Since e > 1, the distance from F to any point in the curve is greater than the distance from the same point to the line DD'; therefore there are points in the curve on the opposite side of DD', which are found in the same way as those to the right of DD', thus: with the half of any of the parallels on the left of DD', as MN, for a radius, and F for a centre, describe an arc cutting MN at P'; this is a point of the curve. For, joining P' and F, and drawing P'L perpendicular to DD', we have

MN (= FP') : MO (= P'L) :: FE : FO; that is, by construction we have FF : P'L :: e : 1.

In the same way, any required number of points may he found. Connecting these points by a line, we have the required hyperbola.

The fixed line DD' is called the Directrix; the fixed point F is called the Focus; OG and OG' are called the

Focal Tangents.

100. To find the equation of the hyperbola.

v

Let F be the focus, and YY' the directrix; through F draw OX perpendicular to YY'; take OX and OY for the coordinate axes. Let (*, y) be any point P on the curve; join FP; draw PM and PD respectively perpendicular to OX and OY.

Represent OF by p, and the ratio of FP to PD by e. Then we have, from definition,

FP = e • PD, or FP* = e? • PD¹;

or    FM⁹ + MP* = e* • PD*;

that is,    {x — p)® + y* = eh?,

which is the required equation.

Coe.—When y = 0, we have x—p = ±ex; therefore,

_P_

1T«’

which shows that the curve cuts the axis of x at the two points A and A, giving

OA =    , and A'O = - z = —

1+ e    e — 1

Since e > 1, 1 — e is a negative quantity; therefore OA' must be measured to the left of the origin.

A and A' are called the Vertices of the hyperbola, and C, the point midway between them, is called the Centre of the hyperbola.

The distance AA' = A'O -f OA

- _Z_ , P _ 2V .

~ e-l^ 1 + e~ e® — 1 ’

it is customary to denote this distance by 2a. Therefore we have

101. Transform equation (1) of Art 100 to A (Fig. 81).

The formulae for this transformation become (Art 100, Cor.),

* = r+~, + *'•    » = »’'

which in (1) gives

(r+i+*' -?)*+»■■ = *    +‘If,

or    +    = Tfl*+*'*);

therefore, rf² = 2epx' — (1 — 6s) x’²

_ /i    , ,A

therefore (Art. 100, Cor.),

y'* = (e* - 1) (2a*' + *'*),

or, suppressing the accents, since the variables are general, we have

>    y⁸ = (e⁸ - 1) (2a* + z⁸).    (1)

Cor.—When x = — a, (1) becomes y⁸ = — (s⁸ — 1) a⁸ = — 4s [by patting (e* — 1) a⁸ = 4*].

Therefore

which in (1) becomes

(2)

(1)

(2)

(3)

y* = (2a* + *»).

102. Transform (2) of Art. 101, Cor., to C (Fig. 81). The formulae for this transformation become * = *' — a, y = y', which in (2) gives, after suppressing accents,

y⁸ = ^(*⁸-a⁸),

or    afy³ —    = — a²#³,

which may be more symmetrically written,

a⁸ 4* ~~ ’

a form analogous to that of the equation of a right line (Art. 22, H).

Coe. 1.—By definition, AF = e-AO, and OA = —

1 + e

(Art. 100, Cor.); therefore,

AF =    = ^\ep = a(e-l) [Art. 100, Cor.];

that is,    AF = a (e — 1).    (4)

that is,

CO = -

(?)

OA =

OA = p -

1 + «

e² — 1 e a(e-l)

a (e — 1)

that is,

CF = CA + AF that is,    CF

CO = CA - AO

[Art. 100, Cor.] ;

(5)

a + a{e — 1) = ae; ae.    (6)

a(e — 1) _ a

OF = p = CF — CO = ae — - ;

/ 6

that is.

/vo    a — 1)

OF = p = -i--

(8)

Cor. 2.—When y = 0, z = curve cuts the axis of $ at two points equally distant from the origin, and on opposite sides of it. When x = 0, y = ± Jv^-^1; hence the curve cuts the axis of y in two imaginary points on opposite sides of ihe origin. We may, however, take two points B and B', on different sides of C, making CB = CB' = 5, as we shall have occasion to u

± a, which shows that the

y

them hereafter.

Cor. 3.—Solving (1) for y, we get

y = ± | Vs? — a?,

which shows that, for every value of x > + a or < — a, there are two real values of y, numerically equal, with contrary signs; hence, for every point P on one side of the axis of x, there is a point P' on the other side of the axis, at the same distance from it; and therefore the curve is symmetrical with respect to the axis of x. When a; is + a or

—    a, y = ± 0; and for every value of x between + a and

—    a, the two values of y are imaginary; therefore the curve is limited towards the centre by two tangents at A and A'.

Similarly, solving (1) for x, we get

which shows that for every value of y from — oo to -f- oo there are two real values of x, numerically equal with contrary signs; hence the curve is symmetrical with respect to the axis of y, and is unlimited in the direction of this axis.

Sen.—Because the curve is symmetrical with respect to the line BB', it follows that if we take CF' = CF (IMg. 82), and CO' = CO, and draw KK' perpendicular to 00', the point F' and the line KK' will form respectively a second focus and directrix.

AA' is called the Transverse axis of the hyperbola; BB' is called the Conjugate axis of the hyperbola. In the ellipse, the conjugate axis is always less than the transverse axis (see Art. 70, Cor.), and therefore the former was called the minor and the latter the major axis. In the hyperbola, the conjugate axis may be greater than the transverse, since i² = a² (e² — 1) (Art. 101, Cor.), and e is > 1; therefore we do not call the axes in the hyperbola the major and minor axes.

The ratio e (Art. 99) is called the Eccentricity of the hyperbola.

The point C is called the Centre of the hyperbola, because it bisects every chord of the hyperbola which passes through it. This may be shown in the same way as in the case of the ellipse (Art. 71, Sch.).

Cob. 4.—To find the latus rectum (Art 53, Cor. 3).

Make x = CF = ae (Cor. 1); denote the corresponding value of y by p\ we have from Eq. (1) (Art. 102),

= j? (a²<? - a*) = i² (e² - 1) = ? (Art, 101, Cor.).

2ft² 4ft²

Therefore,    2p = — = — = latus rectum.

r a 2a

Forming a proportion from this equation, we have

2a : 2ft :: 2ft : 2p.

That is, the latus rectum is a third proportional to the tranverse axis and the conjugate.

Since ft² =    — 1) a² (Art. 101, Cor.), we have

a² -f ft² = d;

that is,    a² + ft² = CF² (Art. 102, Cor. 1).

But    a² + ft² = AB² (see Fig. 82).

Therefore,    AB = CF.

•Hence, the conjugate axis of the hyperbola is a perpendicular to the transverse axis at its centre, and is limited by an arc described with the vertex of the transverse axis as a centre, and with a radius equal to the distance from the focus to the centre.

Cor. 5.—Comparing equation (2), Art. 102, with (2) of Art. 71, we see that the equation of the hyperbola may bo derived from that of the ellipse, by changing + ft² into —ft². Hence, we infer that any function of ft, expressing a property of the ellipse, will be converted into one expressing a corresponding property of the hyperbola, by changing ft into ft V — 1; therefore, in obtaining the properties of the hyperbola that are similar to those which have been proved for the ellipse, we shall, in most cases, either change the sign of ft², or else refer the student to the corresponding demonstration in the ellipse.

By a process similar to that of Art. 71, Cor. 5, the details of which the student must supply, we obtain

y'² : y"² : s (s' + a) (zr — a) : (z'f + a) {x" — a);

that is, the squares of any two ordinates to the transverse axis of an hyperbola are to each other as the rectangles of the segments into which they divide the transverse axis.

Cob. 6.—A point is outside, on, or inside the hyperbola, according as a²y² — tfx? -f a²b² >, =, or < 0. The proof is similar to that given in Art. 71, Cor. 6, for the ellipse.

A point is said to be outside the hyperbola if it lies in the space between the branches, so that no right line can be drawn through it to a focus without cutting the curve.

103. To find the distance of any point in the hyperbola from the focus, in terms of the abscissa of the point.

From the figure we have FP³ = [x — ae)² + y²

= (x - aef +    - a*.

a⁴

(Art. 102.)

= a² — 2aex +    (since a²e² — l? = a²);

therefore,    FP = ex — a.

[We take only the positive value of the root, for the reason given in Art. 72.]

In like manner we find, by writing — ae for + ae,

F'P² — (x + ae)² + if = a² + 2aex + eh?; therefore,    FT = ex + a.

Hence,    FT — FP = 2a;

or, the difference of the distances of any point in an hyperbola from the foci is equal to the transverse axis.

Cor.—This result furnishes two other methods of constructing an hyperbola, having given the axes.

I.    With C as a centre and BA as a radius, describe an arc cutting AA' produced at F and F'; these points are the foci (Art. 102, Cor. 4). Now, with F' as a centre and a radius greater than F'A, describe an arc; then with F as a centre, and a radius equal to that used before, diminished by the transverse axis AA', describe another arc cutting the first at the point P; this will be a point of the curve, since

FP = FP - 2a, or    FT — FP = 2 a.

In the same way, any number of points may be found; joining these points, it will be a branch of the required hyperbola. By using F for the first centre and F' for the second, with the same distances as before, any number of points of the other branch may be found.

II.    Take a ruler, and fasten one end of it at F', so it can revolve about F' as a centre.

Take a string whose length is less than that of the ruler by AA', and fasten one end of it at F and the other end at B, the end of the ruler; then press the string against the edge of the ruler with the point of a pencil P, and revolve the ruler about F', keeping the string tight; the pencil will describe one branch of an hyperbola, since, in every position of it, we shall have

FT — FP = AA'.

104.    A Conjugate Hyperbola ie one having the conjugate avia of a given hyperbola for its transverse axis, and the transverse axis of the given hyperbola for its conjugate axis.

Y

Either of two hyperbolas thus related is conjugate to the other.

Thus, the hyperbola whose transverse axis is BB' (Fig. 86) is the conjugate of the hyperbola whose transverse axis is AA', and conversely, the latter is the conjugate of the former. They are often distinguished as the x Hyperbola and the y Hyperbola, each taking the name of the coordinate axis upon which its transverse axis lies; and when spoken of together are called Conjugate Hyperbolas.

105.    To find the equation of an hyperbola conjlegate to a given hyperbola.

By Art. 102, the equation of the given hyperbola is

!f = | (*-«*),

7vnT2

or, Fig. 86, PM* = ~ (CM³ - CA*)l

Hence, since P' is a point on the conjugate hyperbola, having BB' for its transverse axis and A A' for its conjugate axis, we have,

NF* = £4*(ON* - CB*X CB

or    =    (1)

which is the equation of the conjugate hyperbola, and is the same expression we would obtain from the equation of the given hyperbola by putting —fi² for + J², and —a⁸ for +a\

Sch. 1.—Since the (second hyperbola holds the same relation to the axis of y that the first does to the axis of x, we might have deduced the equation of the y hyperbola at once by changing a to b and b to a, x to y and y to x in the equation of the x hyperbola.

The Bides of the rectangle described on the axes are the tangents to the four branches at the vertices.

Comparing the equation of the conjugate hyperbola with that of the given hyperbola, we have (Art. 101, Cor.),

(e² — 1) J³ = a\

or    a² + J² = Pe² = CF².

(See Art. 102, Cor. 1.) Therefore the foci of the y hyperbola are at the same distance from the centre as the foci of the x hyperbola, but the eccentricity of the former has a different value from that of the latter.

Sch. 2.—The equations of the diagonals CE and CG are respectively

These hyperbolas are called Equilateral hyperbolas, from the equality of the axes. The equilateral hyperbola corresponds to the case in which the ellipse becomes a circle. (See Art. 71, Cor. 7.) The peculiarity in the figure of the equilateral hyperbola is that the curve is identical in form with its conjugate. From Art. 101, Cor., we have

d²

therefore, in the equilateral hyperbola we have e = 162 CONSTRUCTION OF CONJUGATE HYPERBOLAS.

106. To construct a pair of conjugate hyperbolas whose axes are given.

Draw the axes AA' and BB' at right angles to each other; construct the x hyperbola as in Art. 99. Now take CF' = CF, which equals AB (Art. 105,

Sch.), and F' is the focus of the y hyperbola. Take BE = BF', and B'H = B'F'; draw through E and H a right line; it is one of the focal tangents. Through O' draw a line perpendicular to BB'; this is the directrix corresponding to the focus F'of the y hyperbola. The construction is now the same as in Art. 99.

107. To find the equation of the tangent at any point of a,n hyperbola.

To obtain this equation for the hyperbola, we change J2 into — ^ in equations (6), (7), and (11) of Art. 74, and get

apyy' — bhcx' = — a²#²,    (1)

(2)

(3)

W &

y = wx-y'>

y = mx ± Va²m² — S®.

Cor.—To find the point in which the tangent cuts the axis of x, make y = 0 in (1), and get

a²

x = -, = CT, x

which is the same value we found for the abscissa of the point at which the tangent cuts the axis of x in the ellipse.

(Art. 74, Cor. 2). This value of * has the same sign as x'; hence, for the right-hand branch, it is always positive; that is, the tangent to the right-hand branch cuts the axis of a; to the right of the centre.

By Art. 102, Cor. 2, we have F'C = FC = ae ; therefore we have

F'T = os + % = % (ex' + a),

jC    X

and    FT = ae — —, = (ex' — a).

X    X

Hence, F'T : FT :: ex' -f a : ex' — a :: F'P : FP

(by Art. 103). That is, the tangent of an- hyperbola cuts the distance between the foci in segments proportional to the adjacent focal radii of contact; and therefore it bisects the internal angle between these focal radii.

This principle affords a method of drawing a tangent to an hyperbola at a given point.

Ufet P be the given point (see Fig. 88). Draw the focal radii F'P and FP to the given point P. On the longer, F'P, lay off PD = PF, and join DF. Through P draw PT perpendicular to DF; PT will be the trangent required, for it bisects the angle FPF'.

The subtangent MT = CM - CT = x' -That is,

x'²_efi

the subtangent = —^— •

108. To find the equation of the normal at any point of an hyperbola.

We change J² into — i² in (2) of Art 75, and get

y-y' = —!§?-(* “*■)»    (J)

which is the required equation of the normal at (x', tf).

Cob. 1.—To find the point in which the normal cats the axis of x, we make y = 0 in (1), and get, after redaction

x = a-- x' = CN (Fig. 88) = eV (Art 101, Cor.). The subnormal MN = CN — CM

a*+

a?

Sch.—The expression CN = «V enables ns, as in the case of the ellipse (Art. 75, Sch.), to draw a normal at any point P of the hyperbola, or one from any point N of the transverse axis.

Cob. 2.—By Art. 102, Cor. 1,

F'C = FC = ae;

therefore we have

F'N = e (ex' -f a),

and    FN = e (ex' — a).    r

Hence, F'N : FN :: ex' + a : ex' — a :: F'P : FP

(Art. 103). That is, the normal of an hyperbola cuts the distance between the foci in segments proportional to the adjacent focal radii of contact; and hence it bisects the external angle between the focal radii of contact.

109. To find the locus of the intersection of the tangent at any point with the perpendicular on it from either focus.

Changing the sign of V¹ in (3) and (4) of Art. 76, and adding the squares of the resulting equations together, we get

2s + y² = a\

for the required locus, which is therefore a circle described on the transverse axis.

110. To find the co-ordinates of the point of contact of a tangent to an hyperbola from a fixed point.

Let (*', y ) be the required point of contact, and (x", y") the fixed point through which the tangent passes.

Changing + b% to — J² in the results of Art. 77, we get

, _    T a*y"Vay'* —+ aW

x ~    '&x"*-d>y"»

, aWy" =F JVVay,J-iV', + a*i*

V~ T 6V⁷²-ay"³

These values indicate that from any fixed point two tangents can be drawn to an hyperbola, real, coincident; or imaginary, according as

ay"² - 5V'² +    >, =, or < 0 ;

that is, according as the point (x", y") is outside, on, or inside the curve (Art. 102, Cor. 6).

Cor.—It is clear that if any two real tangents be drawn from a given point to touch the same branch, their abscissas of contact will have like signs; and unlike, if they touch different branches. Hence, since the values of a: in the former case must have the same signs, we have, regarding only their numerical values,

a*5V' > a²y'Wafy"% — 5V'² + aW; or squaring, transposing, and reducing, we have

y"    (1)

But (Art. 105, Sch. 2) y = -x is the equation of the diag-

onal of the rectangle formed upon the axes of the hyperbola; therefore, the ordinate of the point from which two tangents can be drawn 'to the same branch of an hyperbola must be less than the corresponding ordinate of the diagonal; that is, the point itself must be somewhere between the diagonals (CE, CG) or (CH, CK) produced, and the adjacent branch of the curve (Fig. 86). These diagonals produced are called Asymptotes of the hyperbola, which we shall consider in Art. 113. Hence, generally, the two tangents which can be drawn to an hyperbola from any external point, will both touch the same branchy if the external point be between that branch and the adjacent portions of the asymptotes; but if the external point be so placed that we cannot pass from it to the curve without crossing an asymptote, the two tangents touch different branches of the curve.

111. Tangents are drawn to an hyperbola from a given external point; to find the equation of the chord of contact (Art. 77).

Change J³ into — J³ in (5) of Art. 78, and get

cPyy' — fixx' = — a²Wy    (1)

which is the equation of the chord of contact.

112- Through any fixed point a chord is drawn to an hyperbola, and tangents to the hyperbola are drawn at the extremities of the chord; to find the equation of the locus of the intersection of the tan-gents, when the chord is turned about the fixed point.

Change J² into — P in (3) of Art. 79, and get

a²yy' — Vhcx' = — aW,    (1)

which is the equation required, and the locus is a right line.

Sch.—The line (1) is called the Polar of the point (V, y') with regard to the hyperbola a²y² — Vh? = — aW> and the point (x'y y') is called the Pole of the line.

The statements in Art. 49 with respect to the circle may all be applied to the hyperbola as they were to the parabola (Art. 61), and the same conclusions arrived at that were reached in Arts. 49 and 61, and referred to in the ellipse (Art. 79, Sch.).

113.    An Asymptote of a curve is a line which continually approaches the curve, and becomes tangent to it only at an infinite distance, while it passes within a finite distance of the origin. We have called the diagonals produced of the rectangle on the axes (Art. 110, Cor.), the asymptotes of the hyperbola; we now proceed to show that they are such, that is, that they meet the curve only at infinity.

114.    To prove that the diagonals of the rectangle on the axes are asymptotes to both the given and conjugate hyperbolas.

Fig. 89.

Produce the ordinate MP of any point P in the given hyperbola, to meet the diagonal CR and the conjugate hyperbola, in the points P' and P" respectively. The distance of the point P from CR = PP' sin PPG, and therefore it

varies as PP'. Now, if CM, the

common abscissa = x, PM = y, P'M = y\ and P"M = y", we have, from the equations of the given hyperbola, the diagonal, and the conjugate hyperbola,

a)

a)

a)

y"’= ?(**+°')-

(2)

(3)

Subtracting (1) from (2), we have

y'* — y² = P, or Subtracting (2) from (3), we have

If now we suppose the abscissa CM to increase continually, and the line MP to move parallel to itself, the ordinates y, y', and y" will increase continually, and therefore, from (4) and (5), y' — y and y"— y' will diminish continually; and when x (CM), and therefore y, y⁷, and y" become infinitely great, y' — y and y" — y' will become infinitely small; that is, as x increases indefinitely, the two curves continually approach the diagonal CR, and become tangent to it and to each other only at infinity. Hence the diagonals are asymptotes to both curves.

Cor. 1.—The equations of CR and CR' are (Art. 105, Sch. 2),

and

y = -x or J a

y =--x or

⁹ a

H—

therefore the equation -a — ^ = 0 includes both asymptotes.

Cor. 2.—Let ACR = 0, ACE' = 6'; then

tan

#=».

tan 0'

__

b

•

a

a

sin

0 —

b

cos 0

a

1

MM    _ •

Va³ + a*'

Va³ + V³

“ e’

sin

0' = —

b

cos 0'

a

1

Va³ + V³’

Vd³ + J*

e

115. To find the equation of any diameter. (Def. of Art. 62.)

Change l³ into — 5s in (2) of Art. 80, and get

i³

y = -₂ cot 0 • x for the required equation.

Since d* and J² are constant for any given hyperbola, and 0 is constant for any given system of parallel chords, (1) is the equation of a right line passing through the origin, that is, through the centre of the hyperbola. Hence, every diameter of the hyperbola passes through the centre. By giving to 0 a suitable value, (1) may be made to represent any right line passing through the centre. Hence, every right line that passes through the centre of an hyperbola is a diameter; that is, it bisects some system of parallel chords.

Sch.—To draw a diameter of an hyperbola, draw any two parallel chords, and bisect them ; the line passing through the points of bisection is a diameter. The intersection of two diameters will be the centre of the hyperbola.

Cob. 1.—Let & = the inclination of the diameter itself to the transverse axis ; then we have

tan ff = -;

x⁷

which in (1) gives

tan 0 tan O' = —, a²

as the relation between 0 and O' when they are the angles which a system of parallel chords and their diameter respectively make with the axis of x.

Cob. 2.—Writing the equation of the diameter in the form

y = tan 0 . x,    (1)

and eliminating y between this equation and that of the given hyperbola, to find the abscissas of the points of intersection of (1) and the curve, we obtain

V& — a³ tan² 0

(2)

Now, eliminating y between (1) and the equation of the conjugate hyperbola (Art. 105), to find the abscissas of the points of intersection of (1) and the conjugate curve, we obtain

ab

— — - - - — — •

Va² tau² 0 — If¹

If a² tan³ 0 <    that is, if tan 0 < ± -, the values of x

a

in (2) are real, showing that (1) intersects the given hyperbola at finite distances from the centre ; while the values of x in (3) are imaginary, showing that (1) does not cut the y hyperbola.

If a³ tan² 0 > 5*, that is, if tan 0 > ± ^, the values of x

in (2) are imaginary, showing that (1) does not cut the given hyperbola; while the values of x in (3) are real, showing that (1) cuts the y hyperbola at finite distances from the centre.

If a² tan² 0 = W, that is, if tan 0 = ± -, the values of x

in (2) and (3) are infinite, showing that (1) does not cut either the x or the y hyperbola. In this case, (1) coincides with the diagonals of the rectangle described on the axes of the two conjugate hyperbolas (Art. 105, Sch. 2), that is, with the asymptotes (Art. 113).

We learn, then, that diameters which cut the given hyperbola in real points, must either make with the transverse axis an angle less than is made by the first of these diagonals, or greater than is made by the second, as DD' and HH'. If they cut the conjugate hyperbola in real points, they must either make with the transverse axis an angle greater than is made by the first of these diagonals, or less than is made by the second, as EE' and KK'. If they

CONJUGATE DIAMETERS IN THE HYPERBOLA. 171

coincide with these diagonals, as LL' and RR⁷, they will intersect the hyperbolas at an infinite distance. Hence, every right line drawn through the centre of an hyperbola must meet the hyperbola or its conjugate, unless it coincides with one of the asymptotes.

116. If one diameter of an hyperbola bisects all chords parallel to a second diameter, the second will bisect all chords parallel to the first.

Let 0 and O' be the respective inclinations of any two diameters to the transverse axis. Then the condition that the first diameter shall bisect all chords parallel to the second diameter (Art. 115, Cor. 1) is

tan 0 tan 0' = ~ •    (1)

a²

But this is also the condition that the second diameter bisects all chords parallel to the first

Sch.—Two diameters are Conjugate when each bisects all chords parallel to the other.

Because the conjugate of any diameter is parallel to the chords which that diameter bisects, therefore the inclinations of two conjugates must be connected in the same way as those of a diameter and its bisected chords. Hence, if 0 and O’ are the inclinations, the equation of condition for conjugate diameters in the hyperbola (Art. 115, Cor. 1) is

tan 0 tan O’ =    (2)

This condition shows that the tangents of inclination of any two conjugate diameters have like signs; therefore it indicates that the angles made with the transverse axis by the two conjugates are either both acute or both obtuse.. Therefore, conjugate diameters of an hyperbola lie on the same side of the conjugate axis, as CD and CE, or CK and CH' (see Fig. 90).

Cob.—From (2), if

tan 0 <    tan 0' > -;

a

and if    tan d > — -    tan 0' < —

a    a

Therefore (Art. 115, Cor. 2), if one of two conjugates, DD', meets an hyperbola, the other, EE', meets the conjugate hyperbola.

117. The tangent at either extremity of any diameter is parallel to its conjugate diameter.

[For demonstration, see Art. 82.]

118. Given the co-ordinates xy' of one extremity of a diameter, to find the co-ordinates x’\ y" of either extremity of the conjugate diameter.

By the Extremities of the conjugate diameter, we mean the points in which the conjugate cuts the conjugate hyperbola.

Let (x', yj be the point D (Fig. 90), and (x", y") the point E or E'. Since the conjugate diameter EE' is parallel to the tangent at (x', y') (Art. 117), and passes through the origin, therefore its equation (Art. 107) is

which, combined with the equation of the conjugate hyperbola (Art 105) gives

x"= ± j>y’

±-x'.

a

We see that the upper signs of the co-ordinates are both positive and the lower signs both negative, while in the ellipse (Art. 83), the upper signs are unlike and the lower also. This agrees with the properties of the two curves developed in Arts. 81 and 116, Sch.

1X9. To express the length of a semi-diameter (a'), and its conjugate (6'), in terms of the abscissa of the extremity of the diameter.

Let (xy') and (x", y") be the extremities D and E of the diameters DD' and EE'; then we have

a'³ = a/³+y² = a'³+^(z'³-a³) (Art. 102)

therefore    a'² = 0V² — i² (Art 101, Cor.). (1)

Also, b'² = at'² + y"² = ^,if²+-iat²( Art. 118)

tr    Q,

= a/*_fl2 ₊(Art. ₁₀2);

CL*

therefore,    b'² = ehd² — a² (Art. 101, Cor.). (2)

Coe. (1) — (2) gives

o'* —y³ = a²-V>;

that is, the difference of the squares of any two conjugate diameters of an hyperbola iB equal to the difference of the squares of the axes.

120. To find the length of the perpendicular from the centre to the tangent at any point.

Let (at, if) be the point, and p the perpendicular. The equation of the tangent at (at, if) is (Art 107),

a²yif — bb^af = — aW.    (1)

Therefore (Art. 24),

a³^

ah

Va*y'² + W²

= (Art 119.)

121. To find the angle between any pair of conjugate diameters.

By the

Let <(> be tbe required angle ECD in Fig. 90. same process as in the ellipse (Art. 86), we find

.    , ab

sin 0 = -rn ab

Cob.—Clearing (1) of fractions, we have

a’V sin 0 = ab,    (2)

which shows that the area of the parallelogram whose sides touch the hyperbola at the ends of any pair of conjugate diameters is constant, and equal to the rectangle of the axes.

Sch.—By Art. 119, Cor., a'¹ — V² = a constant; therefore, a' and V increase or decrease together; hence, by causing D to move along the hyperbola from A, E also will move along from B (Fig. 90). But any diameter CD tends towards an infinite lengthy as its inclination tends towards

the limit 0 = tan⁻¹ ~ (Art. 115, Cor. 2); therefore its semiconjugate CE tends towards infinity; and, as a'b' sin 0 is constant, and a' and V tend towards infinity, sin 0 tends towards 0; or, the angle between two conjugates of an hyperbola diminishes without limit. When the two conjugates approach, infinity in length, they tend to coincide with the diagonals of the rectangle constructed on the axes; but they are never equal, since a'² — V² is always equal to a² — i² (Art. 119, Cor.), unless the curve is equilateral. Therefore, the infinite diameters which form the limit of the conjugates are not equal infinites, and hence we do not, as in the ellipse, have equi-conjugates. We may, however, call these conjugates in their limit, when they coincide with each other and with either of the asymptotes, Self Conjugates,¹ since each is a diameter conjugate, to itself.

The inclinations of the self-conjugate diameters to the transverse axis are determined by the equation

tan 0 = ± -• (Art. 115, Cor. 2.)

The first value corresponds to the angle ACE, and the second value to the angle ACK (Fig. 91).

The inclination of these self-conjugates to each other, as ECK or ECK', is determined by

sin <(> = 2 sin BCE cos BCE a    b

that is, where

. ,    2 ab

mu * =

<p = ECK or ECK'.

122. If a chord and diameter of an hyperbola are parallel, the supplemental chord and the conjugate diameter are parallel. (See Def., Art. 88.)

]A^C

—

K7-* j

Fig. 92.

Let DD' be a diameter of the hyperbola; PD and PD' two supplemental chords, the first parallel to the diameter EE'; then will the supplemental chord PD'

' be parallel to the con j ugate diameter KK'.

Let (x\ y') be the point D, and therefore (— x\ — y') will

be the point D'. Let <p and </>' be the inclinations of the two chords DP and D'P. Then, by the same process as in Art. 89, or simply by changing V* into — J³ in that Art, we get 176 EQUATION REFERRED TO CONJUGATE DIAMETERS.

tan 6 tan <£' = j, cr

as the condition that the two chords DP and D'P shall be supplemental

Now, from Art. 116, Sch., we have

tan 0 tan ff =

a²

as the condition that two diameters shall be conjugate to each other; the rest of the argument of Art 89 applies directly to the hyperbola. Therefore, the supplemental chord PD' is parallel to the conjugate diameter KK'.

123. To find the equation of the hyperbola referred to any pair of conjugate diameters.

To do this we must transform the equation of the hyperbola

aY — W# = — aW,    (1)

from rectangular to oblique axes, having the same origin.

Let DD' and SS' be two conjugate diameters. Take CD for the new axis of x, and CS for the new axis of y. Denote the angle ACD by 0 and ACS by O'. Let cr, y be the co-ordinates of any point P of the hyperbola referred to the old axes, and x'y y' the coordinates of the same point referred to the new axes.

Now we may use the same process employed in Art 90 ; or, we may simply change A² into — i³ in (3) erf* Art. 90 (see Art. 102, Cor. 5), and get

(a*8in²0-&co&0)x,*+(a*sm²0'--PcosW)^*^ —dV. (1)

We pat this latter equal to — b'\ because we have supposed the new axis of ac to meet the given hyperbola, as in Fig. 93; therefore we know (Art. 116, Cor.) that the new axis of y will not meet the given hyperbola; hence

aWg-Pootf*' is*DeB,aTe

From (2)we get From (3) we get

(4)

(5)

a² sin* 9 — Z>* coe* d = — —

a!²

a² sin*®'-J* cos* O' =

Substitute (4) and (5) in (1), divide by — «*£*, omit accents from the variables, and we get

or,    c'*^-&'**»= -o'*y*,    (7)

which is the equation required, and is of the same form aa when referred to the axes of the curve (Art 102).

Similarly, the equation of the conjugate hyperbola referred to the same pair of conjugate diameters is

(8)

(»)

Let a' and i' denote the lengths of the semi-diameters CD and CS. If we make y'= 0 in (1), we get

g>« —    -— = a *.    (2)

o’ sin* 0 — 4* cos* 0 Also, if in (1) we make *'= 0, we get

-a²»

y'* =

a*eina0' — i^cos*#'

= — V\

(3)

?_    $_ - _i

a!²    b'² ~    •

or    c'y — b'h? = a!²#².

[Let the student give the demonstration.]

178 TANGENTS AT EXTREMITIES OF A CHORD.

Cob.—Comparing (7) with (9) of Art. 90, we see that the equation of the hyperbola may be derived from that of the ellipse by changing ft'a into — ft'*. Hence, we infer that any function of ft' expressing a property of the ellipse will be converted into one expressing a corresponding property of the hyperbola by changing ft' into ft'V — 1.

124. To find the equation of a tangent to the hyperbola referred to any pair of conjugate diameters.

By reasoning exactly as in Art 91, using the term “hyperbola” for “ellipse,” or, by changing ft'* into —ft'* in (1) of Art. 91, according to Art. 123, Cor., we get

ci*yf-V*xx! = -a!*b'\    (1)

which is the required equation.

Cob.—To find where the tangent cuts the axis of x, make y = 0 in (1), and get

a'*

x = —-•

125. To prove that tangents at the extremities of any chord of an hyperbola meet on the diameter

which bisects that chord.

r

= — a'* ft'*,    (l)

= - o'* ft'*,    (2)

tangents at the extremities

Take the diameter CD, which bisects the chord PP', for the axis of x, and the conjugate diameter CS for the axis of y.

Now reason as in Art 92, or change ft'* into —ft'* in (1) and (2) of Art. 92, according to Art. 123, Cor., and get

a'² yy' — b'%xx' : and    —a'tyy' — b'²a.sr':

which are the equations of the

of the chord PP' referred to the diameter CD which bisects PP, and the conjugate diameter CS. Now, by Art. 124, Cor^ both of these tangents cut the axis of x at the

point , 0^, which proves the proposition.

126. If tangents are drawn at the extremities of any focal chord of an hyperbola:

I.    The tangents will intersect on the corresponding directrix.

II.    The line drawn from the point of intersection of the tangents to the focus will be perpendicular to the focal chord.

I. Reasoning as in Art. 93, we find for the equation of the chord of contact (Art 111),

dhjy' — &xz' = —    (1)

which, for the right-hand focus (ae, 0), becomes

— Vaex! = — aW\

or    x' =    (2) that is, the point of intersection of the tangents is on the corresponding directrix (Art. 102, Cor. 1), showing that the directrix is the polar of the focus. (Art. 79, Sch.)

II. The equation of the line passing through the right-hand focus and the point («', y') is, by (Art 26),

y = <⁸> Prom (2), x' = -, which in (3) gives

y =    )

= — (x — ae). (Art 101, Cor.) (4)

180 INTERSECTION OF TANGENTS AT RIGHT ANGLES.

The*equation of the chord of contact [see (1) above] La

which jp perpendicular to (4), by Art 27, Cor. 1.

127. Find the locus of the point of intersection) of two tangents to^ an hyperbola at right angles to each other.

Season as in Art. 94, or change P into — P in equation (5) cf that Art., end get

** + 9* = a* — P    (1)

as the required locus. Hence, the locus is a cirole with its centre at C, and with \lp — P for its radius, unless P > a³, in which case the locus is impossible; that is, two tangents cannot be drawn at right angles to each other when P is greater than a*.

128. The rectangle of the focal perpendiculars upon any tangent is constant, and equal to the square of the semi-conjugate axis.

Call p and p' the perpendiculars. The equation of the tangent at any point (x, y') is

By Art 24,

p = +

Pyy' — Pxx' = — PP.

Px'ae — PP l {ex' — a)

'ja'y't+Px'*

IP „ , P

\py'+p

=    (ex' — a). (Art 119.)

,.    , —Hhc'ae—aW    b(ex' + a)

AISO, p' =-----r== = -, V

VaVa + P*a . A³..-, , ^

A    V^y + a²

= (ear' -f a). (Art. 119.)

Hence, pp' — ^ (f-x² — a²) = 5*. (Art. 119.)

129. To find the polar equation of the hyperbola, the left focus being the pole.

Let F'P = r; AF'P = 9; then (Def. of Art 99) wtf have,

F'P = e • PD

= e (F'M - F'O)

= e • MF' — e • F'O = e • F'P cos AF'P — a(&—1) (Art 102, Cor. 1),

or

therefore,

r = er cob 6 — a (e² — 1) ;

_ gfe² — 1) e cos 6 — 1 ’

which is the equation required.

Cor.—When 6 = 0, r = ae + a = F'C -j- CA = F'A, as it should do. (Art. 102, Cor. 1)

When e cos 0 — 1 = 0, that is, when 9 = cos⁻¹-,

e

00.

r - a(e2-¹) 0

But in this case r, or F'K, is parallel to the asymptote CR (See Art. 114, Cor, 2, and Fig. 89). That is, while 0 increases from 0 to cos”¹ -, r increases from ae + a to ao,

e

tracing the branch APP.

182 POLAR EQUATION OF THE HYPERBOLA.

When 0 passes the value cost¹ ~, e cos 8 — 1 becomes

negative, and therefore r becomes negative, and the left-hand branch is generated, the negative end of r tracjng QQ Q"; thus, when 6 = AF'H, r, being negative, is reckoned backwards to Q.

When 0 = 90°, r = — a (e³—1) = — — (Art. 101, Cor.),

CL

which equals the semi latus rectum, p, with a negative sign (Art. 102, Cor. 4), and Q' is located.

When 0 = 180% r = — a (e — 1) = a — ae = — F'A% as it should.

While 0 increases from 90° to 270°, the arc Q'Q"A'Q'" is traced with the negative end of r.

When 0 = 270°, r = — a (a³ — 1) = — j- = — p, and the point Q'" is located.

While 0 increases from 270° to cos"¹ ^, r remains negative, and increases numerically from p to oo, its negative end tracing Q'", Q‘\

At 0 = cos“¹^ in the fourth quadrant, r = oo, and is parallel to the asymptote CR'.

While 0 increases from cos⁻¹^ to 360°, r is positive, and

diminishes from oo to a + ae, and the arc P'", P", A is traced.

130. To find the polar equation of the hyperbola when the pole is at the centre.

Changing a²y² — && = — a²V* into a system of polar co-ordinates (as in Art. 97), we have

a²&    A²

** ⁼ cos* 0 — a² sin² 0 ~ cos² 0^1 (Art 101»Cor*) 0)'

Similarly, the polar equation of the conjugate hyperbola is

f* —

(2)

P

1 — e² cos² 0

Cob.—Equation (1) shows that for every value of 6 between — cos⁻¹ - and + cos⁻¹ -, r has two real values, e    e

numerically equal, with contrary signs. These two values of r, taken together, make the diameter; hence, every diameter of the hyperbola iB bisected at the centre (Art 102, Sch.).

When 0 = cos⁻¹ ^, r = oo; but in this case, r or CB, Fig. 89, coincides with the asymptote.

While 0 increases from cos⁻¹ - to (l80° — cos⁻¹ r is

imaginary, showing that it does not reach either branch of the given hyperbola.

Equation (2) shows that for every value of 0 between cos⁻¹ ^ and ^180° — cos⁻¹ r has two real values, numerically equal, with contrary signs. These two values of r, taken together, make the diameter of the conjugate hyperbola; hence, every diameter of the conjugate hyperbola is bisected at the centre.

When 6 = —cos⁻¹ r = ®; in this case, r coincides with the asymptote.

For every value of 6 between — cos⁻¹ - and + cos⁻¹ -,

6 6

r is imaginary, showing that it does not reach either branch of the given- hyperbola.

In (1), r is least when 0 = 0, giving

PROPERTIES Of HYPERBOLA.

184

which equals a .(Art 101, Cor.). In (2), r is least when 0 = 90°, giving r = b. Hence, in the hyperbola, each axis is the minimum diameter of its own carve.

Also, it is evident from both (1) and (2) that the valne of r is the same for 0 and (tt — 9). Therefore,diameters which make supplemental angles with the transverse axis of an hyperbola are equal.

131. The properties of the hyperbola hitherto established are similar to those of the ellipse. We have now to consider Borne properties peculiar to the hyperbola, arising from the presence of the asymptotes. (See Art. 113.)

132. To prove that the asymptotes are the diagonals of every parallelogram formed on a pair of conjugate diameters.

The equations of the hyperbola and its asymptotes, when referred to the axes of the curve, are respectively

and

(1)

(2)

When we transform the equation of the hyperbola to its conjugate diameters (Art 123), equation (1) becomes

Y

Fig. 96,

therefore we may at once infer that (2) transformed to the same conjugate diameters, becomes

___-l__o •

a'²    b'*-V>

that is, the equations of the asymptotes CE and CE', referred to any pair of conjugate diameters, are

Take CP = a', and CF = V.

y

Equation (3), or y = -x, is the equation of a line passing through the origin and the point (<*', (see Art. 26, Cor. 4), that is, through 0 and D; and (4), or y = — x,

is a line passing through the origin and (a', — b'), that is, through C and E. Hence, (3) and (4), which are the asymptotes, are also the diagonals of the parallelogram EDE'D' on the conjugate diameters PP" and P'P'".

133. To find the equation of the hyperbola referred to its asymptotes as axes.

To do this, we must transform the equation

(1) .

and

(3)

(4)

dy — Vh*=— aW

from rectangular to oblique axes, haring the same origin.

Let CX and CY be the old axes (Fig. 96). Take the lower asymptote CR’ for the new axis of x, and the other, CR, for the new axis of y.

Let x, y be the co-ordinates of any point P in the curve referred to the old axes, and x’, y' the co-ordinates of the same point referred to the new axes. Denote the angles ACR' and ACR by 0 and O' respectively.

The formulas for transformation (Art. 35, Cor. 1) are

x = x' cos 0 -f y' cos O’, y = x' sin 0 + y' sin O'.

Squaring, substituting in (1), and arranging, we have

186 HYPERBOLA REFERRED TO ASYMPTOTES.

From Art. 114, Cor. 2, we have

tan* 0 = ^ = tan’ V; o*    ’

from which we get

a* sin? 0 — & cos’ 0=0, and    a* sin* & — b* cos* ^ = 0.

Also, from Art. 114, Cor. 2, we have

gin 0 sin 6' =--

5*

o» + i*'

and

therefore,

cos 0 oos ff =

<v

a* + i»’

a* sin 6 sin O' — 5* cos 0 cos O' = —

2 a*5*

a* + ft*

Substituting (3), (4), and (5) in (2), we get

4a%*

(5)

x'y' = -aW-,

or suppressing accents from the variables and reducing, we have

a* +

= —4—’    (⁶)

a² 4- J²

and patting m² for —-—, we have,

(7)

xy = m\ which is the equation required.

Cob.—The equation of the conjugate hyperbola, referred to the same axes, is (Art 105)

xy = — m\

If we solve (7) for x, we get

which shows that as y increases, x diminishes, and when y = oo, x £= 0; that is, the curve approaches the axis of y, and finally touches it at an infinite distance from the centre. Similarly, the curve approaches the axis of x, and finally touches it at an infinite distance from the centre.

,    Sch.—The second jmember of (7) is essentially positive,

and of (8) essentially negative; hence, both x and y have the same sign in (7) and contrary signs in (8); therefore one branch of the given hyperbola lies wholly in the first angle and the other in the third; while one branch of the conju gate hyperbola lies wholly in the second and the other in the fourth angle. (See Fig. 96.)

In the case of equilateral hyperbolas (Art 105, Sch. 2), the angle between the asymptotes, which (Art 121, Sch.) is

equal to sin⁻¹ ^== sin"¹1, becomes a right angle;

therefore, the equilateral hyperbola is also called the Rectangular hyperbola. ^ ^

134. To find the equation of the tangent at any point of an hyperbola referred to the asymptotes as axes.

Let (x\ y¹) and (x"₉ y") be any two points, P and P', on the curve. The equation of the secant through these points (Art. 26), is

y — y' = (x - x'). (i)

Since (x¹, yr) and (x", y") are on the

' ~ curve, we have (Art. 133),

^ *    •

C    x'y'= m* — x"y", or

i

which in (1) gives y — y'=—^FT(x — x'), which is the equation of the secant to tfce hyperbola.

• •    > V • - % -

When the points become consecutive, we have *" = *'; hence (2) becomes

y-y’ = -£(*-*>    (³)

Clearing (3) of fractions, transposing, and uniting, we have

+ y'x = 2*V,

or

which is the equation of the tangent required.

Cob. 1.—Making y and x successively = 0 in (4) we get

x = 2x' = CT, and y = 2y' = CT.

Hence, P is the middle point of TT'; therefore, the portion of the tangent included between the asymptotes is bisected at the point of contact.

Cob. 2.—From Cor. 1, we have,

CT x CT' = 4x'y' = a* + 5* (Art. 133).

That is, the rectangle of the intercepts cut off upon the asymptotes by any tangent is constant, and equal to the sum of the squares on the semi-axes.

Cob. 3.—The area of the triangle TCT', Fig. 97, is = JCT x CT' sin TCT'

= 2a;y x f+p (C°r- !» and Art. 121, Sch.),

= ~r~x + » {    133)

%

= ab = constant.

Therefore, the triangle included between any tangent and the asymptotes is constant> and equal to the rectangle of the semi-axes.

135.    To prove that the intercepts of a secant between the hyperbola and its asymptotes are equal.

In equation (2) of Art 134, make y — 0, und get

x = x" + rf    '

= CN (Fig. 97).

Hence,    CN — x' = x",

or    M'N = DP';

therefore,    NP = N'P';

that is, the intercepts of the secant are equal.

Sch.—This proposition affords a convenient method of constructing the curve. If the axes are given, construct the rectangle on them, the diagonals of which are the asymptotes. Then through the extremity of the transverse axis, draw a right line intercepted by the asymptotes; lay off on this line from one asymptote a distance equal to the extremity of the axis from the other asymptote; the point thus found will be a point of the curve. In this manner, find any number of points, and draw a line through them; this will be the required curve.

136.    To prove that the parallelogram formed by drawing lines from any point of an hyperbola parallel to and terminating in the asymptotes, is equal to one-eighth the rectangle on the cutes.

Call <f> the angle TCT' (Fig. 97) ; the area of CM'PD = x'y' sin <f>

=    x    (Art. 133, and Art. 181, Sch.)

*    d T⁻ tr

= ±ab = i (2a • 2b),

which proves the proposition.

EQUATION OF CONJUGATE DIAMETERS.

190

137. To find the equations of two conjugate diameters of an hyperbola referred to its asymptotes.

Fig. 90.

The diameter which passes through the origin and the point P (x'y y') is represented (see Art. 26, Cor. 4) by

The diameter conjugate to this one, CD, is parallel to the tangent at (*', y'), and therefore (Art. 134, Eq. 4) its equation is

or

(2)

Cob.—When the diameters PP' and DD'become the axes, AA' and BB', we have, since the axes bisect the angle between the asymptotes,

CM' = M'A, or    s' = y';

therefore (1) and (2) become

x — y — 0,    and    x + y = 0,

which are the equations of the axes referred to the asymptotes.

138. Given the co-ordinates of the extremity of a diameter, to find those of the extremity of its conjugate.

Let (x, y') be the point P (Fig. 98), and (x”, y") the point D.

The equation of DD' (Art. 137) is

CHORDS PARALLEL TO A DIAMETER.

191

The equation of the conjugate hyperbola (Art. 133) is

xy = —m».    (2)

Eliminating between (1) and (2), we get

x" - T x', y" = ± y'.    (3)

Cob. 1.—The equation of the tangent at P (x\ if) (Art. 134) is

The equation of the tangent at D (x", y"), the extremity of the conjugate diameter (Art. 134) is

or from (3),

Adding (4) and (5), we get x = 0 as the locus of the intersection of the tangents (4) and (5), which is the equation of the axis of y, or the asymptote CR'. Therefore, tangents at the extremities of conjugate diameters meet on the asymptotes.

Cob. 2.—Since T' is a vertex of the parallelogram formed on the conjugate diameters PP' and DD', we have

PT' = CD;

therefore,    TT' = 2PT' = DD';

that is, the portion of the tangent at any point of an hyperbola, included between the asymptotes, is equal to the diameter conjugate to that which passes through the point of contact.

139. If a chord be drawn parallel to any diameter, it will be bisected by the conjugate diameter produced.

Let QQ' be drawn parallel to DD' (Fig. 98); then will it be bisected at M" by CP produced.

Since QQ' is parallel to DD', its equation will differ from that of DD' only by a constant term; therefore [Art. 138, (1)]

¹

See Howiaon’e Analytic Geometry, p. 881.

a)

is the equation of QQ'.

Combine (1) with the equation of PF (Art 137), which is

?-? = »- <²>

and we get x = fax', y = |<y,

as the co-ordinates of M". But from (1) we have CR = at, and CR' = ci</ ; therefore M" is the middle point of RR'. But (Art 135),

RQ = R'Q';

therefore, QM" = M"Q', which proves the proposition.

EXAMPLES.

1.    Find the axes of the hyperbola whose equation is 3yi — 2z* + 12 = 0 ; also the eccentricity of the given and the conjugate hyperbola, and the parameter.

Am. a = V6, 5 = 2; e = Vii t = Vf; 2p = -^=‘

2.    Find the intersection of the hyperbola 3y*—2a?+12 = 0 and the circle a? + y* = 16. AnSt (± 2^3, ± 2).

3.    Find whether the line y = $x cuts the hyperbola 5ya — 2a? = —15, or its conjugate.

Ans. It cuts the conjugate.

4.    Find the equation of an hyperbola of given transverse

axiB, whose vertex bisects the distance between the centre and the focus.    Ans. y* — 3a? = — 3a*.

8. If the ordinate MP (Fig. 95) of an hyperbola be produced to Q, so that MQ = FT, find the locus of Q.

A ns- A right line.

6.    If an ellipse and an hyperbola have the same foci, prove that their tangents at the points of intersection are at right angles. (See Art. 75, Cor. 2, and Art. 107, Cor.)

7.    Find the condition that the line Y— + ~ = l) shall

\?n n /

touch the hyperbola g _ ^ = l). Ans. £a--t = l.

CC 4/

[To obtain this, compare — + ^ = 1 with equation of tangent (Art. 107), which is

and we have

— and m

£

i

which in equation of curve gives the answer.]

8. Find where the tangents from the foot of the directrix will meet the hyperbola, and what angle they will make with the transverse axis.

Ans. The extremity of the lotus rectum ; tan⁻¹ ± e.

9. Find the angle included between the asymptotes of the hyperbola 16^ — 9z² = — 25.    Ans. 73° 44'.

10. Find the perpendicular from the focus of any hyperbola to its asymptotes. Ans. The semi-conjugate axis.

11. If 3AC = 2CF' (Fig. 95), find the inclination of the asymptotes to the transverse axis.    -Ans tan⁻¹

~2~’

12. If the asymptotes of the hyperbola are axes, show that the equation of one directrix is x + y — a — 0.

[See Art. 137, Cor.]

9

13.    Prove that if a circle be described with the focus of an hyperbola for its centre and with a radios equal to the semi-conjugate axis, it will touch the asymptotes in the points where they are cut by the directrix.

14.    Prove that the radius of a circle which touches an hyperbola and its asymptotes is equal to that part of the latus rectum produced which is intercepted between the curve and the asymptote.

15.    Find the length of the normal NP and of BP (Fig. 88).

[See Art. 108, Cor. 1.]    NP=—RP =

16.    Prove that the product of the two perpendiculars let fall from any point of an hyperbola upon the asymptotes is

constant and equal to ,a ^

’ a² + tr

17.    Tangents to an hyperbola are drawn from any point

on either branch of the conjugate curve; prove that their chord of contact touches the opposite branch of the conjugate curve.    - .

[Take the diameter passing through the point for axis of y, and the conjugate diameter for axis of as; equation of chord of contact is

_ ML _

a'2    b'³ “ ’

which soon reduces to y = ± V; .\ etc.]

18. In any equilateral hyperbola, let 0 = the inclination of a diameter, passing through any point P, and </>' = that of the polar of P, the transverse axis being the axis of x; prove that tan <t> tan q>' = 1.

[Equation of diameter is y — polar of Pis ass' — yy¹ = cP; .

1/

y_

x'

a?'    ,    ,,

= tan <p ;

tan <p; etc.]