E. W. (Edward West) Nichols.

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are the co-ordinates of the centre and

2a

COR. 1. If ax 2 + ay* + ex + dy -f m = be the equation
of another circle, it must be concentric with the circle repre-
sented by (1) ; for the co-ordinates of the centre are the same.
Hence, when the equations of circles have the variables in

62 PLANE ANALYTIC GEOAIETliY.

their terms affected with equal coefficients, each to each, the
circles are concentric. Thus

2 * 2 + 2 if + 3 x + 4 y + 25 =

are the equations of concentric circles.

EXAMPLES.

What is the equation of the circle when the origin is
taken.

1. At D, Fig. 21 ? Ans. x 2 + y- 2 ay = 0.

2. At K, Fig. 21 ? ^4ns. x 2 + y 2 + 2 // = 0.

3. At H, Fig. 21 ? 4n*. x- + y 2 + 2 aar = 0.

What are the co-ordinates of the centres, and the values of
the radii of the following circles ?

4. 4a- a + 4y a 8a?-8y + 2 = 0.

Ans. (1, 1), a = Vj.

5. a- 2 + y 2 + 4 a: 6 y 3 = 0.

^4ws. (- 2, 3), a = 4.

6. 2 x 2 + 2 ?/ 2 8 x = 0.

. (2, 0), a = 2.

7. X 2 + 7

4w*. (3, 0), a = 3.

8. x- + // 2 - 4 x + 8 y 5 = 0.

^n. (2, 4) a = 5.

9. x 2 + ?/" nix + w y + c = 0.

10. z s -f 2/ 2 = m -

11. x* 4 or = y 2 wy.

12. or 2 + y 2 = c 2 + d\

13. x 2 + ex -f- ?/ 2 =/.

THE CIRCLE.

63

Write the equations of the circles whose radii and whose
centres are

14. a = 3, (0, 1). 18. a = m, (b, e).
Ans. x*+y*-2y = 8.

15. a = 2, (1, - 2). 19. a = b, (e, - d).
Ans. a: 2 + y 2 -2a;+4y + l=0.

16. a = 5, (- 2, - 2). 20. a = 5, (I, k).
Ans. x 2 + y 2 + 4 y -f- 4 a; = 17.

17. a = 4, (0, 0).

= 16.

21. a = A, (2, ft).

22. The radius of a circle is 5 ; what is its equation if it is
concentric with a* 2 + y 2 4x=2?

Ans. x* -\- y z 4 x = 21.

23. Write the equations of two concentric circles which
have for their common centre the point (2, 1).

24. Find the equation of a circle passing through three
given points.

39. To deduce the polar equation of the circle.
Y

The equation of the circle when referred to OY, OX is
( X _ *')* + (y _ y 'Y = a*.

64 PLANE ANALYTIC GEOMETRY.

To deduce the polar equation let P be any point of the
curve, then

(OA, AP) = (x, y}
(OB, BO') = (x', jO
(OP, POA) = (r, 6)
(00', O'OB) = (/, ff)

From the figure, OA = x = r cos 6, AP = y = r sin 0,
OB = x' = r' cos 6', BO' = y' = r' sin & ;
hence, substituting, we have,

(r cos r' cos 0') 2 + (r sin - r' sin fl') 2 = a 2 .
Squaring and collecting, we have,

r>(cos 2 + sin 2 6) + r' 2 (cos 2 0' + sin 2 6') - 2 rr'(cos cos V
+ sin sin 0') = a 2

i.e., r 2 + r' 2 2 r/ cos (0 - 6'} = a 2 . . . (1)

ts Ae polar equation of the circle.

This equation might have been obtained directly from the
triangle OO'P.

COR. 1. If ff = 0, the initial line OX passes through the cen-
tre and the equation becomes

2

r _|_ _ rr C os = a.

COB. 2. If & = 0, and / = a, the pole lies on the circum-
ference and the equation becomes

r = 2 a cos 6.

COB. 3. If & = 0, and r' = 0, the pole is at the centre and
the equation becomes

r = a.

40. To show that the supplemental chords of the circle are
perpendicular to each other.

The supplemental chords of a circle are those chords which
pass through the extremities of any diameter and intersect each
other on the circumference.

65

FIG. 23.

Let PB, PA be a pair of supplemental chords. We wish to
prove that they are at right angles to each other.
The equation of a line through B ( a, o) is

y = s (x + a}.
For a line through A (a, 0), we have

y = s' (x a).
Multiplying these, member by member, we have

y 2 = ss' (cc 2 - a 2 ) ... (a)

for an equation which expresses the relation between the
co-ordinates of the point of intersection of the lines.

Since the lines must not only intersect, but intersect on the
circle whose equation is

7/ 2 = a 2 x 2 ,

this equation must subsist at the same time with equation (a)
above ; hence, dividing, we have

1 = ss',
or, 1 + ss' = . . . (1)

Hence the supplemental chords of a circle are perpendicular
to each other.

Let the student discuss the proposition for a pair of chords
passing through the extremities of the vertical diameter.

66

PLANE ANALYTIC GEOMETRY.

41. To deduce the equation of the tangent to the circle.

C\|Y

FIG.

Let CS be any line cutting the circle in the points P' (x', y'),
P" (x", y"}. Its equation is

y - y' = 4^-4 (* - *') ( Art - 26 > ( 4 ) )

ic ~""~ *C

Since the points (x'y'), (x" } y") are on the circle, we have
the equations of condition

& + y" = a* . . . (1)

*"' 4. y"2 = a 2 ... (2)

These three equations must subsist at the same time ; hence,
subtracting (2) from (1) and factoring, we have,

(x' + x") (x' - x") + (y' + y") (y' - y"} = ;

y' - y" x' + x"

'' x' - x" ~ y' + y" '
Substituting in the equation of the secant line it becomes

/ *5 f CC / t\ /o\

y -y - = - y' + y" (*-*> < 3 )

If we now revolve the secant line upward about P" the
point P' will approach P" and will finally coincide with it
when the secant CS becomes tangent to the curve. But when

THE CIRCLE. 67

l v coincides with P", x' x" and // = y"; hence, substituting
in (3) we have,

y - y" = - fir (* - *")> ( 4 )

or, after reduction,

or, symmetrically,

^r + ^- = i ... (6)

a* a

for the equation of the tangent.

SCHOL. The SUB-TANGENT for a given point of a curve is the
distance from the foot of the ordiuate of the point of tangency
to the point in which the tangent intersects the X-axis ; thus,
in Fig. 24, AT is the sub-tangent for the point P". To find
its value make y = in the equation of the tangent (5) and
we have,

OT = x = .
x

But AT = OT - OA = ^- - x "

x

.-. sub-tangent = ~. x = ^ .

42. To deduce the equation of the normal to the circle.

The normal to a curve at a given point is a line perpen-
dicular to the tangent drawn at that point.

The equation of any line through the point P" (x", y"} Fig.
24, is y - y" = s (x - x"} ... (1)

In order that this line shall be perpendicular to the tangent
P"T, we must have

1 -f ss' = 0.

x" y"

But Art. 41, (4) s' = ; hence, we must have s = ^ .

V x"

IS PLANE ANALYTIC GEOMETRY.

Therefore, substituting in (1), we have,

y -*"-* <*-*"> < 2 > ;

or, after reduction,

yx" - xy" = ... (3)

for the equation of the normal.

We see from the form of this equation that the normal to
the circle passes through the centre.

SCHOL. The SUB-NORMAL for a given point on a curve is
the distance from the foot of the ordinate of the point to the
point in which the normal intersects the X-axis. In the circle,
we see from Fig. 24 that the

Sub-normal = x".

43. By methods precisely analogous to those developed in
the last two articles, we may prove the equation of the tangent
to

(x - x'Y + (y- y'Y = 2

to be

(x - x') (x" - x') + (y - y') (y" - y'} = a 2 . . . (1)
and that of the normal to be

(y -y"} (*" - *') - ( - *") (y" - sO - o , . . (2)

Let the student deduce these equations.

EXAMPLES.

1. What is the polar equation of the circle ax* + ay 2 -{- ex +
dy +/= 0, the origin being taken as the pole and the X-axis
as the initial line ?

Ans. r* + ( - cos + - sin 6 \r + ^ = 0.
\a a J a

2. What is the equation of the tangent to the circle
a; 8 -f y 2 = 25 at the point (3, 4) ? The value of the sub-
tangent ? Ans. 3 x -f 4 y = 25 ; ^.

3. What is the equation of the normal to the circle
x* 4- y 9 = 37 at the point (1, 6) ? What is the value of the
sub-normal? Ans. y = Gx\ 1.

THE CIRCLE. 69

4. What are the equations of the tangent and normal to the
circle x 2 -\- y 2 = 20 at the point whose abscissa is 2 and ordi-
nate negative ? Give also the values of the sub-tangent and
sub-normal for this point.

Ans. ,2x 4 i/ = 20; 2y + 4.r = <);
Sub-tangent = 8 ; sub-normal = 2.

Give the equations of the tangents and normals, and the
values of the sub-tangents and sub-normals, to the following
circles :

5. .r 2 + 7/ 2 == 12, at (2, + V8).

6. x 2 + if = 25, at (3, - 4).

7. 3.2 + y* = 20, at (2, ordinate +).

8. .r 2 + v/ 2 = 32, at (abscissa +, 4).

9. .r 2 + ?/ 2 = a 2 , at (b, c).

10. .7' 2 + ?/ 2 = w-? at (1, ordinate -f ).

11. x 2 + ?/ 2 = A;, at (2, ordinate ).

12. .r' 2 + //- = 18, at (m. ordinate +)

13. Given the circle a; 2 -|- y 2 = 45 and the line 2y -\- x = 2;
required the equations of the tangents to the circle which are
parallel to the line.

A ( 3 x -\- 6 ?/ = 45.
Ans. < * j r

(3x + 67/= 45.

14. What are the equations of the tangents to the circle
x ~ + y 2 = 45 which are perpendicular to the line 2 y -\- x = 2 ?

^y ~ Qx ==45 '

lGx-3 y = 45.

16. The point (3, 6) lies outside of the circle x- -\- y 2 = 9;
required the equations of the tangents to the circle which
pass through this point.

( x = 3.

Ans. < , o 1 ~

( 4 ?/ 3 x = lo.

70 PLANE ANALYTIC GEOMETRY.

17. What is the equation of the tangent to the circle
( x _ 2) a + (y 3) a = 5 at the point (4, 4) ?

Ans. 2x +y = 12.

18. The equation of one of two supplementary chords of
the circle a; 2 + y 1 = 9 is y = x + 2, what is the equation
of the other ?

Ans. 2 y + 3 x = 9.

19. Find the equations of the lines which touch the circle
(x a) 2 + (y by = r 2 and which are parallel to y = sx -{- c.

20. The equation of a circle is a; 2 -j-y 2 4x-|-4y = 9;
required the equation of the normal at the point whose
abscissa = 3, and whose ordinate is positive.

Ans. 4 x y = 10.

44. To find the length of that portion of the tangent lying
between any point on it and the point of tangency.

Let (xi, yi) be the point on the tangent. The distance of
this point from the centre of the circle whose equation is

(x x') 2 -\- (y y'Y = a 2 is evidently
V(i - x')* + (y, - y'y. See Art. 27, (1).

But this distance is the hypothenuse of a right angled tri-
angle whose sides are the radius a and the required distance
d along the tangent ; hence

rf 2 = (^ - a-') 2 + (jf! - y'Y - a 2 ... (1)
COR. 1. If x' = and //' = 0, then (1) becomes

</ 2 = V + yx 2 - 2 ... (2)
as it ought.

45. To deduce the equation of the radical axis of two given
circles.

The RADICAL AXIS OF TWO CIRCLES is the locus of a point
from which tangents drawn to the two circles are equal.

THE CIRCLE.

71

FIG. 25.

Let

(x - X 'Y + (y - y/) 2 = a 2 ,

(x a:") 2 + (y ?/") = b 2 be the given circles.

Let P (iCj, yj be cm?/ point on the radical axis ; then from
the preceding article, we have,

d 2 = (xj -x'} 2 + (T/J ?/) 2 a 2
<f 2 = (arj - x"Y + (yi - ?/") 2 - * 2

.-. by definition (x l x') 2 + (?/i y')' 2 a 2 = (ccj x") 2

+ (yi ~~ y"} 1 W", hence, reducing, we have,

2 (x" x'} x l + 2 (y" v/') ?/! = x" 2 x'- + y" 2 y' 2
-\- a~ b 2 .

Calling, for brevity, the second member m, we see that
(a*!, T/J) will satisfy the equation.

2 (x" - x >)x + 2 (y" -y}y = m . . . (1)

But (a?!, 7/i) is any point on the radical axis ; hence every
point on that axis will satisfy (1). It is, therefore, the re-
quired equation.

COK. 1. If c = and c' = be the equation of two circles,
then, c c' =

is the equation of their radical axis.

72 PLANE ANALYTIC GEOMETRY.

COR. 2. From the method of deducing (1) it is easily seen
that if the two circles intersect, the co-ordinates of their points
of intersection must satisfy (1) ; hence the radical axis of two
intersecting circles is the line joining their points of intersection,
PA, Fig. 25.

Let the student prove that the radical axis of any two
circles is perpendicular to the line joining their centres.

46. To show that the radical axes of three given circles in-
tersect in a common point.

Let c = 0, c = 0, and c" =

be the equations of the three circles.

Taking the circles two and two we have for the equations of

c - c' = . . . (1)

c - c" = . . . (2)

c' - c" = . . . (3)

It is evident that the values of x and y which simultaneously
satisfy (1) and (2) will also satisfy (3) ; hence the proposition.

The intersection of the radical axes of three given circles is
called THE RADICAL CENTRE of the circles.

EXAMPLES.

Find the lengths of the tangents drawn to the following
circles :

1. (a- _ 2) 2 + (// - 3)' J = 1() from (7, 2).

Ans. d = VlO.

2. a- 8 + (// + 2)-' = 10 from (3, 0).

Ans. d = V3.

3. (x a)- + y- = 12 from (b, c).

4. 3.8 _|_ ? / _ o x + 4 y = 2 from (3, 1).

5. a- 2 + IT = 25 from (6, 3).

Ans. d = V20.

THE CIRCLE. 73

6. x 2 + y 2 - 2 x = 10 from (5, 2).

Ans. d = 3.

7-. (x _ a ) + (,, _ by = c from (d, /).
8. ^ + if 4 y = 10 from (0, 0).

Give the equations of the radical axis of each of the follow-
ing pairs of circles :

Ans. 5x-\-oy-\-2=0.

10. ^ 2 + r-4y = 0.

11. f(a;+3;) + 3f-2y-8 = 0.

( x 2 + if 2 y = 0. Ans. x = .

12. f (* + a) 2 + y 2 - c 2 = 0.
| ^ _|_ ( ?/ _ 3)2 _ 16 = 0.

13. ( x 2 + y 2 = 16.

} (a - I) 2 + y 2 = a 2 .

14. \x z + (y - a) 2 = c 2 .
| (cc 2) 2 + y 2 = c? 2 .

Find the co-ordinates of the radical centres of each of the
following systems of circles :

15. ( (x - 3) 2 + y 2 = 16.

16. ( x- + y 2 - 4 x + 6 y 3 = 0.

17. (x 2 + y z = a.^

( x * 4. ,f 2 x + 4 y = 10.

\ I / /

18. ( a; 2 + y 2 ^ = c -

a; 2 -j- y 2 = m.

x 2 -\- y 2 ay = d.

74

PLANE ANALYTIC GEOMETRY.

47. To find the condition that a straight line ij = sx -\- b
must fulfil in order that it man touch the circle x? -\- y 2 = a 2 .

In order that the line may touch the circle the perpendicu-
lar let fall from the centre on the line must be equal to the

From Art. 21, Fig. 13, we have

p = b cos y = =. -= == ;
sec Y VI -f tan. 2 y

b

: VIT^ ;

hence, r 2 (1 + s 2 ) = 6 2 . . . (1)

is the required condition.

COR. 1. If we substitute the value of b drawn from (1) in
the equation y = sx + b, we have

y=sxr Vf^M 2 .-. (2)
for the equation of the tangent in terms of its slope.

48. Two tangents are drawn from a point without the circle ;
required the equation of the chord joining the points oftangency.

FIG. 26.

Let P' (x', y 1 ) be the given point, and let P'P", PT, be the
tangents through it to the circle.

THE CIRCLE. 75

It is required to deduce the equation of PP".
The equation of a tangent through P" (x", y"} is

Since P' (a;', 2/)'is on this line, its co-ordinates must satisfy
the equation ; hence

The point (x", y"), therefore, satisfies the equation

.-. it is a point on the locus represented by (1). A similar
source of reasoning will show that P is also a point of this
locus. But (1) is the equation of a straight line ; hence, since
it is satisfied for the co-ordinates of both P" and P, it is the
equation of the straight line joining them. It is, therefore,
the required equation.

49. A chord of a given circle is revolved about one of its
points ; required the equation of the locus generated by the
point of intersection of a pair of tangents drawn to the circle at
the points in which the chord cuts the circle.

Let P / (x', ?/), Fig. 27, be the point about which the chord
P'AB revolves. It is required to find the equation of the
locus generated by P! (x^ y^, the intersection of the tangents
AP t , BP 1} as the line P'AB revolves about P'.

From the preceding article the equation of the chord AB is

gys i y\y _ 1

a 2 a 2 "
Since P' (x f , y'} is on this line, we have

r ,
r

a" or

x ' x i y'y _ 1

hence ;r ~\ r J-

a a

76

PLANE ANALYTIC GEOMETRY.

is satisfied for the co-ordinates of P, (a;,, y,) ; hence P, lies on
the locus represented by (1). Hut P t is the intersection of
any pair of tangents drawn to the circle at the points in

FIG. 27.

which, the chord, in any position, cuts the circle ; hence (1)
will be satisfied for the co-ordinates of the points of intersec-
tion of every pair of tangents so drawn.

Equation (1) is, therefore, the equation of the required
locus. We observe that equation (1) is identical with (1) of
the preceding article ; hence the chord PP" is the locus whose
equation we sought.

The point P' (x f , y f ) is called THE POLE of the line PP"

** + */!/ = A and the HnePP" (*?+& = l\ is called
a 2 a 2 J \ a 2 a 2 )

THE POLAR of the point P' (x' t y>) with regard to the circle

THE CIRCLE. 77

As the principles here developed are perfectly general, the
pole may be without, on, or within the circle.

Let the student prove that the line joining the pole and the
centre is perpendicular to the polar.

no connection with the same terms used in treating of polar
co-ordinates, Chapter I.

50. If the polar of the point P' (x'y 1 ), Fig. 27, passes through
P\ ( #i> 2/i) > then the polar of P v (x lt y^ will pass through P'
(*', !/')

The equation of the polar to P' (x', y') is
x x . y y -.

~Q~ ' ~2~ =

a 2 Of

In order that P x (a^, y^) may be on this line, we must have,
x'x l _^_ y'y l _ i

But this is also the equation of condition that the point
P' (x r , y'} may lie on the line whose equation is

But this is the equation of the polar of P t (a^, y^) ; hence
the proposition.

51. To ascertain the relationship between the conjugate diam-
eters of the circle.

A pair of diameters are said to be conjugate when they are
so related that when the curve is referred to them as axes its
equation will contain only the second powers of the variables.

Let x- + y* = a* . . . (1)

be the equation of the circle, referred to its centre and axes.
To ascertain what this equation becomes when referred to
OY', OX', axes making any angle with each other, we must
substitute in the rectangular equation the values of the old

78

PLANE ANALYTIC GEOMETRY.

co-ordinates in terms of the new. From Art. 33, Cor. 1, we
have

x = x' cos + y' cos 9

y = x' sin -}- y' sin 9

for the equations of transformation. Substituting these
values in (1) and reducing, we have,

y' 2 + 2x'y cos (g> 6) + x" 2 = a 2 . . . (2)

Now, in order that OY', OX' may be conjugate diameters
they must be so related that the term containing x'y' in (2)
must disappear ; hence the equation of condition,

cos (<p 6) ;

.-. <p - e = 90, or g> - = 270.

The conjugate diameters of the circle are therefore perpen-
dicular to each other. As there are an infinite number of
pairs of lines in the circle which satisfy the condition of being
at right angles to each other, it follows that in the circle there
are an infinite number of conjugate diameters.

THE CIRCLE. 79

EXAMPLES.

1. Prove that the line y = V3 x + 10 touches the circle
x 2 -j- y 2 = 25, and find the co-ordinates of the point of tangency.

/ K K\

Ans. Point of tangency ( _. -^/3, )

V - 2 /

2. W T hat must be the value of b in order that the line
?/ = 2 x -f- b may touch the circle x 2 -+- 7/ 2 = 16 ?

^Ircs. ft = V80.

3. What must be the value of s in order that the line
y = sx 4 may touch the circle x 2 -\- y* = 2 ?

Ans. * = V 7.

4. The slope of a pair of parallel tangents to the circle
x 2 -+- y 2 = 16 is 2 ; required their equations.

= 2 x + V80.
= 2x- V80.

Two tangents are drawn from a point to a circle ; required
the equation of the chord joining the points of tangency in
each of the following cases :

5. From (4, 2) to x 2 + if = 9.

Ans. 4 x -{- 2 y = 9.

6. From (4, 3) to x* + if = 8.

Ans. 3 x + 4 y = 8.

7. From (5, 1) to x 2 + if = 16.

^4ws. a; -j- 5 y = 16.

8. From (a, Z>) to x 2 -\- y 2 = c*.

Ans. ax -\- by = <?.

W T hat are the equations of the polars of the following points :

9. Of (2, 5) with regard to the circle x z + if = 16 ?

A ^ I I 1

T6 + l6 =

10. Of (3, 4) with regard to the circle x 2 + y 2 = 9 ?

s. 3 x + 4 v = 9.

N> PLANE ANALYTIC GEOMETRY.

11. Of (</, f>) with regard to the circle x- + if = m ?

Ans. ax -}- by = m.

\Vhat are the poles of the following lines :

12. Of 2 x + 3 y = "> with regard to the circle x 2 + y* = 25 ?

Ans. (10, 15).

13. Of * + y = 4 with regard to the circle

f-0 + l7r 1? An,. (2,4).

14. Of y = sx -f b with regard to the circle

3-

-2 ..2

+ #- = 1 ?

2 I 9

a*

Ans. (
b > b

15. Find the equation of a straight line passing through
(0, 0) and touching the circle x- -\- y 2 3 x -f- 4 y = 0.

Ans. y = ' x.

GENERAL EXAMPLES.

1. Find the equation of that diameter of a circle which
bisects all chords drawn parallel to y = sx -}- b.

Ans. sy -\- x = 0.

2. Required the co-ordinates of the points in which the
line 2 y cc-)-l=0 intersects the circle

Q O

T + -'-

3. Find the co-ordinates of the points in which two lines
drawn through (3, 4) touch the circle

[The points are common to the chord of contact and the
circle.]

THE CIRCLE. 81

4. The centre of a circle which touches the Y-axis is at
(4, 0) ; required its equation.

Ans. (x 4) 2 -f- y- = 16.

5. Find the equation of the circle whose centre is at the
origin and to which the line y = x -\- 3 is tangent.

Ans. 2 y? + - y~ = \$

6. Given x 2 -\- y z = 16 and (x 5)'- -f- y" = 4 ; required the
equation of the circle which has their common chord for a
diameter.

7. Required the equation' of the circle which has the dis-
tance of the point (3, 4) from the origin as its diameter.

Ans. a; 2 + ?/ 2 3 x 4 y = 0.

8. Find the equation of the circle which louches the lines
represented by x = 3, y = 0, and y = x.

9. Find the equation of the circle which passes through the
points (1, 2), (- 2, 3), (- 1, - 1).

10. Required the equation of the circle which circumscribes
the triangles whose sides are represented by y = 0, 3 y = 4 x,
and 3 ?/ = 4 # + G.

Ans. x 1 + if - f x f y = 0.

11. Required the equation of the circle whose intercepts
are a and b, and which passes through the origin.

Ans. x~ -\- y- ax by = 0.

12. The points (1, 5) and (4, 6) lie on a circle whose centre
is in the line y = x 4 ; required its equation.

Ans. 2 a- 2 + 2 if 17 x y = 30.

13. The point (3. 2) is the middie point of a chord of the
circle a- 2 + y- = 16 ; required the equation of the chord.

14. Given x"- -\- y- = 16 and the chord y 4 x = 8. Show
that a perpendicular from the centre of the circle bisects the
chord.

15. Find the locus of the centres of all the circles which
pass through (2, 4), (3, 2).

82 PLANE ANALYTIC GEOMETRY.

16. Show that if the polars of two points meet in a third
point, then that point is the pole of the line joining the first
two points.

17. Required the equation of the circle whose sub-tangent
= 8. and whose sub normal = 2.

Ans. X* + if = 20.

18. Required the equation of ihe circle whose sub-normal
= 2, the distance of the point in which the tangent intersects
the X-axis from the origin beinr = 8.

Ans. x~ -\- y* = 16.

19. Required the conditions in order that the circles
* 2 + y 2 + ex + dy -f e = and ax- + aif + kx + ly + m =
may be concentric.

Ans. c = k, d = I.

20. Required the polar co-ordinates of the centre and the

r 2 2 r (cos 6 + V 3 sin *) = 5 -

Ans. (2, 60) ; r = 3.

21. A line of fixed length so moves that its extremities
remain in the co-ordinate axes ; required the equation of the
circle generated by its middle point.

22. Find the locus of the vertex of a triangle having given
the base = 2 a and the sum of the squares of its sides = 2 b*.

Ans. x~ -f- y 2 = b* a a .

23. Find the locus of the vertex of a triangle having given
the base = 2 a and the ratio of its sides

= . Ans. A circle.

n

24. Find the locus of the middle points of chords drawn
from the extremity of any diameter of the circle

** .'/ 2 _ 1

? T ^ ~ I-

THE PARABOLA.

83

CHAPTER VI.
THE PARABOLA.

52. THE parabola is the locus generated by a point moving
in the same plane so as to remain always equidistant from a
fixed point and a fixed line.

The fixed point is called the Focus ; the fixed line is called
the DIRECTRIX ; the line drawn through the focus perpendic-
ular to the directrix is called the Axis ; the point on the axis
midway between the focus and directrix is called the VERTEX
of the parabola.

53. To find the equation of the parabola, given the focus and
directrix.

FUi. 29.

Let RC be the directrix and let F be the focus. Let OX,
the axis of the curve, and the tangent OY drawn at the vertex

84 PLANE ANALYTIC GEOMETRY.

O, be the co-ordinate axes. Take any point P on the curve
and draw PA || to OY, PB || to OX and join P and F. Then
(OA, AP) = (x, i/) are the co-ordinates of P.
From the right angled triangle FAP, we have

? / 2 = AP 2 = FP 2 - FA 2 ; ... (1)
Hut from the mode of generating the curve, we have

FP 2 = BP 2 = (AO + OD) 2 = (x + OD) 2 ,
and from the figure, we have

FA 2 = (AO - OF)- = (x - OF) 8 .
Substituting these values in (1), we have

y 2 = (x + OD) 2 - (x - OF) 2 . . . (2)

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