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S. L. (Sidney Luxton) Loney.

The elements of coordinate geometry online

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/oF e
If e > 1, then F lies on one or
other of the two straight lines SU
and SU' inclined to KK' at an angle



t\H


K


^


^-^


IVI


^


zp^v




t


\


^


X




K'


^


\y*



^'-■©-



GENERAL EQUATION OF THE SECOND DEGREE. 323

If e = 1, then PSM is a right angle, and the locus
becomes two coincident straight lines coinciding with SX.

If e< 1, the z PSM is imaginary, and the locus consists
of two imaginary straight lines.

If, again, both KK' and S be at infinity and S be on
KK\ the lines SU and SU' of the previous figure will be
two straight lines meeting at infinity, i.e. will be two
parallel straight lines.

Finally, it may happen that the axes of an ellipse may
both be zero, so that it reduces to a point.

Under the head of a conic section we must therefore
include :

(1) An Ellipse (including a circle and a point).

(2) A Parabola.

(3) A Hyperbola.

(4) Two straight lines, real or imaginary, inter-
secting, coincident, or parallel.

349. To shew that the general equation of the second
degree

aoi? + 2hxy + hy^ + 2gx + 2/?/ + c = (1)

always represents a conic section.

Let the axes of coordinates be turned through an angle
6, so that, as in Art. 129, we substitute for x and y the
quantities x cos 6 — y sin 6 and x sin 6 + y cos 9 respec-
tively.

The equation (1) then becomes

a (x cos — y sin Oy + 2h (x cos — y sin 0) {x sin + y cos 6)

+ h(x sin + y cos Oy + 2g (x cos — y sin 6)

+ 2/{x sin 6 + y cos 6) + c = 0,

i. e. x^ (a cos^ + 2h cos sin + b sin^ 0)

+ 2xy {h (cos^ - sin^ 0)-(a — h) cos 6 sin 6]

+ 2/^ {a sin^ 6 - 2h cos OsinB + h cos^ 6) + 2x (g cos +/sin 0)

+ 2y{fcosO-gsine) + c = (2) .

21—2



324 COORDINATE GEOMETRY.

Now choose the angle 6 so that the coefficient of xy in
this equation may vanish,

i. e. so that h (cos^ 6 — sin^ 0) — (a — h) sin 6 cos ^,

i. e. 2h cos 20 = (a — b) sin 20,

2h
i. e. so that tan 20 = -, .



a



Whatever be the values of a, b, and h, there is always
a value of satisfying this equation and such that it lies
between — 45° and + 45°. The values of sin and cos are
therefore known.

On substituting their values in (2), let it become

Ax'' + JBy'' + 2Gx + 2Fy + c = (3).

First, let neither A nor B be zero.

The equation (3) may then be written in the form

Gy „/ i<^V ^' ^'



,/ Gy „/ ry g' r'

A[x^-)^B[y^-j^)=-^^^-c.

— -7 , ~ n)



The equation becomes



G^ F^
Ax^ + Bf = -j +-^-c = K (say) (4),



B

K ' K
B



i-e. Tr^^,-^ • (5).



K K

If — and -^ be both positive, the equation represents an

ellipse. (Art. 247.)

K K

If -J and — be one positive and the other negative, it
An

represents a hyperbola (Art. 295). If they be both

negative, the locus is an imaginary ellipse.

If X be zero, then (4) represents two straight lines,
which are real or imaginary according as A and B have
opposite or the same signs.



CENTRE OF A CONIC SECTION. 325

Secondly^ let either J. or ^ be zero, and let it be ^.
Then (3) can be written in the form

G F^



-(-3



F\2



^ "^ 2G 2BG



y



0.

Transform the origin to the point whose coordinates
are

__c_ J^ _F
~2G'^2M' ~B
This equation then becomes

By^+2Gx^0,

2G
^.e. r^ - ^^'

which represents a parabola, (Art. 197.)

If, in addition to A being zero, we also have G zero, the
equation (3) becomes

By^ + 2Fy + c = 0,



F F^ G

'^^B^^\/w-r

and this represents two parallel straight lines, real or
imaginary.

Thus in every case the general equation represents one
of the conic sections enumerated in Art. 348.

350. Centre of a Conic Section. Def. The

centre of a conic section is a point such that all chords of
the conic which pass through it are bisected there.

When the equation to the conic is in the form

ax^ + 2hxy + by'^ + g-0 (1),

the origin is the centre.

For let (x, y) be any point on (1), so that we have

aa;'2 + 2Aa;y + 62/'' + c = (2).

This equation may be written in the form

a (- ^f + 2A (- a;') (- y') + 6 (- yj + c = 0,

and hence shews that the point (— a:', — y'^ also lies on (1).



326 COORDINATE GEOMETRY.

But the points {x\ y) and {—x\ —y) lie on the same
straight line through the origin, and are at equal distances
from the origin.

The chord of the conic which passes through the origin
and any point {x, y) of the curve is therefore bisected at
the origin.

The origin is therefore the centre.

351. When the equation to the conic is given in the
form

aa? + 27t£C2/ + hy"^ + 2gx + 2fy + c = (1),

the origin is the centre only when both f and g are zero.

For, if the origin be the centre, then corresponding
to each point {x, y) on (1), there must be also a point
(—a;', —y') lying on the curve.

Hence we must have

ax^ + 'i.hxy + hy'^ + 2gx +'ify'-\-c = (2),

and ax'^ + Ihx'y' + hy'^^ - 2gx' -2fy' + c=^0 (3).

Subtracting (3) from (2), we have

gx +fy' - 0.

This relation is to be true for all the points {og\ y')
which lie on the curve (1). But this can only be the case
when g — and f= 0.

352. To obtain the coordinates of the centre of the
conic given hy the general equation, and to obtain the
equation to the curve referred to axes through the centre
parallel to the original axes.

Transform the origin to the point (S, y)j so that for x
and y we have to substitute x + x and y + y. The equation
then becomes

a{x + xf + 2h{x + x){y + y) + b (y + yY + 2g(x + x)

i. e. ax^ + 2hxy + by^ + 2x (ax + hy + g) + 2y (Jix + by +f)
+ ax^'}-2hxy + by'^ + 2gx + 2fy + c=^0 (2).



EQUATION REFERKED TO THE CENTRE. 327

If the point (x, y) be the centre of the conic section, the
coefficients of x and y in the equation (2) must vanish, so
that we have

ax + hy + g = (3),

and hx + hy +f = (4).

Solving (3) and (4), we have, in general,

With these values the constant term in (2)

= a^ + 2hxy + hy^ + 2gx + Ify + c

= X {ax + hy + g) + y (hx + hy +/) + gx+/y + c

= gx + fy + c (6),

by equations (3) and (4),

abc + 2 fgh — af^ — bi/^ — ch^ , ,. _,

= ^^ — — , by equations (5),

A

~ab^^^'

where A is the discriminant of the given general equation
(Art. 118).

The equation (2) can therefore be written in the form

aaf + 2hxy + by^ + -7 — ^ = 0.

This is the required equation referred to the new axes
through the centre.

Ex. Find the centre of the conic section

2x^ - 5xy - Sy^ ~x-iy + Q = 0,
and its equation when transformed to the centre.

The centre is given by the equations 2x~^y-^ = 0, and
-^x-By -2 = 0, sothatS=-y, and ?/= -|.
The equation referred to the centre is then
2x^-5xy-Sy^ + c' = 0,
where c'= -|. S-2.?7 + 6 = ^+f + 6 = 7. (Art. 352.)

The required equation is thus

2x^-5xy-3y^ + 7 = 0.



328 COORDINATE GEOMETRY.

353. Sometimes the equations (3) and (4) of the last
article do not give suitable values for x and y.

For, if ah — 1^ be zero, the values of x and y in (5) are
both infinite. When ah — h^ is zero, the conic section is a
parabola.

The centre of a parabola is therefore at infinity.

Again, if - = - = ^ , the result (5) of the last article is

of the form ~ and the equations (3) and (4) reduce to the
same equation, viz.,

ax + hy + g = 0.

We then have only one equation to determine the
centre, and there is therefore an infinite number of centres
all lying on the straight line

ax -{-hy -{-g — O.

In this case the conic section consists of a pair of
parallel straight lines, both parallel to the line of centres.

354. The student who is acquainted with the Dif-
ferential Calculus will observe, from equations (3) and (4)
of Art. 352, that the coordinates of the centre satisfy the
equations that are obtained by difierentiating, with regard
to x and y, the original equation of the conic section.

It will also be observed that the coefficients of x, y, and
unity in the equations (3), (4), and (6) of Art. 352 are the
quantities (in the order in which they occur) which make
up the determinant of Art. 118.

This determinant being easy to write down, the student
may thence recollect the equations for the centre and the
value of c.

The reason why this relation holds will appear from the
next article.

355. Ex. Find the condition that the general equation of the
second degree may represent two straight lines.

The centre {x, y) of the conic is given by

ax-\-hy-\-g = (1),

and hx + by+f=0 (2).



EQUATION TO THE ASYMPTOTES. 329

Also, if it be transformed to the centre as origin, the equation
becomes

ax^+2hx7j + hif + c' = (3),

where c' = gx+fy + c.

Now the equation (3) represents two straight lines if c' be zero,
i.e. if gx+fy + c—Q (4).

The equation therefore represents two straight lines if the relations
(1), (2), and (4) be simultaneously true.

Eliminating the quantities ^ and y from these equations, we have,
by Art. 12,

a, h, g

h, b, f =0.

This is the condition found in Art. 118.

356. To find the equation to the asymptotes of the conic
section given hy the general equation of the second degree.

Let the equation be

a'3(? + 2hxy + hy'^ + 2gx + ^fy + c-O... (1).

Since the equation to the asymptotes has been shewn to
differ from the equation to the curve only in its constant
term, the required equation must be

ax^ + '2hxy + hy^ + 2gx + 2fy + c + A, = (2).

Also (2) is to be a pair of straight lines.
Hence
ah (c + X) + ^fgh - af - 6/ - (c + X) K- = 0. (Art. 116.)

ahG + 2fgh-af^-hg^-c¥ A

ineretore A = — ■ ; ,„ — = -^ — j-^.

ao — a^ ab — A"

The required equation to the asymptotes is therefore

ax^ + 2hxy + hy^ + 2gx + 2/y + c j — 7^= . . .(2).

Cor. Since the equation to the hyperbola, which is
conjugate to a given hyperbola, differs as much from the
equation to the common asymptotes as the original equation
does, it follows that the equation to the hyperbola, which is
conjugate to the hyperbola (1), is

ax^ + 2hxy + hy"^ + 2gx + 2fy + c— 2 - — y^ = 0.



330 COORDINATE GEOMETRY.

357. To determine hy an examination of the general
equation what hind of conic section it represents.

[On applying the method of Art. 313 to the ellipse and
parabola, it would be found that the asymptotes of the
ellipse are imaginary, and that a parabola only has one
asymptote, which is at an infinite distance and perpen-
dicular to its axis.]

The straight lines as(? + Ihxy + hy'^ = (1)

are parallel to the lines (2) of the last article, and hence
represent straight lines parallel to the asymptotes.

Now the equation (1) represents real, coincident, or
imaginary straight lines according as J^ is >= or <a6,
i.e. the asymptotes are real, coincident, or imaginary,
according as A^ > = or < ab^ i. e. the conic section is a hyper-
bola, parabola, or ellipse, according as 1^ > = or < ah.

Again, the lines (1) are at right angles, i.e. the curve is
a rectangular hyperbola, if a ^-h — ^.

Also, by Art. 143, the general equation represents a
circle if a = h, and h = 0.

Finally, by Art. 116, the equation represents a pair of
straight lines if A = ; also these straight lines are parallel
if the terms of the second degree form a perfect square, i.e.
if h^ = ah.

358. The results for the general equation

ao(? + 2hxy + hy^ + 2gx + 2fy + c =

are collected in the following table, the axes of coordinates
being rectangular.



Curve.

Ellipse.
Parabola.
Hyperbola.
Circle.

Rectangular hyperbola.
Two straight lines, real or
imaginary.

Two parallel straight lines.



Condition.

h^ < ah.

A^ = ah.

W- > ah.

a = h, and A = 0.

a + h = 0.

A-O,

i.e.

ahc+2fgh-af^-hg''-ch''^0.
A = 0, and h^ ~ ah.



EXAMPLES. 331

If the axes of coordinates be oblique, the lines (1) of Art. 356 are
at right angles if a + 6- 2/icos a;=0 (Art. 93); so that the conic
section is a rectangular hyperbola ii a + b-2h cos w = 0.

Also, by Art. 175, the conic section is a circle if & = a and

h= a cos la.

The conditions for the other cases in the previous article are the
same for both oblique and rectangular axes.



EXAMPLES. XL.

What conies do the following equations represent? "When
possible, find their centres, and also their equations referred to the
centre.

1. 12x^ - 2Sxy + 10y^-25x + 2Qy = U.

2. lSx^-18xy + S7y^ + 2x + Uy -2 = 0.

3. y^-2^3xy + 3x^ + 6x-4:y + 5 = 0.

4. 2x^-72xy + 2Sy^-4:X-2Sy-^8 = 0.

5. 6x^-.5xy-Qy'^ + 14:X + 5y + 4:=0.

6. Sx^-8xy-Sy^ + 10x-lSy + 8 = 0.

Find the asymptotes of the following hyperbolas and also the
equations to their conjugate hyperbolas.

7. 8x^-\-10xy-Sy^-2x + 4:y = 2. 8. y^-xy -2x'^-5y + x-6 = 0.
9. 55x^-120xy + 20y^ + G4:X-48y = Q.

10. 19a;2 + 24:xy + %/- 22a; - 6?/ = 0.

11. If (S, y) be the centre of the conic section

f{x, y) = ax^ + 2hxy + hij'^ + 2gx 4- 2fy + c = 0,
prove that the equation to the asymptotes is/ (a;, y)=f[x, y).

If t be a variable quantity, find the locus of the point {x, y) when



12. a; = a f i + - j and ?/=« ( * - )



13. x = at + U' and y = bt + at^.

14. x = l + t + t'^ and y = l-t + t\

If ^ be a variable angle, find the locus of the point {x, y) when

15. a; = a tan (^ + a) and y = btsin{d + ^).

16. a; = acos(^ + a) and ?/ = &cos (^ + i8).
What are represented by the equations

17. {x-y)^ + (x-a)^=0. 18. xy + a^=a{x + 2j).



332



COOKDINATE GEOMETRY.



[Exs. XL.]



19. x^-7f = {y-a){x'^- 2/2).

20. x^ + y^-xy{x + y) + a^{y-x) = Q.
22. x^ + y'^ + {x + y) {xy-ax-ay) = 0.
24. {r cos 6-a){r-a cos 6) = 0.

26. r+-=3cos^ + sin0.
r

28. r (4 - 3 sin'-^ d) = 8a cos d.



21. (a;2-a2)2_ 2/4=0.
23. a;2 + a;r/ + 2/2=0.
25. rsin2^ = 2acos^.

27. -=l + cos^ + ^3sin^.



359. To trace the parabola given by the general equa-
tion of the second degree

a^ + 2hxy + b'lf' + 2^cc + yy + c = (1),

and to find its latus rectum.

First Method. Since the curve is a parabola we
have A^ = ab, so that the terms of the second degree form
a perfect square.

Put then a — c^ and 6 = yS^, so that h = a(B, and the
equation (1) becomes

{ax + Pyf + 2gx-v 2fy + c = (2).

Let the direction of the axes be changed so that the

straight line ax + Py = 0, i.e. y = — -^x, may be the new-
axis of -Z".




We have therefore to turn the axes through an angle

/8



such that tan ^ = — — , and therefore



sin = —



Va2 + ^2



and cos 6 =



sJa^ + P"'



TRACING OF PARABOLAS. 333

For X we have to substitute

JT cos — Fsin 0, i. e. ,

Ja' + ft'
and for y the quantity

JTsin^ + Tcos^, i.e. -" - ^+1^ ^ (Art. 129.)

For ax + Py we therefore substitute Y \J{a? + y8^).
The equation (2) then becomes

T' (a? + yS') + ^jj^[9 (P^ + « Y) +/(lir - aX)] + = 0,



^.e.



<^-^)^^'(£^f^-^j (^)'

where /i = - ~^ — Bl (4)

and ^2^l=J^xff=K^- ^



I.e.



H=



{a'+0 " " a^ + yS^'






The equation (3) represents a parabola whose latus

rectum is 2 , , whose axis is parallel to the new axis

(a' + JS'f
of X, and whose vertex referred to the new axes is the
point (R, K).

360. Equation of the axis, and coordinates of the
vertex, referred to the original axes.

Since the axis of the curve is parallel to the new axis of
X, it makes an angle 9 with the old axis of x, and hence
the perpendicular on it from the origin makes an angle
90° + ^.

Also the length of this perpendicular is K.



384 COORDINATE GEOMETRY.

The equation to the axis of the parabola is therefore
X cos (90° + 6)+y sin (90° + 6) = K,
i.e. —X sin + y cos = IC,

i.e. ax^Py = Kj^TW=^-"^^ (6).

Again, the vertex is the point in which the axis (6)
meets the curve (2).

We have therefore to solve (6) and (2), i.e. (6) and

^ilT^+^^^+^-^2/ + '^ = o (7).

The solution of (6) and (7) therefore gives the required
coordinates of the vertex.

361. It was proved in Art. 224 that if PK be a
diameter of the parabola and Q V the ordinate to it drawn
through any point Q of the curve, so that ^ F is parallel to
the tangent at P, and if 6 be the angle between the diameter
P V and the tangent at P, then

^F2 = 4acosec2^.PF (1).

If QL be perpendicular to P F and QL' be perpendicular
to the tangent at P, we have

QL^QVBm.e, and ^i:' = PFsin6i,

so that (1) is QL^ - ia cosec . QL'.

Hence the square of the perpendicular distance of any
point Q on the parabola from any diameter varies as the
perpendicular distance of Q from the tangent at the end of
the diameter.

Hence, if Ax + By + C — be the equation of any
diameter and A'x + B'y + C — be the equation of the
tangent at its end, the equation to the parabola is

{Ax + By-hCf = X{A'x-\-B'y^C') (2),

where X is some constant.

Conversely, if the equation to a parabola can be reduced
to the form (2), then

Ax-\-By+G^O (3)



TRACING OF PARABOLAS. 335

is a diameter of the parabola and the axis of the parabola is
parallel to (3).

We shall apply this property in the following article.

362. To trace the parabola given hy the general equa-
tion of the second degree

ax^+ ^hxy + hy"^ + 2gx + %fy + g=^ (1).

Second Method. Since the curve is a parabola, the
terms of the second degree must form a perfect square
and A^ = ah.

Put then a — a^ and h ~ /3^, so that h = ayS, and the
equation (1) becomes

{ax + ^yf = -{2gx + 2/y + c) (2).

As in the last article the straight line ax + jSy = is a
diameter, and the axis of the parabola is therefore parallel
to it, and so its equation is of the form

ax + /3y + \ = (3).

The equation (2) may therefore be written

{ax + l3y + Xy = -{2gx+ 2fy + c) + X^+2\ (ax + fy)

= 2x{Xa-g) + 2y((3\-f) + X'-G (4).

Choose A, so that the straight lines

ax + /3y + \ = (5)

and 2x{Xa-g) + 2y(j3X-f)+X^-G = (6)

are at right angles, i.e. so that

a{Xa-g) + pH3X-/) = 0,

i.e. so that X= „ „ „ (7).

a? + ^^ ^ '

The lines (5) and (6) are now, by the last article, a
diameter and a tangent at its extremity ; also, since they
are at right angles, they must be the axis and the tangent
at the vertex.



336 COORDINATE GEOMETRY.

The equation (4) may now, by (7), be written



z e.






where PiV is the perpendicular from any point F of the
curve on the axis, and A is the vertex.

Hence the axis and tangent at the vertex are the lines
(5) and (6), where X. has the value (7), and the latus rectum



= 2



f •



(a? + 13^)

363. Ex. Trace the parabola

9a;2 - 'iixy + 16r/2 - 18a; - 101?/ + 19 = 0.

The equation is

(3a;-42/)2-18a;-101^ + 19 = (1).

First Method. Take ^x-^y = as the new axis of x, i.e. turn
the axes through an angle ^, where tan^ = |, and therefore sin^ = |
and cos^=|.

For X we therefore substitute Xcos^-Fsin^, i.e. — = ; for

5

3X + 4Y
y we put Xsin^+Fcos^, i.e. = — , and hence for ^x-Ay the

quantity -57.

The equation (1) therefore becomes

2572_i[72Z-54Y]-i[303Z+404r| + 19 = 0,

i.e. 25r2-75Z-70r+19=0 (2).

This is the equation to the curve referred to the axes OX and OY.
But (2) can be written in the form

147
T



r2_i_=3Z-^|,



i.e. (7-^)2=3Z-i| + f|=3(Z+f).



TRACING OF PARABOLAS.



337



Take a point A whose coordinates referred to OX and OY are -|
and I, and draw AL and AM parallel to OX and OY respectively.




lA

Referred to AL and AM the equation to the parabola is Y^=SX.
It is therefore a parabola, whose vertex is A, whose latua rectum is 3,
and whose axis is AL.

Second Metliod. The equation (1) can be written

(3a;-4?/ + X)2=(6\ + 18)a; + ?/{101-8X) + X2-19 (3).

Choose X so that the straight lines

Sx-4y + \=0
and (6X + 18)a; + y (101- 8X) + X2- 19 =

may be at right angles.
Hence X is given by

3 (6X + 18) - 4 (101 - 8X) = (Art. 69),
and therefore X = 7.

The equation (3) then becomes

(8a; -4:y + If = 15 {4.x + 3i/ + 2),

(^^r - ^^^ <^)-

Let AL be the straight line

3a;-4^ + 7 = (5),

and blithe straight line 4a; + 3z/ + 2 = (6).

These are at right angles.

If P be any point on the parabola and FN be perpendicular to
AL, the equation (4) gives PN^=S . AN.

Hence, as in the first method, we have the parabola.

The vertex is found by solving (5) and (6) and is therefore the
point (-If, ft).

L. 22



338 COORDINATE GEOMETRY.

In drawing curves it is often advisable, as a verification, to find
whether they cut the original axes of coordinates.

Thus the points in which the given parabola cuts the axis of x
are found by putting 2/ = in the original equation. The resulting
equation is Occ^- 18a; + 19 = 0, which has imaginary roots.

The parabola does not therefore meet Ox.

Similarly it meets Oy in points given by 16?/2- 101i/ + 19 = 0, the
roots of which are nearly %\ and y\.

The values of OQ and OQ' should therefore be nearly ^^ and 6^.

364. To find the direction and magnitude of the axes
of the central conic section

aQ(? + 2Axy + hy"^ =1 (1).

First Method. We know that, when the equation to
a central conic section has no term containing xy and the
axes are rectangular, the axes of coordinates are the axes of
the curve.

Now in Art. 349 we shewed that, to get rid of the term
involving xy^ we must turn the axes through an angle 6
given by

tan2^ - ^ (2).

a — ^ '

The axes of the curve are therefore inclined to the axes
of coordinates at an angle 6 given by (2).
Now (2) can be written

2 tan ^ _ 2A _ 1
l-tan^^"^^6-X^'^^^'

.-. tan2^ + 2A.tan(9-l =0 (3).

This, being a quadratic equation, gives two values for ^,
which differ by a right angle, since the product of the two
values of tan ^ is — 1. Let these values be 6^ and 6^, which
are therefore the inclinations of the required axes of the
curve to the axis of x.

Again, in polar coordinates, equation (1) may be written
r"" (a cos2 e + 2h cos (9 sin ^ + 6 sin' $)=! = cos^ 6 + sin' (9,
i.e.

cos2(9 4-sin2^ 1+tan'^

^'' = _ — — —

a cos^ + 2h cos sin $ + b sin^ a + 2h tan 6 + b tan^ 6

(4)..



AXES OF A CENTRAL CONIC SECTION. 839

If in (4) we substitute either value of tan 6 derived
from (3) we obtain the length of the corresponding
semi-axis.

The directions and magnitudes of the axes are therefore
both found.

Second Method. The directions of the axes of the
conic are, as in the first method, given by

tan 2^=.-^.
a —

When referred to the axes of the conic section as the
axes of coordinates, let the equation become

- + ■^-1 rm

Since the equation (1) has become equation (5) by a
change of axes without a change of origin, we have, by
Art. 135,

1 1

-2 + ^2 = ^ + ^ (6),

and -l^^ah-h' (7).

arp-'

These two equations easily determine the semi-axes a
and /5. [For if from the square of (6) we subtract four

times equation (7) we have (-^ — 02) , and hence —^— -^^'j

1 In

hence by (6) we get —^ and — .

The difficulty of this method lies in the fact that we
cannot always easily determine to which direction for an
axis the value a belongs and to which the value yS.

If the original axes be inchned at an angle w, the equa-
tions (6) and (7) are, by Art. 137,

1 1 a + h — 2h cos w

, \ ah — //

and ~¥7yi= - 2 '

22—2



340 COORDINATE GEOMETRY.

Cor. 1. The reciprocals of the squares of the semi-
axes are, by (3) and (4), the roots of the equation

2^ - {ci + h) Z + ah -h^ = 0.
Cor. 2. From equation (4) we have
Area of an ellipse = 7ra/5 =



\lah — h^
365. Ex. 1. Trace the curve

14a;2 - 4a;2/ + 112/2- 44a; -58i/ + 71 = (1).

Since ( - 2)^ - 14 . 11 is negative, the curve is an ellipse. [Art. 358.]
By Art. 352 the centre (S, y) of the curve is given by the equations

145-2^-22=0, and -25 + 11^-29 = 0.
Hence ic= 2, and ^=3.

The equation referred to parallel axes through the centre is
therefore l^x^ - ^xy + lly'^ + c' = 0,

where c' = - 225 - 29^ + 71 = - 60,

so that the equation is

Ux^-4:xy + lly'^ = m (2).

The directions of the axes are given by

,. „„ 2h -4

*"^2^=^36=i43ri= -

2tan^ .

so that



l-tan2^ ^'
and hence 2 tan^ ^ - 3 tan - 2 = 0.

Therefore tan ^i=2, and tan 6^= -\.
Referred to polar coordinates the equation (2) is

r2 (14 cos2 ^ - 4 cos ^ sin ^ + 11 sin^ d) = 60 (cos^ 9 + sin^ 0),

l + tan2^



r''^ = 60



14-4tan6' + lltan2 6'*



When tan ^1 = 2, »'i' = 60x ^^^^=6.
When tan 6^=-^ r^^=GO x j^j:j:|^_ =4.



TRACING A CENTRAL CONIC SECTION.



341



The lengths of the semi-axes are therefore fJ6 and 2.

Hence, to draw the curve,
take the point C, whose coordi-
nates are (2, 3).

Through it draw A'CA in-
clined at an angle tan~i 2 to the
axis of X and mark off

A'C = GA=>J6.
Draw BCB' at right angles
to AC A' and take B'G=CB=:2.

The required ellipse has A A'
and BB' as its axes.

It would be found, as a veri-
fication, that the curve does not meet the original axis of x, and
that it meets the axis of y at distances from the origin equal to
about 2 and 3 J respectively.




Ex. 2. Trace the curve

x'^~Bx2j + if + 10x-10y + 21 =

1 . 1 is positive, the curve is a hyperbola.


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