. Two parabolas have a common axis and concavities in oppo-

site directions ; if any line parallel to the common axis meet the

parabolas in P and P', prove that the locus of the middle point of PP'

is another parabola, provided that the latera recta of the given para-

bolas are unequal.

22. A parabola is drawn to pass through A and P, the ends of

a diameter of a given circle of radius a, and to have as directrix a

tangent to a concentric circle of radius h ; the axes being AB and

a perpendicular diameter, prove that the locus of the focus of the

parabola IS - + ,^^=1,

203. To find the points of intersection of any straight

line with the parabola

2/^ = 4acc (1).

The equation to any straight line is

y = 7nx + c .(2).

The coordinates of the points common to the straight

line and the parabola satisfy both equations (1) and (2),

and are therefore found by solving them.

12â€”2

180 COORDINATE GEOMETRY.

Substituting the value of y from (2) in (1), we have

{mx + cf â€” 4:ax,

i.e. m^oc^ + 2x (mc - 2a) + G^ - (3).

This is a quadratic equation for x and therefore has two

roots, real, coincident, or imaginary.

The straight line therefore meets the parabola in two

points, real, coincident, or imaginary.

The roots of (3) are real or imaginary according as

{2(mc-2a)f-47^iV

is positive or negative, i.e. according as â€” aTnc + a^ is

positive or negative, i.e. according as mc is ^ ia.

204. To find the length of the chord intercepted by the parabola on

the straight line

y â€” mx + c (1).

If (o^i , y-^ and {x^, y^) he the common points of intersection, then,

as in Art. 154, we have, from equation (3) of the last article,

{^x^ â€” x^j = (ajj + x^) â€” ^x^x^

_4(mc-2a)^ 4:C^ _l&a{a-mc)

~ m* m^ wi'* '

and yj^-y^ = m{x^-x^).

Hence the required length = \/(l/i - 2/2)^ + (^1 ~ ^"2)^

= Jl + m'-^ (x^ - ajg) = ^2 Jl + mP J a (a - mc).

205. To find the equation to the tangent at any point

(^'3 y) 9f i^^^ parabola y^ = 4:ax.

The definition of the tangent is given in Art. 149.

Let P be the point [x, y) and Q a point (x", y") on the

parabola.

The equation to the line PQ is

2/-y' = f5f'(^-Â»'') (!)â€¢

Since P and Q both lie on the curve, we have

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