Online Library → S. L. (Sidney Luxton) Loney → The elements of coordinate geometry → online text (page 21 of 26)

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.(1).

Since

(^7

[Art. 358.]

The centre {x, y) is given by

- 3_ ^ ^

and â€”x + y -5 = 0,

so that ^=-2, and ^ = 2.

The equation to the curve, referred to parallel axes through the

centre, is then

a;2 - 3a;i/ + 2/2 + 5 ( - 2) - 5 X 2 + 21 = 0,

i.e. x^-3xy + y^=-l (2).

The direction of the axes is given by

2h -3

tan 20 =

:0O,

a-b 1-1

so that 20=90Â° or 270Â°,

and hence 0^= 45Â° and 02=135Â°.

The equation (2) in polar coordinates is

r2 (cos2 0-3 cos sin + sin^ 0) = - (sin^ + cos^ 0),

l + tan2

I.e.

l-3tan0 + tan2

342

COORDINATE GEOMETRY.

When ^1 = 45Â°, r^^^ - ^-^â€” =2, so that i\ = sl'^

-2

When 6'2=135Â°, r^- -

2

, so that 7

â– ^v

-2

5

2 - " > '2- 1 + 3 + 1" 5

To construct the curve take the point G whose coordinates are - 2

and 2. Through G draw a straight line AG A' inclined at 45Â° to the

axis of X and mark off A'C=GA = J2.

Also through A draw a straight line KAK' perpendicular to GA

and take AK=^K'A = ^^. By Art. 315, GK and GK' are then the

asymptotes.

The curve is therefore a hyperbola whose centre is G, whose

transverse axis is ^'^, and whose asymptotes are GK and GK'.

On putting a;=0 it will he found that the curve meets the axis of

y where y â€” Z or 7, and, on putting ?/ = 0, that it meets the axis of x

where x= - 3 or - 7.

Hence

0*3=3, 0(9' = 7, Oi? = 3, and 0R' = 1.

366. Tojind the eccentricity of the central conic section

ax^-\- 2hxy â– \-hy'^= 1 (1).

First, let 7i^ â€” ah be negative, so that the curve is

ECCENTEICITY OF A CONIC SECTION. 343

an ellipse, and let the equation to the ellipse, referred to

its axes, be

By the theory of Invariants (Art. 135) we have

^^+|-^ = ^ + ^ (2),

and a^^"''^"'^'' (^)'

Also, if e be the eccentricity, we have, if a be > ^,

^ = 2â€”-

a.

<2'

" 2-6" a' + ZS^

But, from (2) and (3), we have

Hence

ab â€” h^

e" _ s/(a-bf + 4:h^

2-e^ ' a + b ^^^'

This equation at once gives e^.

Secondly^ let h- â€” ab be positive, so that the curve is

a hyperbola, and let the equation referred to its principal

axes be

^ _ ^^ = 1

so that in this case

-2- o2 = ^* + ^'a^^-^2^2 = ^^-'^^' = -(^'-^^)â€¢

Hence a^ â€” B^ = â€” T^ r s-nd a^B^ = -r^ j ,

' /r â€” ab hr â€” ab

so that a^ + /3^ = + x/(a^ - ^J + 4a^/3^ ^ + h?-]l ^^'^ '

344 COORDINATE GEOMETRY.

In this case, if e be the eccentricity, we have

'"^- j:^^~a''^'~~ a + b ^^'

This equation gives e^.

In each case we see that e is a root of the equation

Â«2 N2 ^a-bY + 4:h^

\2-e')

(a + hf '

i.e. of the equation

e^ {ab - h') + {{a-bY + ih'} (e^ - 1 ) - 0.

367. To obtain the foci of the central conic

aoi? + Ihxy + by^ â€” 1.

Let the direction of the axes of the conic be obtained as

in Art. 364, and let 6-^ be the inclination of the major axis

in the case of the ellipse, and the transverse axis in the case

of the hyperbola, to the axis of x.

Let r/ be the square of the radius corresponding to ^i,

and let r^ be the square of the radius corresponding to the

perpendicular direction. [In the case of the hyperbola r^

will be a negative quantity.]

The distance of the focus from the centre is ^jr^ â€” r^

(Arts. 247 and 295). One focus will therefore be the point

(vri^ â€” T^ cos ^1 , Jr^ â€” r^ sin ^J,

and the other will be

(- sjr^ â€” ri cos 6^ , - Jr^ - r/ sin 6^).

"Ex.. Find the foci of the ellipse traced in Art. 365.

2 1

Here tan 6-^ = 2, so that sin 6^=-j^ and cos ^j = -^ .

Also ri2 = 6, and 7-22=4, so that sjrf^=^2.

The coordinates of the foci referred to axes through C are therefore

FOCI OF A CONIC.

345

Their coordinates referred to the original axes OX and OY are

2Â±

V2

x/5'

2V2\

368. The method of obtaining the coordinates of the

focus of a parabola given by the general equation may be

exemplified by taking the example of Art. 363.

Here it was shewn that the latus rectum is equal to 3,

so that, if aS' be the focus, ^aS' is J.

It was also shewn that the coordinates of A referred to

OX and Y are â€” ^ and ^.

The coordinates of S referred to the same axes are

+

f and -I,

^.e.

2V and -I

Its coordinates referred to the original axes are therefore

_7_

20

W COS

i.e.

7

2 J)

Z- sin 6 and -J-pr sin + ~ cos

+

i-i and

2 5

^* a

7 3

20*5

nd 1-^3

7_

5

I. 1

5*5'

^,

100'

In Art. 393 equations will be found to give the foci of

any conic section directly, so that the conic need not first

be traced.

369. Ex. 1. Trace the curve

The equation may be written

.(l).

(^^^T-K'^^T- <^'-

Now the straight lines 3a;- 2?/ + 4 = and 2x + Sy-5 = are at

right angles. Let them be C3I and

GN, intersecting in C which is the

point (-t\, H)-

If P be any point on the curve

and PM and PN the perpendiculars

upon these lines, the lengths of P3I

and PN are

3x-2y + 4: ^ 2x + 3y-5

-^w ^^^ - ^w- â–

Hence equation (2) states that

3PM2 + 2P^â– 2=:3,

PM^

I.e.

PN^

346 COORDINATE GEOMETRY.

The locus of P is therefore an ellipse whose semi-axes measured

along CM and CN are a/I and 1 respectively.

Ex. 2. What is represented by the equation

{x'^-a^f + iy^-a^)^=a*?

The equation may be written in the form

a;4 + t/-i - 2a2 (a;2 + ?/2) + a* = 0,

i.e. {x^ + y^f-2a^{x^ + y^)+a*=2xY,

i.e. {x^ + y^-a^)^-{^2xyf=0,

i.e. {x^ + y/^^y +y^- Â«^) (^^ - fj^^y +y^- ^^) = O-

The locus therefore consists of the two ellipses

x^ + /ij2xy+y^-a^=0, and x^-fj2xy + y'^-a^=0.

These ellipses are equal and their semi-axes would be found to be

asj2 + ^2 and a ^2 -^2.

The major axis of the first is inclined at an angle of 135Â° to the

axis of X, and that of the second at an angle of 45Â°.

EXAMPLES. XLI.

Trace the parabolas

1. {x-4:ij)^=51y. 2. {x-y)^=x + y + l.

3. {5x-12yf=2ax + 2day + a\

4. {4a; + 3?/ + 15)2= 5 (3a; -4?/).

5. 16a;2 + 24a;2/ + 92/2-5a;-10?/ + l = O.

6. 9x^+24^y + 16y^-4:y-x + 7 = 0.

7. 144a;2 - 120xy + 25y^ + 619a; - 272^ + 663 = 0, and find its focus.

8. 16a;2-24a;2/ + 9?/2 + 32a; + 86?/-39 = 0.

9. 4a;2-4a;2/ + 2/2-12a; + 62/ + 9 = 0.

Find the position and magnitude of the axes of the conies

10. 12a;2-12a;i/ + 7?/2=48. 11. 3a;2+2a;?/ + 32/2=8.

12. x'^-xy~6y^=Q.

Trace the following central conies.

13. a;2-2a;2/cos2a + ^2â€” 2a2. 14. a;2-2a;2/cosec2a-f2/^ = a2.

15. xy = a{x + y). 16. xy-y^ = a^.

17. 2/^-2a;?/ + 2a;2 + 2a;-22/ = 0. 18. x'^ + xy + y^ + x+y = l.

LEXS. XLI.] TRACING OF CONIC SECTIONS. EXAMPLES. 347

19. 2x^ + Zxii-2y'^-lx + y-2 = Q.

20. 40a;2 + 36a;?/ + 25?/2- 196a; -122?/ + 205 = 0.

21. x^-%xy + y^ + 10x-l(iy + 21=^0.

22. x'^-xy + 2^2 _ 2ax - Qay + 7a^ = 0.

23. 10a;2 - ASxy - lOif + dSx + Uy - 5^ = 0.

24. 4:x^ + 27xy + S5y^~14M-Bly-&:=0.

25. {3x-4y + a){4:X + Sy + a) = a\

26. 3 (2a; -3^ + 4)2 + 2 (3a; + 2?/ -5)2 = 78.

27. 2 (3a; -4^/ + 5)2 -3 (4a; + 3?/ -10)2 = 150.

Find the products of the semi-axes of the conies

28. 2/2-4a;i/ + 5a;2 = 2. 29. 4(3a; + 42/- 7)2 + 3 (4a;-3t/ + 9)2=3.

30. lla;2 + 16a;?/ - ?/2- 70a; -40y + 82 = 0.

Find the foci and the eccentricity of the conies

31. a;2-3a;i/ + 4aa; = 2a2. 32. 4a;?/ - 3a;2 - 2a?/ = 0.

33. 5a;2 + 6a;?/ + 5?/ + 12a; + 4?/ + 6 = 0.

34. a;2 + 4a;?/ + ?/2-2a; + 2/y-6 = 0.

35. Shew that the latus rectum of the parabola

(a2 + 62) (^^2 + ^2) ^ (5^ + ay _ ahf

is 2ab-hs/a^~+^.

36. Prove that the lengths of the semi-axes of the conic

aa;2 + 2hxy + ay^ = d

are \/ â€” â€” r and

respectively, and that their equation is a;2_y2_o_

37. Prove that the squares of the semi-axes of the conic

ax2 + 2/?a;?/ + &?/2 + 2(;a; + 2/?/ + c =

are 2A-^ { {ab -h^)(a + h^ J {a - hf + 4/i2)} ,

where A is the discriminant.

38. If X be a variable parameter, prove that the locus of the

vertices of the hyperbolas given by the equation x^-y'^ + \xy = a^ is

the curve {x^+y^)^=a^{x^-y^).

39. If the point {at^^, 2at-^) on the parabola ?/2=4aa; be called the

point t^, prove that the axis of the second parabola through the four

points ^1, t^, ^3, and t^ makes with the axis of the first an angle

oot-(*lÂ±^Â±iÂ±^A.

Prove also that if two parabolas meet in four points the distances

of the centroid of the four points from the axes are proportional to the

latera recta.

348 COORDINATE GEOMETRY. [ExS. XLI.]

40. If the product of the axes of the conic x^ + 2xy + ny'^=8 be

unity, shew that the axes of coordinates are inclined at an angle

sin-i 1.

41. Sketch the curve Qx^-7xy-5y^~4:X + lly = 2, the axes being

inclined at an angle of 30Â°.

42. Prove that the eccentricity of the conic given by the general

equation satisfies the relation

e^ , (a + & - 2/1 cos w)2

h 4 = - â€”

1 - e^ " (a6 - h^) sin^ w '

where w is the angle between the axes.

43. The axes being changed in any way, without any change of

origin, prove that in the general equation of the second degree the

P + 9^ - 2fg cos 0} ap+bg^-2fgh , A

quantities c, '' â€¢^ . ^^ , -^â€” â€” â€” â€” =^^ , and -.-^- are

sm^w sm^w sin^w

invariants, in addition to the quantities in Art. 137.

[On making the most general substitutions of Art. 132 it is clear

that c is unaltered; proceed as in Art, 137, but introduce the condition

that the resulting expressions are equal to the product of two linear

quantities (Art. 116); the results will then follow.]

CHAPTER XVL

THE GENERAL CONIC.

370. In the present chapter we shall consider proper-

ties of conic sections which are given by the general equation

of the second degree, viz.

aoi? + '2Jnxy + hy^ + Igx + ^fy + c = (1).

For brevity, the left-hand side of this equation is often

called ^ (tc, 3/), so that the general equation to a conic is

^ (a;, y) = 0.

Similarly, <^(aj', y) denotes the value of the left-hand

side of (1) when x and y are substituted for x and y.

The equation (1) is often also written in the form aS'^O.

371. On dividing by c, the equation (1) contains five

independent constants - , - , - , - , and - .

c G G G c

To determine these five constants, we shall therefore

require five conditions. Conversely, if five independent

conditions be given, the constants can be determined.

Only one conic, or, at any rate, only a finite number of

conies, can be drawn to satisfy five independent conditions.

372. To find the equation to the tangent at any foint

{x\ y') of the conic section

cf> {x, y) = ax^ + 2hxy + hy"^ + 2gx + 2fy + c = 0. . .(1).

Let {x", y") be any other point on the conic.

350 COORDINATE GEOMETRY.

The equation to the straight line joining this point to

(x\ y) is

2/-2/' = |^(^-^') (2).

Since both (ic', 3/') and (a?", 2/") lie on (1), we have

aaj'2 + "Ihxy' + hy'^ + 2^cc' + 2/2/' + c - (3),

and aÂ£c"2 + 2^a;'y ' + hy'"" + 2^Â£c" + Ify" + c = (4).

Hence, by subtraction, we have

a {x^ - x'") + 2^ {x'y' - xy) + 5 {y"" - 3/"^;

+ 2^(x'-a.") + 2/(2/'-2/'') = (5).

But

2 (o^y - x"y") = {x' + x") {y' - y") + {x' - x") {y' + y"\

so that (5) can be written in the form

{^ - x") [a {x' + x") + h(y'+ y") + 2^]

+ ky' - y") [h {^' + x") + h{y' + y") + 2/] = 0,

y"-y' ^ a{x' + x") + h(y' + y") + 2g

*â– ^' x" -x' k {x' + x") + b{y' + y") + 2/ "

The equation to any secant is therefore

y y~ h{x' + x") + h{y'+y") + 2f^ a:;...^

To obtain the equation to the tangent at {x', y'), we put

x" = x' and y" = y' in this equation, and it becomes

, ax + hy' + <7 . ,,

y-y =- -7-7 â€” 1^, â€” > (^ - ^\

^ ^ hx +hy +f^ '

i. e. {ax + liy' + g) x + (hx + by +f)y

= ax'^ + 2hxy' + hy'^ + gx + fy

â€” â€” gx â€”fy â€” c, by equation (3).

The required equation is therefore

axx' + h (xy + x'y) +l>yy' + g (x + x) + f (y + y')

+ c = (7).

Cor. 1. The equation (7) may be written down, from

the general equation of the second degree, by substituting

XX foi- a^, yy' for 3/^, xy' + xy for 2xy, x + od for 2a?, and

y-vy iovly. (Cf. Art. 152.)

THE GENEKAL CONIC. POLE AND POLAR. 351

Cor. 2. If the conic pass through the origin we have

c = 0, and then the tangent at the origin (where x =0 and

y = 0) is gx +fy = 0,

i. e. the equation to the tangent at the origin is obtained by

equating to zero the terms of the lowest degree in the

equation to the conic.

373. The equation of the previous article may also be obtained

as follows ; If {x', y') and {x", y") be two points on the conic section,

the equation to the line joining them is

a{x-xf){x-x") + hl{x-x'){y-y") + {x-x"){ij-y')] + h{y-y'){y-y")

= ax^ + 2hxy + by^ + 2gx + 2fy + c (1).

For the terms of the second degree on the two sides of (1) cancel,

and the equation reduces to one of the first degree, thus representing

a straight line.

Also, since {x', y') lies on the curve, the equation is satisfied by

putting x=x' and y=y'.

Hence {x', y') is a point lying on (1).

So {x'\ y") lies on (1).

It therefore is the straight line joining them.

Putting x" â€” x' and y" =y' we have, as the equation to the tangent

at {x', y'),

a{x- x'f + 27i {x - x') [y - if) + & {y - y'f

= ax^ + ^hxy + 'by'^ + 2gx + 2fy + c,

i.e. 2axx' + 2}i {x'y + xy') + 2\)yy' + 2gx + 2fy + c

= ax"^ + 2hx'y' + hy"^

â€” - 2gx' - 2fy' - c, since {x', y') lies on the conic.

Hence the equation (7) of the last article.

374. To find the condition that any straight line

lx + my + n = (1),

may touch the conic

ax^ + 2hxy + by^ + 2gx + 2fy + c = (2).

Substituting for y in (2) from (1), we have for the equation giving

the abscissae of the points of intersection of (1) and (2),

x^ {am? - 2hlm + hP) - 2x (hmn - bin - gm^ +flm)

+ bn^-2fmn + cm^=0 (3).

If (1) be a tangent, the values of x given by (3) must be equal.

The condition for this is, (Art. 1,)

{hmn - bin - gm^+fhrif={am? - 2hlm + bl^) {bn^ - 2fmn + cm^).

352 COORDINATE GEOMETRY.

On simplifying, we have, after division by iii^,

p {jic -p) + m2 {ca - p2) + n^ [ah - li') + 2mn [gh - af) + 2wZ {lif - hg)

+ 2lm{fg-ch) = 0.

Ex. Find the equations to the tangents to the conic

x^ + 4:xy + 3y^-5x-Gy + S = (1),

ivhich are parallel to the straight line a; + 4i/ = 0.

Tlie equation to any such tangent is

a; + 4^ + c = (2),

where c is to be determined.

This straight line meets (1) in points given by

3x2 _ 2x (5c + 28) + 3c2 + 24c + 48 = 0.

The roots of this equation are equal, i.e. the line (2) is a tangent,

if {2(5c + 28)}2 = 4. 3.(3c2 + 24c+48), i.e. if c=-5 or -8.

The required tangents are therefore

a; + 4?/-5 = 0, and x + iij -Q â€” 0.

375. As in Arts. 214 and 274 it may be proved that

the polar of {x\ y) with respect to ^ (cc, y) = is

{ax' + hy -{â– g) X + {hx + by' +/) y + gx +/y' + c = 0.

The form of the equation to a polar is therefore the

same as that of a tangent.

Just as in Art. 217 it may now be shewn that, if the

polar of F passes through T, the polar of T passes through

P.

The chord of the conic which is bisected at {x\ y),

being parallel to the polar of (x, y) [Arts. 221 and 280],

has as equation

{ax + hy +g){x- x) + {hx' + by +/){y â€” y) = 0.

376. To find the equation to the diameter bisecting all

chords parallel to the straight line y = mx.

Any such chord is y = 7nx->rK (1).

This meets the conic section

ax^ + 2hxy + by"^ + 2gx + 2fy + c =

in points whose abscissae are given by

ax^ + ^hx {mx + K) + b {mx + Kf + 2gx + 2f{mx + K)+ g=^0,

i. e. by x^ {a + 2hm + bm^) + 2x {hK + bmK + ^ + fm)

+ blO + 2fK-\-c = 0.

THE GENERAL CONIC. CONJUGATE DIAMETERS. 353

If fl?i and x^ be the roots of this equation, we therefore

have

X ^x - cy{ h + hm)K + g+fm

^ ^ a + 2hm + bm^

Let {X, Y) be the middle point of the required chord,

so that

â€ž X1 + X2 (h + bm) K+ gf+/7n , .

2 a + 2hm + b7)iF ' ^ ^'

Also, since (X, Y) lies on (1) we have

Y-^mX+K (3).

If between (2) and (3) we eliminate K we have a

relation between X and Y.

This relation is

â€” (a + 2hm + bm^) X = (h + bm) {Y â€” mX) + g +fm,

i. e. X [a + hm) + Yih+ bm) + g -Â¥fm = 0.

The locus of the required middle point is therefore the

straight line whose equation is

x{a + hm) + y(h + bm) + g â– \-fm == 0.

If this be parallel to the straight line y - m'x, we

have

a + Am

m =

â€¢w,

h + bm

i.e. a + h (m + m') + bmm' = (5).

This is therefore the condition that the two straight

lines y â€” tnx and y â€” m!x may be parallel to conjugate

diameters of the conic given by the general equation.

377. To find the condition that the pair of straight lines, whose

equation is

Ax'^ + 2Hxy+By^=0 (1),

may be parallel to conjugate diameters of the general conic

ax'^ + 2hxy + by^ + 2gx + 2fy + c = .'. (2).

Let the equations of the straight lines represented by (1) hey = mx

and y = m'x, so that (1) is equivalent to

B [y - mx) [y - m'x) = 0,

and hence m-\-m'= - â€”, and mm =^ .

L. 23

354 COORDINATE GEOMETRY.

By the condition of the last article it therefore follows that the

lines (1) are parallel to conjugate diameters if

i.e. if Ah-2Hh + Ba = 0.

378. To prove that tioo concentric conic sections always have a

pair, and only one pair, of common conjugate diameters and to find

their equation.

Let the two concentric conic sections be

ax^ + 2hxy + hy'^=l (1),

and a'x^^-2h'xy + l)Y = l (2).

The straight lines

Ax^ + 2Hxy + By^ = (3),

are conjugate diameters of both (1) and (2) if

Ab-2Hh + Ba = 0,

and Ab'-2Hh' + Ba' = 0.

Solving these two equations we have

A -2H__ B

ha' - h'a ~ W^afb ~ hh' - b'h '

Substituting these values in (3), we see that the straight lines

x^ha'-h'a)-xy{ab'-a'b)4-y^{bh'-b'h) = (4)

are always conjugate diameters of both (1) and (2).

Hence there is always a pair of conjugate diameters, real, coinci-

dent, or imaginary, which are common to any two concentric conic

sections.

EXAMPLES. XLII.

1. How many other conditions can a conic section satisfy when

we are given (1) its centre, (2) its focus, (3) its eccentricity, (4) its

axes, (5) a tangent, (6) a tangent and its point of contact, (7) the

position of one of its asymptotes?

2. Find the condition that the straight line lx + niy = l may

touch the parabola {ax-bijf-2 {a^ + &-) (ax + by) + {a^ + b'^f = 0, and

shew that if this straight line meet the axes in P and Q, then PQ

will, when it is a tangent, subtend a right angle at the point {a, b).

3. Two parabolas have a common focus ; prove that the perpen-

dicular from it upon the common tangent passes through the

intersection of the directrices.

[EXS. XLII.] INTERSECTIONS OF TWO CONICS. 355

4. Shew that the conic -^ H r- cos a + i-:-, = sin^ a is inscribed in

a^ ab h^

the rectangle, the equations to whose sides are x^ = a^ and y^ = h^, and

that the quadrilateral formed by joining the points of contact is of

constant perimeter 4 sja^ + 6^, whatever be the value of a.

5. A variable tangent to a conic meets two fixed tangents in two

points, P and Q ; prove that the locus of the middle point of PQ is a

conic which becomes a straight line when the given conic is a parabola.

6. Prove that the chord of contact of tangents, drawn from an

external point to the conic ax^ + 2hxy + hy'^=l, subtends a right angle

at the centre if the point lie on the conic

a;2 (a2 + 7i2) + 2h {a + h) xy+y'^ {h'^ + h^) = a + b.

7. Given the focus and directrix of a conic, prove that the polar

of a given point with respect to it passes through another fixed point.

8. Prove that the locus of the centres of conies which touch the

axes at distances a and b from the origin is the straight line ay = bx.

9. Prove that the locus of the poles of tangents to the conic

ax'^ + 2hocy + by^=l with respect to the conic a'x"^ + 2h'xy + b'y^ = l is

the conic

a {Ji'x + b'yf - 27i [a'x + h'y) {h'x + b'y) + b {a'x + h'yf=db - h\

10. Find the equations to the straight lines which are conjugate

to the coordinate axes with respect to the conic Ax^-{-2Hxy + By^ = l.

Find the condition that they may coincide, and interpret the

result.

11. Find the equation to the common conjugate diameters of the

conies (1) a;2 + 4a;y + 6!/2=l and 2x'^ + &xy + ^y^ â€” l,

and (2) 2x^-^xy + ^y'^ = l and 2x'^ + ^xy -^y^=l.

12. Prove that the points of intersection of the conies

ax^ ^^hxy + by^=l and a'x^ + 2h'xy-\-b'y'^ = l

are at the ends of conjugate diameters of the first conic, if

ab'-{-a'b-2hh' = 2{ah-U'^).

13. Prove that the equation to the equi-conjugate diameters of

â€ž ^, , â€ž ^. ax^ + 2hxy + by^ 2(x^ + y'^)

the conic ax^ + 2kxy + by"^ = 1 is râ€” f^ â€” â€” = â€” ^ â€” rr-^ â€¢

^ ^ ab-lv- a + b

379. Two conies, in general, intersect in four points,

real or imaginary.

For the general equation to two conies can be written

in the form

aay^ + 2x Qiy + g) + hy^ + 2fy + c = 0,

and a'x^ + 2x (Ky + g) + 6^' + %f'y + C - 0.

23â€”2

356 COORDINATE GEOMETRY.

Eliminating x from these equations, we find that the

result is an equation of the fourth degree in y, giving

therefore four values, real or imaginary, for y. Also, by

eliminating x^ from these two equations, we see that there

is only one value of x for each value of y. There are there-

fore only four points of intersection.

380. Equation to any conic passing through the inter-

section of two given conies.

Let S^ ax"" + 2hxy + hy"" + 2gx + ^fy + c = (1),

and S' E ax" + 2h'xy + h'y^ + 2g'x + If'y + c' = . . . (2)

be the equations to the two given conies.

Then ^-_X.S" = (3)

is the equation to any conic passing through the inter-

sections of (1) and (2).

For, since S and aS^' are both of the second degree in x

and 2/, the equation (3) is of the second degree, and hence

represents a conic section.

Also, since (3) is satisfied when both ^S' and S' are zero,

it is satisfied by the points (real or imaginary) which are

common to (1) and (2).

Hence (3) is a conic which passes through the intersec-

tions of (1) and (2).

381. To find the equations to the straight lines passing

through the intersections of two conies given hy the general

equations.

As in the last article, the equation

{a - \a) 0^2 + 2 (A - Xh') xy + (h - \h') ^ + 2 (^ - Xg') x

+ 2(f-\r)y + {c-Xc') = (1),

represents some conic through 4}he intersections of the given

conies.

Now, by Art. 116, (1) represents straight lines if

{a - \a') (b - Kb') (c - Xc') + 2 (f - Xf) (g - Xg') (h - Xh')

-{a- Xa') (/- Xf'f -{b- Xb') (g - Xg'f - (c - Xc') (h - XhJ

= (2).

INTERSECTIONS OF TWO CONIGS. 357

Now (2) is a cubic equation. The thi-ee values of \

found from it will, when substituted successively in (1),

give the three pairs of straight lines which can be drawn

through the (real or imaginary) intersections of the two

conies.

Also, since a cubic equation always has one real root,

one value of A. is real, and it could be shown that there can

always be drawn one pair of real straight lines through the

intersections of two conies.

382. All conies tvhich pass through the intersections of

two rectangular hyperbolas are themselves rectangular hyper-

bolas.

In this case, \i S = and S' â€” be the two rectangular

hyperbolas, we have

Â« + 6 = 0, and a +b' = 0. (Art. 358.)

Hence, in the conic S â€” X;S" â€” 0, the sum of the co-

efficients of x^ and y"^

= (a- Xa') + {b- Xb') =:(a + b)-X{a' + b') - 0.

Hence, the conic S ~ XS' ^ 0, i.e. any conic through the

intersections of the two rectangular hyperbolas, is itself a

rectangular hyperbola.

Cor. If two rectangular hyperbolas intersect in four points

A, B, G, and D, the two straight lines AD and BC, which are a conic

through the intersection of the two hyperbolas, must be a rectangular

hyperbola. Hence AD and BC must be at right angles. Similarly,

BD and GA, and GD and AB, must be at right angles. Hence D is

the orthocentre of the triangle ABG.

Therefore, if two rectangular hyperbolas intersect in four points,

each point is the orthocentre of the triangle formed by the other

three.

383. 7/* Z = 0, i/== 0, ^= 0, and 11 = be the equations

to the four sides of a quadrilateral taken in order, the

equation to any conic passing through its angular points is

LN=X.MR (1).

For L = passes through one pair of its angular points

and N=0 passes through the other pair. Hence LN = is

the equation to a conic (viz. a pair of straight lines) passing

through the four angular points.

358 COORDINATE GEOMETRY.

Similarly MR - is the equation to another conic

passing through the four points.

Hence LN = X . MR is the equation to any conic through

the four points.

Geometrical meaning. Since L is proportional to the perpen-

dicular from any point {x, y) upon the straight line Z/ = 0, the

relation (1) states that the product of the perpendiculars from any

point of the curve upon the straight lines JL = and N=0 is propor-

tional to the product of the perpendiculars from the same point upon

Since

(^7

[Art. 358.]

The centre {x, y) is given by

- 3_ ^ ^

and â€”x + y -5 = 0,

so that ^=-2, and ^ = 2.

The equation to the curve, referred to parallel axes through the

centre, is then

a;2 - 3a;i/ + 2/2 + 5 ( - 2) - 5 X 2 + 21 = 0,

i.e. x^-3xy + y^=-l (2).

The direction of the axes is given by

2h -3

tan 20 =

:0O,

a-b 1-1

so that 20=90Â° or 270Â°,

and hence 0^= 45Â° and 02=135Â°.

The equation (2) in polar coordinates is

r2 (cos2 0-3 cos sin + sin^ 0) = - (sin^ + cos^ 0),

l + tan2

I.e.

l-3tan0 + tan2

342

COORDINATE GEOMETRY.

When ^1 = 45Â°, r^^^ - ^-^â€” =2, so that i\ = sl'^

-2

When 6'2=135Â°, r^- -

2

, so that 7

â– ^v

-2

5

2 - " > '2- 1 + 3 + 1" 5

To construct the curve take the point G whose coordinates are - 2

and 2. Through G draw a straight line AG A' inclined at 45Â° to the

axis of X and mark off A'C=GA = J2.

Also through A draw a straight line KAK' perpendicular to GA

and take AK=^K'A = ^^. By Art. 315, GK and GK' are then the

asymptotes.

The curve is therefore a hyperbola whose centre is G, whose

transverse axis is ^'^, and whose asymptotes are GK and GK'.

On putting a;=0 it will he found that the curve meets the axis of

y where y â€” Z or 7, and, on putting ?/ = 0, that it meets the axis of x

where x= - 3 or - 7.

Hence

0*3=3, 0(9' = 7, Oi? = 3, and 0R' = 1.

366. Tojind the eccentricity of the central conic section

ax^-\- 2hxy â– \-hy'^= 1 (1).

First, let 7i^ â€” ah be negative, so that the curve is

ECCENTEICITY OF A CONIC SECTION. 343

an ellipse, and let the equation to the ellipse, referred to

its axes, be

By the theory of Invariants (Art. 135) we have

^^+|-^ = ^ + ^ (2),

and a^^"''^"'^'' (^)'

Also, if e be the eccentricity, we have, if a be > ^,

^ = 2â€”-

a.

<2'

" 2-6" a' + ZS^

But, from (2) and (3), we have

Hence

ab â€” h^

e" _ s/(a-bf + 4:h^

2-e^ ' a + b ^^^'

This equation at once gives e^.

Secondly^ let h- â€” ab be positive, so that the curve is

a hyperbola, and let the equation referred to its principal

axes be

^ _ ^^ = 1

so that in this case

-2- o2 = ^* + ^'a^^-^2^2 = ^^-'^^' = -(^'-^^)â€¢

Hence a^ â€” B^ = â€” T^ r s-nd a^B^ = -r^ j ,

' /r â€” ab hr â€” ab

so that a^ + /3^ = + x/(a^ - ^J + 4a^/3^ ^ + h?-]l ^^'^ '

344 COORDINATE GEOMETRY.

In this case, if e be the eccentricity, we have

'"^- j:^^~a''^'~~ a + b ^^'

This equation gives e^.

In each case we see that e is a root of the equation

Â«2 N2 ^a-bY + 4:h^

\2-e')

(a + hf '

i.e. of the equation

e^ {ab - h') + {{a-bY + ih'} (e^ - 1 ) - 0.

367. To obtain the foci of the central conic

aoi? + Ihxy + by^ â€” 1.

Let the direction of the axes of the conic be obtained as

in Art. 364, and let 6-^ be the inclination of the major axis

in the case of the ellipse, and the transverse axis in the case

of the hyperbola, to the axis of x.

Let r/ be the square of the radius corresponding to ^i,

and let r^ be the square of the radius corresponding to the

perpendicular direction. [In the case of the hyperbola r^

will be a negative quantity.]

The distance of the focus from the centre is ^jr^ â€” r^

(Arts. 247 and 295). One focus will therefore be the point

(vri^ â€” T^ cos ^1 , Jr^ â€” r^ sin ^J,

and the other will be

(- sjr^ â€” ri cos 6^ , - Jr^ - r/ sin 6^).

"Ex.. Find the foci of the ellipse traced in Art. 365.

2 1

Here tan 6-^ = 2, so that sin 6^=-j^ and cos ^j = -^ .

Also ri2 = 6, and 7-22=4, so that sjrf^=^2.

The coordinates of the foci referred to axes through C are therefore

FOCI OF A CONIC.

345

Their coordinates referred to the original axes OX and OY are

2Â±

V2

x/5'

2V2\

368. The method of obtaining the coordinates of the

focus of a parabola given by the general equation may be

exemplified by taking the example of Art. 363.

Here it was shewn that the latus rectum is equal to 3,

so that, if aS' be the focus, ^aS' is J.

It was also shewn that the coordinates of A referred to

OX and Y are â€” ^ and ^.

The coordinates of S referred to the same axes are

+

f and -I,

^.e.

2V and -I

Its coordinates referred to the original axes are therefore

_7_

20

W COS

i.e.

7

2 J)

Z- sin 6 and -J-pr sin + ~ cos

+

i-i and

2 5

^* a

7 3

20*5

nd 1-^3

7_

5

I. 1

5*5'

^,

100'

In Art. 393 equations will be found to give the foci of

any conic section directly, so that the conic need not first

be traced.

369. Ex. 1. Trace the curve

The equation may be written

.(l).

(^^^T-K'^^T- <^'-

Now the straight lines 3a;- 2?/ + 4 = and 2x + Sy-5 = are at

right angles. Let them be C3I and

GN, intersecting in C which is the

point (-t\, H)-

If P be any point on the curve

and PM and PN the perpendiculars

upon these lines, the lengths of P3I

and PN are

3x-2y + 4: ^ 2x + 3y-5

-^w ^^^ - ^w- â–

Hence equation (2) states that

3PM2 + 2P^â– 2=:3,

PM^

I.e.

PN^

346 COORDINATE GEOMETRY.

The locus of P is therefore an ellipse whose semi-axes measured

along CM and CN are a/I and 1 respectively.

Ex. 2. What is represented by the equation

{x'^-a^f + iy^-a^)^=a*?

The equation may be written in the form

a;4 + t/-i - 2a2 (a;2 + ?/2) + a* = 0,

i.e. {x^ + y^f-2a^{x^ + y^)+a*=2xY,

i.e. {x^ + y^-a^)^-{^2xyf=0,

i.e. {x^ + y/^^y +y^- Â«^) (^^ - fj^^y +y^- ^^) = O-

The locus therefore consists of the two ellipses

x^ + /ij2xy+y^-a^=0, and x^-fj2xy + y'^-a^=0.

These ellipses are equal and their semi-axes would be found to be

asj2 + ^2 and a ^2 -^2.

The major axis of the first is inclined at an angle of 135Â° to the

axis of X, and that of the second at an angle of 45Â°.

EXAMPLES. XLI.

Trace the parabolas

1. {x-4:ij)^=51y. 2. {x-y)^=x + y + l.

3. {5x-12yf=2ax + 2day + a\

4. {4a; + 3?/ + 15)2= 5 (3a; -4?/).

5. 16a;2 + 24a;2/ + 92/2-5a;-10?/ + l = O.

6. 9x^+24^y + 16y^-4:y-x + 7 = 0.

7. 144a;2 - 120xy + 25y^ + 619a; - 272^ + 663 = 0, and find its focus.

8. 16a;2-24a;2/ + 9?/2 + 32a; + 86?/-39 = 0.

9. 4a;2-4a;2/ + 2/2-12a; + 62/ + 9 = 0.

Find the position and magnitude of the axes of the conies

10. 12a;2-12a;i/ + 7?/2=48. 11. 3a;2+2a;?/ + 32/2=8.

12. x'^-xy~6y^=Q.

Trace the following central conies.

13. a;2-2a;2/cos2a + ^2â€” 2a2. 14. a;2-2a;2/cosec2a-f2/^ = a2.

15. xy = a{x + y). 16. xy-y^ = a^.

17. 2/^-2a;?/ + 2a;2 + 2a;-22/ = 0. 18. x'^ + xy + y^ + x+y = l.

LEXS. XLI.] TRACING OF CONIC SECTIONS. EXAMPLES. 347

19. 2x^ + Zxii-2y'^-lx + y-2 = Q.

20. 40a;2 + 36a;?/ + 25?/2- 196a; -122?/ + 205 = 0.

21. x^-%xy + y^ + 10x-l(iy + 21=^0.

22. x'^-xy + 2^2 _ 2ax - Qay + 7a^ = 0.

23. 10a;2 - ASxy - lOif + dSx + Uy - 5^ = 0.

24. 4:x^ + 27xy + S5y^~14M-Bly-&:=0.

25. {3x-4y + a){4:X + Sy + a) = a\

26. 3 (2a; -3^ + 4)2 + 2 (3a; + 2?/ -5)2 = 78.

27. 2 (3a; -4^/ + 5)2 -3 (4a; + 3?/ -10)2 = 150.

Find the products of the semi-axes of the conies

28. 2/2-4a;i/ + 5a;2 = 2. 29. 4(3a; + 42/- 7)2 + 3 (4a;-3t/ + 9)2=3.

30. lla;2 + 16a;?/ - ?/2- 70a; -40y + 82 = 0.

Find the foci and the eccentricity of the conies

31. a;2-3a;i/ + 4aa; = 2a2. 32. 4a;?/ - 3a;2 - 2a?/ = 0.

33. 5a;2 + 6a;?/ + 5?/ + 12a; + 4?/ + 6 = 0.

34. a;2 + 4a;?/ + ?/2-2a; + 2/y-6 = 0.

35. Shew that the latus rectum of the parabola

(a2 + 62) (^^2 + ^2) ^ (5^ + ay _ ahf

is 2ab-hs/a^~+^.

36. Prove that the lengths of the semi-axes of the conic

aa;2 + 2hxy + ay^ = d

are \/ â€” â€” r and

respectively, and that their equation is a;2_y2_o_

37. Prove that the squares of the semi-axes of the conic

ax2 + 2/?a;?/ + &?/2 + 2(;a; + 2/?/ + c =

are 2A-^ { {ab -h^)(a + h^ J {a - hf + 4/i2)} ,

where A is the discriminant.

38. If X be a variable parameter, prove that the locus of the

vertices of the hyperbolas given by the equation x^-y'^ + \xy = a^ is

the curve {x^+y^)^=a^{x^-y^).

39. If the point {at^^, 2at-^) on the parabola ?/2=4aa; be called the

point t^, prove that the axis of the second parabola through the four

points ^1, t^, ^3, and t^ makes with the axis of the first an angle

oot-(*lÂ±^Â±iÂ±^A.

Prove also that if two parabolas meet in four points the distances

of the centroid of the four points from the axes are proportional to the

latera recta.

348 COORDINATE GEOMETRY. [ExS. XLI.]

40. If the product of the axes of the conic x^ + 2xy + ny'^=8 be

unity, shew that the axes of coordinates are inclined at an angle

sin-i 1.

41. Sketch the curve Qx^-7xy-5y^~4:X + lly = 2, the axes being

inclined at an angle of 30Â°.

42. Prove that the eccentricity of the conic given by the general

equation satisfies the relation

e^ , (a + & - 2/1 cos w)2

h 4 = - â€”

1 - e^ " (a6 - h^) sin^ w '

where w is the angle between the axes.

43. The axes being changed in any way, without any change of

origin, prove that in the general equation of the second degree the

P + 9^ - 2fg cos 0} ap+bg^-2fgh , A

quantities c, '' â€¢^ . ^^ , -^â€” â€” â€” â€” =^^ , and -.-^- are

sm^w sm^w sin^w

invariants, in addition to the quantities in Art. 137.

[On making the most general substitutions of Art. 132 it is clear

that c is unaltered; proceed as in Art, 137, but introduce the condition

that the resulting expressions are equal to the product of two linear

quantities (Art. 116); the results will then follow.]

CHAPTER XVL

THE GENERAL CONIC.

370. In the present chapter we shall consider proper-

ties of conic sections which are given by the general equation

of the second degree, viz.

aoi? + '2Jnxy + hy^ + Igx + ^fy + c = (1).

For brevity, the left-hand side of this equation is often

called ^ (tc, 3/), so that the general equation to a conic is

^ (a;, y) = 0.

Similarly, <^(aj', y) denotes the value of the left-hand

side of (1) when x and y are substituted for x and y.

The equation (1) is often also written in the form aS'^O.

371. On dividing by c, the equation (1) contains five

independent constants - , - , - , - , and - .

c G G G c

To determine these five constants, we shall therefore

require five conditions. Conversely, if five independent

conditions be given, the constants can be determined.

Only one conic, or, at any rate, only a finite number of

conies, can be drawn to satisfy five independent conditions.

372. To find the equation to the tangent at any foint

{x\ y') of the conic section

cf> {x, y) = ax^ + 2hxy + hy"^ + 2gx + 2fy + c = 0. . .(1).

Let {x", y") be any other point on the conic.

350 COORDINATE GEOMETRY.

The equation to the straight line joining this point to

(x\ y) is

2/-2/' = |^(^-^') (2).

Since both (ic', 3/') and (a?", 2/") lie on (1), we have

aaj'2 + "Ihxy' + hy'^ + 2^cc' + 2/2/' + c - (3),

and aÂ£c"2 + 2^a;'y ' + hy'"" + 2^Â£c" + Ify" + c = (4).

Hence, by subtraction, we have

a {x^ - x'") + 2^ {x'y' - xy) + 5 {y"" - 3/"^;

+ 2^(x'-a.") + 2/(2/'-2/'') = (5).

But

2 (o^y - x"y") = {x' + x") {y' - y") + {x' - x") {y' + y"\

so that (5) can be written in the form

{^ - x") [a {x' + x") + h(y'+ y") + 2^]

+ ky' - y") [h {^' + x") + h{y' + y") + 2/] = 0,

y"-y' ^ a{x' + x") + h(y' + y") + 2g

*â– ^' x" -x' k {x' + x") + b{y' + y") + 2/ "

The equation to any secant is therefore

y y~ h{x' + x") + h{y'+y") + 2f^ a:;...^

To obtain the equation to the tangent at {x', y'), we put

x" = x' and y" = y' in this equation, and it becomes

, ax + hy' + <7 . ,,

y-y =- -7-7 â€” 1^, â€” > (^ - ^\

^ ^ hx +hy +f^ '

i. e. {ax + liy' + g) x + (hx + by +f)y

= ax'^ + 2hxy' + hy'^ + gx + fy

â€” â€” gx â€”fy â€” c, by equation (3).

The required equation is therefore

axx' + h (xy + x'y) +l>yy' + g (x + x) + f (y + y')

+ c = (7).

Cor. 1. The equation (7) may be written down, from

the general equation of the second degree, by substituting

XX foi- a^, yy' for 3/^, xy' + xy for 2xy, x + od for 2a?, and

y-vy iovly. (Cf. Art. 152.)

THE GENEKAL CONIC. POLE AND POLAR. 351

Cor. 2. If the conic pass through the origin we have

c = 0, and then the tangent at the origin (where x =0 and

y = 0) is gx +fy = 0,

i. e. the equation to the tangent at the origin is obtained by

equating to zero the terms of the lowest degree in the

equation to the conic.

373. The equation of the previous article may also be obtained

as follows ; If {x', y') and {x", y") be two points on the conic section,

the equation to the line joining them is

a{x-xf){x-x") + hl{x-x'){y-y") + {x-x"){ij-y')] + h{y-y'){y-y")

= ax^ + 2hxy + by^ + 2gx + 2fy + c (1).

For the terms of the second degree on the two sides of (1) cancel,

and the equation reduces to one of the first degree, thus representing

a straight line.

Also, since {x', y') lies on the curve, the equation is satisfied by

putting x=x' and y=y'.

Hence {x', y') is a point lying on (1).

So {x'\ y") lies on (1).

It therefore is the straight line joining them.

Putting x" â€” x' and y" =y' we have, as the equation to the tangent

at {x', y'),

a{x- x'f + 27i {x - x') [y - if) + & {y - y'f

= ax^ + ^hxy + 'by'^ + 2gx + 2fy + c,

i.e. 2axx' + 2}i {x'y + xy') + 2\)yy' + 2gx + 2fy + c

= ax"^ + 2hx'y' + hy"^

â€” - 2gx' - 2fy' - c, since {x', y') lies on the conic.

Hence the equation (7) of the last article.

374. To find the condition that any straight line

lx + my + n = (1),

may touch the conic

ax^ + 2hxy + by^ + 2gx + 2fy + c = (2).

Substituting for y in (2) from (1), we have for the equation giving

the abscissae of the points of intersection of (1) and (2),

x^ {am? - 2hlm + hP) - 2x (hmn - bin - gm^ +flm)

+ bn^-2fmn + cm^=0 (3).

If (1) be a tangent, the values of x given by (3) must be equal.

The condition for this is, (Art. 1,)

{hmn - bin - gm^+fhrif={am? - 2hlm + bl^) {bn^ - 2fmn + cm^).

352 COORDINATE GEOMETRY.

On simplifying, we have, after division by iii^,

p {jic -p) + m2 {ca - p2) + n^ [ah - li') + 2mn [gh - af) + 2wZ {lif - hg)

+ 2lm{fg-ch) = 0.

Ex. Find the equations to the tangents to the conic

x^ + 4:xy + 3y^-5x-Gy + S = (1),

ivhich are parallel to the straight line a; + 4i/ = 0.

Tlie equation to any such tangent is

a; + 4^ + c = (2),

where c is to be determined.

This straight line meets (1) in points given by

3x2 _ 2x (5c + 28) + 3c2 + 24c + 48 = 0.

The roots of this equation are equal, i.e. the line (2) is a tangent,

if {2(5c + 28)}2 = 4. 3.(3c2 + 24c+48), i.e. if c=-5 or -8.

The required tangents are therefore

a; + 4?/-5 = 0, and x + iij -Q â€” 0.

375. As in Arts. 214 and 274 it may be proved that

the polar of {x\ y) with respect to ^ (cc, y) = is

{ax' + hy -{â– g) X + {hx + by' +/) y + gx +/y' + c = 0.

The form of the equation to a polar is therefore the

same as that of a tangent.

Just as in Art. 217 it may now be shewn that, if the

polar of F passes through T, the polar of T passes through

P.

The chord of the conic which is bisected at {x\ y),

being parallel to the polar of (x, y) [Arts. 221 and 280],

has as equation

{ax + hy +g){x- x) + {hx' + by +/){y â€” y) = 0.

376. To find the equation to the diameter bisecting all

chords parallel to the straight line y = mx.

Any such chord is y = 7nx->rK (1).

This meets the conic section

ax^ + 2hxy + by"^ + 2gx + 2fy + c =

in points whose abscissae are given by

ax^ + ^hx {mx + K) + b {mx + Kf + 2gx + 2f{mx + K)+ g=^0,

i. e. by x^ {a + 2hm + bm^) + 2x {hK + bmK + ^ + fm)

+ blO + 2fK-\-c = 0.

THE GENERAL CONIC. CONJUGATE DIAMETERS. 353

If fl?i and x^ be the roots of this equation, we therefore

have

X ^x - cy{ h + hm)K + g+fm

^ ^ a + 2hm + bm^

Let {X, Y) be the middle point of the required chord,

so that

â€ž X1 + X2 (h + bm) K+ gf+/7n , .

2 a + 2hm + b7)iF ' ^ ^'

Also, since (X, Y) lies on (1) we have

Y-^mX+K (3).

If between (2) and (3) we eliminate K we have a

relation between X and Y.

This relation is

â€” (a + 2hm + bm^) X = (h + bm) {Y â€” mX) + g +fm,

i. e. X [a + hm) + Yih+ bm) + g -Â¥fm = 0.

The locus of the required middle point is therefore the

straight line whose equation is

x{a + hm) + y(h + bm) + g â– \-fm == 0.

If this be parallel to the straight line y - m'x, we

have

a + Am

m =

â€¢w,

h + bm

i.e. a + h (m + m') + bmm' = (5).

This is therefore the condition that the two straight

lines y â€” tnx and y â€” m!x may be parallel to conjugate

diameters of the conic given by the general equation.

377. To find the condition that the pair of straight lines, whose

equation is

Ax'^ + 2Hxy+By^=0 (1),

may be parallel to conjugate diameters of the general conic

ax'^ + 2hxy + by^ + 2gx + 2fy + c = .'. (2).

Let the equations of the straight lines represented by (1) hey = mx

and y = m'x, so that (1) is equivalent to

B [y - mx) [y - m'x) = 0,

and hence m-\-m'= - â€”, and mm =^ .

L. 23

354 COORDINATE GEOMETRY.

By the condition of the last article it therefore follows that the

lines (1) are parallel to conjugate diameters if

i.e. if Ah-2Hh + Ba = 0.

378. To prove that tioo concentric conic sections always have a

pair, and only one pair, of common conjugate diameters and to find

their equation.

Let the two concentric conic sections be

ax^ + 2hxy + hy'^=l (1),

and a'x^^-2h'xy + l)Y = l (2).

The straight lines

Ax^ + 2Hxy + By^ = (3),

are conjugate diameters of both (1) and (2) if

Ab-2Hh + Ba = 0,

and Ab'-2Hh' + Ba' = 0.

Solving these two equations we have

A -2H__ B

ha' - h'a ~ W^afb ~ hh' - b'h '

Substituting these values in (3), we see that the straight lines

x^ha'-h'a)-xy{ab'-a'b)4-y^{bh'-b'h) = (4)

are always conjugate diameters of both (1) and (2).

Hence there is always a pair of conjugate diameters, real, coinci-

dent, or imaginary, which are common to any two concentric conic

sections.

EXAMPLES. XLII.

1. How many other conditions can a conic section satisfy when

we are given (1) its centre, (2) its focus, (3) its eccentricity, (4) its

axes, (5) a tangent, (6) a tangent and its point of contact, (7) the

position of one of its asymptotes?

2. Find the condition that the straight line lx + niy = l may

touch the parabola {ax-bijf-2 {a^ + &-) (ax + by) + {a^ + b'^f = 0, and

shew that if this straight line meet the axes in P and Q, then PQ

will, when it is a tangent, subtend a right angle at the point {a, b).

3. Two parabolas have a common focus ; prove that the perpen-

dicular from it upon the common tangent passes through the

intersection of the directrices.

[EXS. XLII.] INTERSECTIONS OF TWO CONICS. 355

4. Shew that the conic -^ H r- cos a + i-:-, = sin^ a is inscribed in

a^ ab h^

the rectangle, the equations to whose sides are x^ = a^ and y^ = h^, and

that the quadrilateral formed by joining the points of contact is of

constant perimeter 4 sja^ + 6^, whatever be the value of a.

5. A variable tangent to a conic meets two fixed tangents in two

points, P and Q ; prove that the locus of the middle point of PQ is a

conic which becomes a straight line when the given conic is a parabola.

6. Prove that the chord of contact of tangents, drawn from an

external point to the conic ax^ + 2hxy + hy'^=l, subtends a right angle

at the centre if the point lie on the conic

a;2 (a2 + 7i2) + 2h {a + h) xy+y'^ {h'^ + h^) = a + b.

7. Given the focus and directrix of a conic, prove that the polar

of a given point with respect to it passes through another fixed point.

8. Prove that the locus of the centres of conies which touch the

axes at distances a and b from the origin is the straight line ay = bx.

9. Prove that the locus of the poles of tangents to the conic

ax'^ + 2hocy + by^=l with respect to the conic a'x"^ + 2h'xy + b'y^ = l is

the conic

a {Ji'x + b'yf - 27i [a'x + h'y) {h'x + b'y) + b {a'x + h'yf=db - h\

10. Find the equations to the straight lines which are conjugate

to the coordinate axes with respect to the conic Ax^-{-2Hxy + By^ = l.

Find the condition that they may coincide, and interpret the

result.

11. Find the equation to the common conjugate diameters of the

conies (1) a;2 + 4a;y + 6!/2=l and 2x'^ + &xy + ^y^ â€” l,

and (2) 2x^-^xy + ^y'^ = l and 2x'^ + ^xy -^y^=l.

12. Prove that the points of intersection of the conies

ax^ ^^hxy + by^=l and a'x^ + 2h'xy-\-b'y'^ = l

are at the ends of conjugate diameters of the first conic, if

ab'-{-a'b-2hh' = 2{ah-U'^).

13. Prove that the equation to the equi-conjugate diameters of

â€ž ^, , â€ž ^. ax^ + 2hxy + by^ 2(x^ + y'^)

the conic ax^ + 2kxy + by"^ = 1 is râ€” f^ â€” â€” = â€” ^ â€” rr-^ â€¢

^ ^ ab-lv- a + b

379. Two conies, in general, intersect in four points,

real or imaginary.

For the general equation to two conies can be written

in the form

aay^ + 2x Qiy + g) + hy^ + 2fy + c = 0,

and a'x^ + 2x (Ky + g) + 6^' + %f'y + C - 0.

23â€”2

356 COORDINATE GEOMETRY.

Eliminating x from these equations, we find that the

result is an equation of the fourth degree in y, giving

therefore four values, real or imaginary, for y. Also, by

eliminating x^ from these two equations, we see that there

is only one value of x for each value of y. There are there-

fore only four points of intersection.

380. Equation to any conic passing through the inter-

section of two given conies.

Let S^ ax"" + 2hxy + hy"" + 2gx + ^fy + c = (1),

and S' E ax" + 2h'xy + h'y^ + 2g'x + If'y + c' = . . . (2)

be the equations to the two given conies.

Then ^-_X.S" = (3)

is the equation to any conic passing through the inter-

sections of (1) and (2).

For, since S and aS^' are both of the second degree in x

and 2/, the equation (3) is of the second degree, and hence

represents a conic section.

Also, since (3) is satisfied when both ^S' and S' are zero,

it is satisfied by the points (real or imaginary) which are

common to (1) and (2).

Hence (3) is a conic which passes through the intersec-

tions of (1) and (2).

381. To find the equations to the straight lines passing

through the intersections of two conies given hy the general

equations.

As in the last article, the equation

{a - \a) 0^2 + 2 (A - Xh') xy + (h - \h') ^ + 2 (^ - Xg') x

+ 2(f-\r)y + {c-Xc') = (1),

represents some conic through 4}he intersections of the given

conies.

Now, by Art. 116, (1) represents straight lines if

{a - \a') (b - Kb') (c - Xc') + 2 (f - Xf) (g - Xg') (h - Xh')

-{a- Xa') (/- Xf'f -{b- Xb') (g - Xg'f - (c - Xc') (h - XhJ

= (2).

INTERSECTIONS OF TWO CONIGS. 357

Now (2) is a cubic equation. The thi-ee values of \

found from it will, when substituted successively in (1),

give the three pairs of straight lines which can be drawn

through the (real or imaginary) intersections of the two

conies.

Also, since a cubic equation always has one real root,

one value of A. is real, and it could be shown that there can

always be drawn one pair of real straight lines through the

intersections of two conies.

382. All conies tvhich pass through the intersections of

two rectangular hyperbolas are themselves rectangular hyper-

bolas.

In this case, \i S = and S' â€” be the two rectangular

hyperbolas, we have

Â« + 6 = 0, and a +b' = 0. (Art. 358.)

Hence, in the conic S â€” X;S" â€” 0, the sum of the co-

efficients of x^ and y"^

= (a- Xa') + {b- Xb') =:(a + b)-X{a' + b') - 0.

Hence, the conic S ~ XS' ^ 0, i.e. any conic through the

intersections of the two rectangular hyperbolas, is itself a

rectangular hyperbola.

Cor. If two rectangular hyperbolas intersect in four points

A, B, G, and D, the two straight lines AD and BC, which are a conic

through the intersection of the two hyperbolas, must be a rectangular

hyperbola. Hence AD and BC must be at right angles. Similarly,

BD and GA, and GD and AB, must be at right angles. Hence D is

the orthocentre of the triangle ABG.

Therefore, if two rectangular hyperbolas intersect in four points,

each point is the orthocentre of the triangle formed by the other

three.

383. 7/* Z = 0, i/== 0, ^= 0, and 11 = be the equations

to the four sides of a quadrilateral taken in order, the

equation to any conic passing through its angular points is

LN=X.MR (1).

For L = passes through one pair of its angular points

and N=0 passes through the other pair. Hence LN = is

the equation to a conic (viz. a pair of straight lines) passing

through the four angular points.

358 COORDINATE GEOMETRY.

Similarly MR - is the equation to another conic

passing through the four points.

Hence LN = X . MR is the equation to any conic through

the four points.

Geometrical meaning. Since L is proportional to the perpen-

dicular from any point {x, y) upon the straight line Z/ = 0, the

relation (1) states that the product of the perpendiculars from any

point of the curve upon the straight lines JL = and N=0 is propor-

tional to the product of the perpendiculars from the same point upon

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