Maths hyperbola notes theory

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1. DEFINITIONThe hyperbola is the locus of a point which moves such that its distance from a fixed point called focusis always e times (e > 1) its distance from a fixed line called directrix.

2. VARIOUS FORMS OF HYPERBOLA2.1 Standard form

Let S be the focus and ZM the directrix of a hyperbola.Since e > 1, we can divide SZ internally and externally in the ratio e : 1, let the points of division be A

and A as in the figure. Let AA = 2a and is bisected at C.

Then, SA = e. AZ, SA = e. ZA SA + SA e (AZ ZA ) 2ae i.e., 2SC = 2ae or SC = ae

Similarly by subtraction, SA SA

= e(ZA ZA) 2e.ZC

y

B

B’

M P

Sx

AZCZ’A’

M’

S’

2a 2eZC ZC a / e. Now, take C as the origin, CA as the x-axis, and the perpendicular line CY as the y-axis. Then, S isthe point (ae, 0) and ZM the line x = a/e. Let P(x, y) be any point on the hyperbola.Then the condition PS2 = e2. (distance of P from ZM)2 gives (x – ae)2 + y2 = e2 (x – a/e)2

or x2(1 – e2) + y2 = a2 (1 – e2) i.e.,2 2

2 2 2

x y1

a a (e 1)

... (i)

Since e > 1, e2 – 1 is positive. Let a2 (e2 – 1) = b2. Then the equation (i) becomes2 2

2 2

x y1.

a b

The eccentricity e of the hyperbola2 2

2 2

x y1

a b is given by the relation

22

2

be 1

a

.

Since the curve is symmetrical about the y-axis, it is clear that there exists another focus

S at (–ae, 0) and a corresponding directrix Z M with the equation x = –a/e, such that the samehyperbola is described if a point moves so that its distance from S is e times its distance from Z M .

(i) Foci: S = (ae, 0) & S (–ae, 0)

(ii) Equation of directories: x =a a

& xe e

(iii) Vertices: A = (a, 0) & A = (–a, 0)(iv) Transverse Axis: The lines segment AA of length 2a in which the foci S & S both called

Transverse axis of the Hyperbola.

(v) Conjugate Axis: The line segment BB ( B (0, b)) and ( B (0, –b)) is called theConjugate axis of the hyperbola. The Transverse axis & the Conjugate axis of the hyperbolaare together called principal axes of the hyperbola.

Theory Notes - Hyperbola

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(vi) Length of latus rectum =

a

b,aeL,

a

b,aeL&

a

b2 222

Illustration 1:Show that the equation x2 – 2y2 – 2x + 8y – 1 = 0 represents a hyperbola. Find the coordinates ofthe centre, length of the axes, eccentricity, latus rectum, coordinates of foci and vertices and equationsof directories of the hyperbola.

Solution :x2 – 2y2 – 2x + 8y – 1 = 0 (x2 – 2x) –2 (y2 – 4y) = 1 (x2 – 2x + 1) – 2(y2 – 4y + 4) = –6

(x – 1)2 – 2(y2 – 4y + 4) = –6 1)3(

)2y(

6

)1x(2

2

2

2

Shifting the origin at (1, 2) without rotating the coordinate axes and denoting the new coordinateswith respect to these axes by X and Y, we have

X = (x – 1) and Y = (y – 2) .... (i)

Using these relations, equation (i) is reduced to 13

Y

)6(

x2

2

2

2

... (ii)

This equation is of the form 2

2

2

2

b

Y

a

X = –1, where a2 = 26 and b2 = 23

Illustration 2:

Find the equation of the hyperbola whose foci are (8, 3) (0, 3) and eccentricity =3

4.

Solution :The centre of the hyperbola is the midpoint of the line joining the two foci. So the coordinates of the

centre are8 0 3 3

,2 2

i.e., (4, 3).

Let 2a and 2b be the length of transverse and conjugate axes and let e be the eccentricity. Then the

equation of the hyperbola is 2

2

2

2

b

)3y(

a

)4x(

= 1 ... (i)

Now, distance between the two foci = 2ae

22 )33()08( = 2ae ae = 4 a = 3

3

4e

Now, b2 = a2 (e2 – 1) b2 = 9

9

161 = 7.


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Thus, the equation of the hyperbola is

7

)3y(

9

)4x( 22

= 1 [Putting the values of a and b in (i)]

7x2 – 9y2 – 56x + 54y – 32 = 0

Drill Exercise - 1

1. Find the coordinates of the vertices, foci, eccentricity and the equations of the directrix of the hyperbola4x2 – 25y2 = 100.

2. Find the eccentricity of the hyperbola whose latus-rectum is 8 and conjugate axis is equal to half thedistance between the foci.

3. Find the coordinates of the centre of the hyperbola, x2 + 3xy + 2y2 + 2x + 3y + 2 = 0.

4. Find the eccentricity of the hyperbola 2

2

2

2

b

y

a

x = 1 which passes through (3, 0) and (3 2 , 2)

5. The foci of the ellipse16

x2

+ 2

2

b

y = 1 and the hyperbola

25

1

81

y

144

x 22

coincide, then find the value

of b2.

2.2 Focal distance: The focal distance of any point (x, y) on the hyperbola 1b

y

a

x2

2

2

2

are ex – a and

ex + a

2.3 Another Definition of the HyperbolaThe difference of the focal distances of a point

on the hyperbola is constant. PM and PM are

perpendiculars to the directrices MZ and M Z .

PS PS e (PM PM) .

y

M P

Sx

ZCZ’

M’

S’

eMM e(2a / e) 2a = constant.

2.4 Auxiliary CircleA circle drawn with centre C and T.A. as a diameter is called the Auxiliary Circle of the hyperbola.Equation of the auxiliary circle is x2 + y2 = a2.


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Note from the figure that P and Q are called the “Corresponding Points” on the hyperbola and theauxiliary circle. ' ' is called the eccentric angle of the point ‘P’ on the hyperbola 0 2 b g .

2.5 Parametric Coordinates

The equations x = a sec and y = b tan together represents the hyperbolaxa

yb

2

2

2

2 1 where

is a parameter. n other words, (a sec , b tan ) is a point on the hyperbola for all values of

. (2n 1) , n I2

The point (a sec , b tan ) is briefly called the point .

Note : Equation of a chord joining 1 & 2 is 1 2 1 2 1 2x ycos sin cos

a 2 b 2 2

.

General Note :Since the fundamental equation to the hyperbola only differs from the that to the ellipse in having- b2 instead of b2 it will be found that many proposition for the hyperbola are derived from thosefor the ellipse by simply changing the sign of b2.

2.6 General Form

The equation of hyperbola, whose focus is point (h, k), directrix is lx + my + n = 0

& centricity ‘e’ is given by (x – h)2 + (y – k)2 =)m(

)nmyx(e22

22

l

l

Illustration 3:

Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 2) and eccentricity 3 .

Solution :Let S(1, 2) be the focus and P (x, y) be a point on the hyperbola. Draw PM perpendicular from P onthe directrix.

Then by definition. SP = ePM


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2 2

2 2

2x y 1(x 1) (y 2) 3

2 1

S(1, 2)Focusz’

z

M2x+y=1

(x – 1)2 + (y – 2)2 = 35

)1yx2( 2

(x – 1)2 + (y – 2)2 = 3 {2x + y – 1}2

7x2 – 2y2 + 12xy – 2x + 14 – 22 = 0

This is the required equation of the hyperbola.

Drill Exercise - 2

1. Find the equation of the set of all points such that the difference of their distances from (4, 0) and(–4, 0) is always equal to 2.

2. Find the locus of the point which satisfies 10y)5x(y)5x( 2222 .

3. Find the equation of a hyperbola with coordinate axes as principal axes and the distances of oneof its vertices from the foci are 3 & 1.

4. Find the parametric equation of the hyperbola 125

)3y(

36

)5x( 22

.

5. The foci of a hyperbola coincide with the foci of the ellipse25

x2

+9

y2

= 1, then find the equation of

the hyperbola if its eccentricity is 2.

2.7 Conjugate HyperbolaThe hyperbola whose transverse and conjugate axes are respectively the conjugate and transverseaxes of a given hyperbola is called the conjugate hyperbola of the given hyperbola.

e.g.,2 2 2 2

2 2 2 2

x y x y1 & 1

a b a b

are conjugate hyperbolas of each.

Note: If e1 & e

2 are the eccentricities of the hyperbola & its conjugate then

2 21 2

1 11.

e e

3. ASYMPTOTESDefinition: If the length of perpendicular drawn from a point on the hyperbola to a straight line tends


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to zero as the point on moves to infinity. The straight line is called asymptotes.

Let y = mx + c is the asymptote of the hyperbola 1b

y

a

x2

2

2

2

. Solving these two we get the quadratic

as (b2 – a2m2)x2 – 2a2mcx–a2 (b2 + c2) = 0In order that y = mx + c be an asymptote, both roots of equation (1) must approach infinity.which are: coefficient of x2 = 0 & coefficient of x = 0

or m =a

b & a2mc = 0 c = 0 ... (1)

Y

Q B P

A

CR S

XA

B equation of asymptote are 0b

y

a

x&0

b

y

a

x .

Obviously angle between the asymptotes is 2tan–1(b/a).

If we draw lines through B, B parallel to the transverse axis and through A, A parallel to theconjugate axis, then P (a, b), Q (–a, b), R(–a, –b) and S(a, –b) all lie on the asymptotesx2/a2–y2/b2 = 0 so asymptotes are diagonals of the rectangle PQRS. This rectangle is called associatedrectangle.

Note:

1

b

y

a

xC,1

b

y

a

xH

2

2

2

2

2

2

2

2

& A

2

2

2

2

b

y

a

x = 0

clearly C + H = 2A{ H = hyperbola

C = Conjugate hyperbolaA = Asymptotes.}

Particular Case :When b = a the asymptotes of the rectangular hyperbola .x2 - y2 = a2 or y = x which are at right angles.

Note :(i) Equilateral hyperbola rectangular hyperbola.

(ii) If a hyperbola is a equilateral than the conjugate is also equilateral.

(iii) A hyperbola and its conjugate have the same asymptote.

(iv) The equation of the pair of asymptotes differ the hyperbola and the conjugate hyperbolaby the same constant only.

(v) The asymptotes pass through the centre of the hyperbola and the bisectors of the anglesbetween the asymptotes are the axes of the hyperbola.

(vi) The asymptotes of a hyperbola are the diagonals of the rectangle formed by the linesdrawn through the extremities of each axis parallel to the other axis.

(vii) Asymptotes are the tangent to the hyperbola from the centre.


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(viii) A simple method to find the coordinates of the centre of the hyperbola expressed as ageneral equation of degree 2 should be remembered as let f(x, y) = 0 represents a

hyperbola. Findfx and

fy . Then the point of intersection of

fx = 0 and

fy = 0 gives

the centre of the hyperbola.

Illustration 4:

Show that the acute angle between the asymptotes of the hyperbola 1b

y

a

x2

2

2

2

(a2 > b2) is

2cos–11

e

, where e is the eccentricity of the hyperbola.

Solution :

Equation of the asymptotes of the given hyperbola is 0b

y

a

x2

2

2

2

b2x2 – a2y2 = 0

If is an angle between the asymptotes, then tan = 22

22

ab

ba

= 22 ba

ab

so that tan = 22 ba

ab

cos )2/( )ba/(a 222 = 1/e.

Drill Exercise - 3

1. Find the equations of the asymptotes of the hyperbola, 3x2 + 10xy + 8y2 + 14x + 22y + 7 = 0

2. Find the equation of the conjugate hyperbola of the hyperbola, 3x2 – 5xy – 2y2 + 5x + 11y – 8 = 0.

3. The asymptotes of the hyperbola are parallel to 2x + 3y = 0 and 3x + 2y = 0. Its centre is (1, 2) andits passes through (5, 3). Find the equation of hyperbola.

4. The ordinate of any point P on the hyperbola 25x2 – 16y2 = 400 is produced to cut its asymptotes inpoints Q and R. Prove that QP.PR = 25.

5. Prove that the product of the perpendiculars from any point on the hyperbola2 2

2 2

x y1

a b to its

asymptotes is equal to2 2

2 2

a b

a b.


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4. RECTANGULAR OR EQUILATERAL HYPERBOLAA hyperbola is called rectangular if its asymptotes are at right angles. The asymptotes ofx2/a2 –y2/b2 =1 are y = (b/a) x so they are perpendicular if –b2/a2 = – 1 i.e., b2 = a2, i.e., a = b.Hence equation of a rectangular hyperbola can be written as x2–y2 = a2

We give below some important observations of rectangular hyperbola.

(i) a2 = a2 (e2 – 1) gives e2 = 2 i.e., e = 2 .

(ii) Asymptotes are y = x.

(iii) Rotating the axes by an angle – 4/ about the same origin, equation of the rectangular hy-perbola x2 – y2 = a2 is reduced to xy = a2/2 or xy = c2, (c2 = a2/2).In xy = c2, asymptotes are coordinate axes.

(iv) Rectangular hyperbola is also called equilateral hyperbola.

4.1 Rectangular Hyperbola referred to its asymptotes as axis of coordinates

(i) Equation is xy = c2 with parametric representation x = ct, y =t

c, Rt ~ (0}.

(ii) Equation of a chord joining the points P (t1) & Q (t

2), x + t

1 t

2y = c (t

1 + t

2)

(iii) Equation of the tangent at P(x1, y

1) is

xx

yy1 1

2 and at P(t) isxt

ty c 2 .

(iv) Chord with a given middle point as (h, k) is kx + hy = 2hk.

(v) Equation of the normal at P(t) is x t3 - yt = c(t4 - 1).

(vi) Vertex of this hyperbola is (c, c) and (-c, -c) ; focus is 2 2c c,d i and

2 2c c,e j ,the directories are x + y = 2 c and L R c. .b g 2 2 = T.A.= C.A.

Drill Exercise - 4

1. Find the lengths of transverse and conjugate axes, eccentricity and coordinate of foci and vertices,length of the latusrectum, equation of the directrices of the rectangular hyperbola xy = 36.

2. The distance between the directrices of a rectangular hyperbola is 10 units, then find the distancebetween its foci.

3. If a circle cuts a rectangular hyperbola xy = c2 in A, B, C, D and the parameters of these four pointsbe t

1, t

2, t

3 and t

4 respectively. Then show that t

1 = t

2.

4. If the tangent and normal to a rectangular hyperbola cut off intercepts a1 and a

2 on one axis and b

1 and


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b2 on the other show that a

1a

2 + b

1b

2 = 0.

5. If a rectangular hyperbola circumscribes a triangle, prove that it also passes through its orthocentre.

5. POSITION OFA POINT P w.r.t HYPERBOLALet S = 0 be the hyperbola and P (x

1, y

1)

be the point and S1 S(x

1, y

1).

ThenS

1 < 0 P is in the exterior region

S1 > 0 P is in the interior region

Interiorregion

Exteriorregion

Interiorregion

xO

y

S1 = 0 P lies on the hyperbola

6. LINEANDA HYPERBOLA

The straight line y = mx + c is a secant, a tangent or passes outside the hyperbola2 2

2 2

x y1

a b

according as : c2 > = < a2 m2 – b2.

Drill Exercise - 5

1. Find the positions of the points (7, –3) and (2, 7) relative to the hyperbola 9x2 – 4y2 = 36.

2. Find the equation of the tangent to the hyperbola x2 – 4y2 = 36 which is perpendicular to the linex – y + 4 = 0.

3. Find the equation of the tangent to the hyperbola 2x2 – 3y2 = 6 which is parallel to the line y = 3x + 4.

4. Find the point of contact of the line y = x – 1 with hyperbola 3x2 – 4y2 = 12.

5. Find the value of m for which y = mx + 6 is a tangent to the hyperbola100

x2

–49

y2

= 1.

7. TANGENTAND NORMAL7.1 Tangent

(i) Equation of the tangent to the hyperbola2 2

2 2

x y1

a b at the point (x

1 y

1) is 1

b

yy

a

xx21

21 .

(ii) In general two tangents can be drawn from an external point (x1, y

1) to the hyperbola and

they are y - y1 = m

1(x - x

1) and y - y

1 = m

2(x - x

2) , where m

1 and m

2 are roots of the

equation (x12 - a

2)m2 - 2 x

1y

1m + y

12 + b2 = 0. If D < 0, then no tangent can be drawn from

(x1 y

1) to the hyperbola.

(iii) Equation of the tangent to the hyperbola 1b

y

a

x2

2

2

2

at the point ( )tanb,seca is


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1b

tany

a

secx

.

(iv) Point of intersection of the tangents at 1 and 2

is

x = a y bcos

cos,

sin

cos

1 2

1 2

1 2

1 2

2

2

2

2

FHGIKJ

FHGIKJ

FHGIKJ

.

(v) Equation of a chord joining 1 & 2 is 1 2 1 2 1 2x ycos sin cos

a 2 b 2 2

.

(vi) y = mx 222 bma can be taken as the tangent to the hyperbola 1b

y

a

x2

2

2

2

.

Illustration 5:Find the equations of the tangents to the hyperbola 3x2 – y2 = 3, which are perpendicular to the linex + 3y = 2.

Solution :Let m be the slope of the tangent to the given hyperbola. Then,m × (slope of the line x + 3y = 2) = –1

m1

3

= – 1 m = 3

Now, 3x2 – y2 = 3 13

y

1

x 22

This is of the form 1b

y

a

x2

2

2

2

, where a2 = 1 and b2 = 3.

So, the equations of the tangents are y = mx 222 bma

y = 3x 39 y = 3x 6

Drill Exercise - 6

1. If the tangent at the point (h, k) to the hyperbola x2/a2 – y2/b2 = 1 cuts the auxiliary circle in pointswhose ordinates are y

1 and y

2 then prove that 1/y

1 + 1/y

2 = 2/k.

2. Show that the area of the triangle formed by the lines x – y = 0, x + y = 0 and any tangent to thehyperbola x2 – y2 = a2 is a2.

3. The tangent at any arbitrary point ‘P’ on2 2

2 2

x y

a b = 1 meets the line bx – ay = 0 at point ‘Q’, then find

the locus of mid point of PQ.

4. Find the equation of the tangent to the hyperbola 2x2 - 3y2 = 6 which is parallel to the line y = 3x + 4.


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5. Find the conditions that a straight line with slope m will be normal to parabola y2 = 4ax as well as atangents to rectangular hyperbola x2 - y2 = a2.

7.2 Normal

(i) The equation of the normal to the hyperbola 1b

y

a

x2

2

2

2

at point P(x1, y

1) on the curve

1

2

1

2

y

yb

x

xa = a2+ b2 a2e2

(ii) The equation of the normal at the point P (a sec , b tan )

hyperbola

tan

by

sec

axis1

b

y

a

x2

2

2

2

= a2 + b2 = a2 e2.

(iii) In general, four normals can be drawn to a hyperbola from any point and if ,,, be the

concentric angles of these four co-normal points, then is an odd multiple of .

Drill Exercise - 7

1. Find the equation of normal to the hyperbola16

x2

–9

y2

= 1 at (–4, 0).

2. The normal to the hyperbolaxa

yb

2

2

2

2 1 drawn at an extremity of its latus rectum is parallel to an

asymptote. Show that the eccentricity is equal to the square root of1 5

2

d i.

3. If the tangent and the normal to the rectangular hyperbola xy = c2, at a point, cuts off interceptsa

1 and a

2 on the x-axis and b

1, b

2 on the y-axis, then find the value of a

1a

2 + b

1b

2 .

4. The normal at P to a hyperbola of eccentricity e, intersects its transverse and conjugate axes atL and M respectively. If locus of the mid point of LM is hyperbola , then find the eccentricity of thehyperbola.

5. If the normal at P to the hyperbola 2

2

2

2

b

y

a

x =1 meets the transverse axis in G and conjugate axis in

g and CF be perpendicular to the normal, from the centre then prove thatPF.PG = CB2 = b2, PF. Pg = CA2 = a2. Also prove that SG = e. SP (where S is the focus)

7.3. Chord of Contact of Tangents Drawn from a Point Outside the HyperbolaChord of contact of tangents drawn from a point outside the hyperbola isT = 0 i.e., (xx

1/a2) – (yy

1/b2) = 1.

Illustration 6:


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From any point on the hyperbola 1b

y

a

x2

2

2

2

tangents are drawn to the hyperbola 2b

y

a

x2

2

2

2

.

Then show that the area cutoff by the chord of contact on the asymptotes is 4 ab

Solution :

Let P (x1, y

1) be a point on the hyperbola 1

b

y

a

x2

2

2

2

. Then, 1b

y

a

x2

21

2

21 .

The chord of contact of tangent from P to the hyperbola 2b

y

a

x2

2

2

2

is

2b

yy

a

xx21

21 ...(i)

The equations of the asymptotes are 0b

y

a

x and 0

b

y

a

x

The points of intersection of (i) with the two asymptotes are given by

x1 =

by

ax

b2y,

by

ax

a2x,

by

ax

b2y,

by

ax

a2

112

112

111

11

Area of the triangle = ab4

by

ax

ab8

2

1|yxyx|

2

1

2

21

2

21

1221

8. CHORD OF HYPERBOLA WITH SPECIFIED MIDPOINTChord of hyperbola with specified midpoint (x

1, y

1) is

T = S1 , where S

1 and T have usual meanings.

Illustration 7:Find the equation of the chord of the hyperbola 25x2 – 16y2 = 400, which is bisected at the point (5, 3).

Solution :

Equation of the given hyperbola can be written as 125

y

16

x 22

Therefore, equation of the chord of this hyperbola in terms of the middle point (5, 3) is (T = 1S )25x 3y 5 0

1 116 25 16 25 125x – 48y = 481

Illustration 8:Find the locus of the midpoints of the chords of the circle x2 + y2 = 16 which are tangents to thehyperbola. 9x2 – 16y2 = 144

Solution :Let (h, k) be the middle point of a chord of the circle x2 + y2 = 16Then its equation is hx + ky –16 = h2 + k2 – 16 i.e.,

hx + ky = h2 + k2 ... (i)


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Let (i) touch the hyperbola

9x2 – 16y2 = 144 i.e.,2 2x y

116 9 ... (ii)

at the point ( , ) say, then (i) is identical with 19

y

16

x

... (iii)

Thush16

= 22 kh

1

k9

22 kh

h16

and 22 kh

k9

Since ( , ) lies on the hyperbola (ii), 1kh

k9

9

1

kh

h16

16

12222

16h2 – 9k2 = (h2 + k2)2

Hence the required locus of (h, k) is (x2 + y2)2 =16x2 – 9y2.9. PAIR OFTANGENTS

Equation of pair of tangents from point (x1, y

1) to the hyperbola 2

2

2

2

b

y

a

x = 1 is

SS1 = T2 i.e.,

1

b

y

a

x2

2

2

2

1

b

y

a

x2

21

2

21

=2

21

21 1

b

yy

a

xx

10. DIRECTOR CIRCLEThe locus of the point of intersection of two perpendicular tangents to a hyperbola is called its directorcircle. Its equation is x2 + y2 = a2 – b2.

Equation of any tangent to x2/a2– y2/ b2 = 1 is y = mx )bma( 222 ... (i)

Tangent perpendicular to (i) is y = – 222 bm/axm

1 ... (ii)

Locus of point of intersection of these perpendicular tangents i.e., equation of the director circle canbe obtained by eliminating m between (i) and (ii).

(y – mx)2 + (my + x)2 = a2m2 – b2 + a2 – b2 m2 or (m2 + 1) x2 + (m2 + 1) y2

= (a2 – b2) (m2 + 1)Cancelling (m2 + 1), we get the equation of director circle as x2 + y2 = a2 – b2.

Drill Exercise - 8

1. Find the number of point(s) outside the hyperbola2 2x y

125 36 from where two perpendicular tangents

can be drawn to the hyperbola.

2. Find the equation to the chords of the hyperbola, 25x2 16y2 =400 which is bisected at the point (6 , 2).


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3. If m1 and m

2 are the slopes of the tangents to the hyperbola

x y2 2

25 161 which pass through the

point (6, 2) then find the value of m1m

2 and m

1 + m

2.

4. If the chord through the points (a sec, b tan) and (a sec, b tan) on the hyperbola 2

2

2

2

b

y

a

x = 1

passes through a focus, then prove that tan2

tan

2

=

)0,ae(focusfor,e1

e1

)0,ae(focusfor,e1

e1

.

5. Tangents are drawn from any point on the hyperbola x2 – y2 = a2 + b2 to the hyperbola 2

2

2

2

b

y

a

x = 1,

then prove that they meet the axes in concyclic points.

11. HIGHLIGHTS

(i) Equation of the tangent at P (t) ist

x + yt = 2c where p is the point on the curve xy = c2

(ii) Equation of the normal at P(t) is xt3 – yt = c (t4 – 1). where p is the point on the curvexy = c2.

(iii) Locus of the feet of the perpendicular drawn from focus of the hyperbola2 2

2 2

x y

a b = 1 upon

any tangent is its auxiliary circle i.e., x2 + y2 = a2 and the product of the feet of theseperpendiculars is b2 .

(iv) The portion of the tangent between the point of contact and the directrix subtends a rightangle at the corresponding focus.

(v) Locus of the feet of the perpendicular drawn from focus of the hyperbolaxa

yb

2

2

2

2 1 upon

any tangent is its auxiliary circle i.e. x2 + y2 = a2 and the product of the feet of theseperpendiculars is b2. (semi C.A.)2.

(vi) The foci the hyperbola and the points P and Q in which any tangent meets the tangents at thevertices are concyclic with PQ as diameter of the circle.

(vii) Perpendicular from the foci on either asymptote meet it in the same points as thecorresponding directrix and the common points of intersection lie on the auxiliary circle.

(viii) The tangent at any point P on a hyperbolaxa

yb

2

2

2

2 1 with centre C, meets the

asymptotes in Q and R and cuts off a CQR of constant area equal to ab from the


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asymptotes and the portion of the tangent intercepted between the asymptote is bisectedat the point of contact. This implies that locus of the centre of the circle circumscribing the

CQR in case of a rectangular hyperbola is the hyperbola itself and for a standardhyperbola the locus would be the curve, 4(a2x2 - b2y2) = (a2 + b2)2.

(ix) The tangent and normal at any point of a hyperbola bisect the angle between the focal radii.This spells is reflection property of the hyperbola as an incoming light ray aimed towards onefocus is reflected from the outer surface of the hyperbola towards the other focus.

(x) If from any point on the asymptote a straight line be drawn perpendicular to the transverseaxis, the product of the segments of this line intercepted between the point and the curve isalways equal to the square on the semi conjugate axis.

(xi) If the angle between the asymptote of a hyperbola2 2

2 2

x y1

a b is 2 , then the eccentricity

of the hyperbola is sec .

(xii) If a circle intersects a rectangular hyperbola at four points, then the mean value of the pointsof intersection is the midpoint of the line joining the centres of circle and hyperbola.

(xiii) A rectangular hyperbola circumscribing a triangle passes through the orthocenter of this triangle.

If ii

cct ,

t

= i = 1, 2, 3 be the angular points P, Q, R then orthocenter is 1 2 3i 2 3

c, ct t t

t t t

.

(xiv) If a circle and the rectangular hyperbola xy = c2 meet in the four points t1 , t

2, t

3 & t

4 , then

(a) t1t2t3 t

4 = 1

(b) the centre of the mean position of the four points bisects the distance between thecentre of the two curves.

(c) the centre of the circle through the points t1, t

2 & t

3 is

1 2 3 1 2 31 2 3 1 2 3

c 1 c 1 1 1t t t , t t t

2 t t t 2 t t t

.

Drill Exercise - 9

1. The chord of the hyperbola x2/a2 – y2/b2 = 1 whose equation is x cos + y sin= p subtends a rightangle at the centre. Prove that it always touches a circle.

2. Find the product of the length of the perpendiculars drawn from foci on any tangent to the hyperbola(x2/a2) - (y2/b2) = 1.

3. Find the locus of the point, tangents from which to the rectangular hyperbola x2 – y2 = a2 contain anangle of 45º.

4. Show that the normal to the rectangular hyperbola xy = c2 at the point ‘t’ meets the curve again at thepoint ‘t

1’ such that t

1.t3 = –1.


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Answer KeyDrill Exercise - 1

1.29 25

e , ( 5,0),( 29,0), x5 29

2.4

3

3. (- 1 , 0) 4.3

135. 7

Drill Exercise - 2

1. 15x2 – y2 = 15 2. pair of rays 3. x2 3y2 + 3 = 0

4. x = 5 + 6 sec, y = –3 + 5 tan 5. 112

y

4

x 22

Drill Exercise - 3

1. 3x + 4y + 5 = 0 and x + 2y + 3 = 0 2. 3x2 – 5xy – 2y2 + 5x + 11y – 16 = 0

3. 6x2 + 13xy + 6y2 – 38x – 37y – 98 = 0

Drill Exercise - 4

1. (LR) = (TA) = (CA)= 12 2 ; e = 2 ; foci )26,26( , )26,26( ;

vertex = (6, 6), (–6, –6); equation of directrices = x + y = ± 26 . 2. 20

Drill Exercise - 5

1. point (–7, 3) lies inside ; (2, 7) lies outside 2. x y 3 3 0

3. y = 3x + 5, y = 3x – 5 4. 4, 3 5.20

17

Drill Exercise - 6

3.2 2

2 2

x y 3

4a b 4. y = 3x + 5 5. m6 - 2m = 0

Drill Exercise - 7

1. y = 0 3. 0 4. e

Drill Exercise - 8

1. 0 2. 75x – 16y = 418 3. m1 + m

2 = 24/11, m

1m

2 = 20/11

Drill Exercise - 9


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2.12

2b 3. (x2 + y2)2 + 4a2 (x2 – y2) = 4a4 5. 6x2 + 6y2 + 13xy – 38x – 37y – 98 = 0

SOLVED SUBJECTIVE EXAMPLESExample 1 :

A tangent to the hyperbola 1b

y

a

x2

2

2

2

cuts the ellipse 1b

y

a

x2

2

2

2

in points P and Q. Find the locus

of the midpoint of PQ.

Solution :

Let M(x1, y

1) be the midpoint of the chord PQ of the ellipse 1

b

y

a

x2

2

2

2

.

Equation of PQ is2 2 2 2 22

1 1 1 1 1 1 12 2 2 2 2 2 2

1 1

xx yy x y b xx x yby

a b a b a y y a b

This is tangent to the hyperbola 1b

y

a

x2

2

2

2

if 221

4

21

42

2

2

21

2

21

21

4

bya

xba

b

y

a

x

y

b

2

21

2

21

2

2

21

2

21

b

y

a

x

b

y

a

x

Hence locus of (x1, y

1) is

2

2

2

22

2

2

2

2

b

y

a

x

b

y

a

x

Example 2 :

A straight line is drawn parallel to the conjugate axis of the hyperbola 1b

y

a

x2

2

2

2

to meet it and the

conjugate hyperbola respectively in the point P and Q. Show that the normals at P and Q to the curvesmeet on the x-axis.

Solution :Let P(a sec , b tan ) be a point on the hyperbola, and Q(a tan , b sec ) be a point on theconjugate hyperbola.

a sec = atan sec = tan

Equation of the normal to the hyperbola

1b

y

a

x2

2

2

2

at P is y – b tan =

secb

tana (x – b sec )

Equation of the normal to the conjugate hyperbola at Q is

y – b sec = – )tanax(tanb

seca


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Eliminate x and use sec = tan

We get y (sec – tan ) = 0 y = 0

Hence the normals meet on the x-axis.Example 3 :

From a point G on the transverse axis of the hyperbola 1b

y

a

x2

2

2

2

, GL is drawn perpendicular to

one of its asymptotes. Also Gp is a normal to the curve at P. Prove that LP is parallel to the conjugateaxis.

Solution :Let P(a sec , btan ) be any point on the hyperbola

Equation of the normal at P is ax cos + by cot = a2 + b2.It meets the x-axis (transverse axis) at y = 0

x =

seca

ba 22

2 2a b

G sec ,0a

The equation of line perpendicular to the asymptote bx – ay = 0 and passing through G, i.e., equation

of GL is y = –

sec

a

bax

b

a 22

ax + by = (a2 + b2) secIts intersection with the asymptote bx – ay = 0 gives x = a sec . So the x coordinate of

L is a sec , which is equal to the x-coordinate of the point P LP is parallel to the y-axis LP is parallel to the conjugate axis.

Example 4 :A variable straight line of slope 4 intersects the hyperbola xy = 1 at two points. Find the locus of thepoint which divides the line segment between these points in the ratio 1 : 2.

Solution :Let the line be y = 4x + c. It meets the curve xy = 1 atx (4x + c) = 1 4x2 + cx –1 x

1 + x

2 = –c/4

Also y (y – c) = 4 y2 – cy – 4 = 0 y1 + y

2 = c

Let the point which divides the line segment in the ratio 1 : 2 be (h, k)

h3

x2x 21

x2 = 3h + c/4 x

1 = –c/2 – 3h

Also k3

y2y 21

y2 = 3k – c y

1 = –3k + 2c

Now (h, k) lies on the line y = 4x + c k = 4h + c c = k – 4h x

1 = –k/2 + 2h – 3h = –h – k/2 and y

1 = –3k + 2k – 8h = –k – 8h

16h2 + k2 + 10hk = 2. Hence locus of (h, k) is16x2 + y2 + 10 xy = 2

Example 5 :Prove that if normal to the hyperbola xy = c2 at point t meets the curve again at a point t

1 then

t3 t1 + 1 = 0.


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Solution :Equation of normal at point t i.e., (ct, c/t) is

y – xt2 =t

c (1 – t4) ... (1)

It meets the curve again at t1 then (ct

1, c/t

1) must satisfy (1)

32

11

421

1

tt

1tt

t

1)t1(

t

ctct

t

c

0)tt(tt

1

t

11

2

1

1

1

tt

)tt( (1 + t3t

1) = 0

Clearly 1tt t3 t1 + 1 = 0.

Example 6 :The angle between a pair of tangents drawn from a point P to the parabola y2 = 4ax is 45°. Show thatthe locus of the point P is a hyperbola.

Solution :Let P ( , ) be any point on the locus. Equation of pair of tangents from P( , ) to the parabolay2 = 4ax is

[ )ax4y()a4(])x(a2y 222 [T2 = SS

1] ... (i)

A = coefficient of x2 = 4a2

2H = coefficient of xy = –4and B = coefficient of y2 = 2 – ( 2 – 4a ) = 4a .Since the angle between the two lines of (1) is 45°, we have

1 = tan45° =22 H AB

A B

(A + B)2 = 4 (H2 –AB) (4a2 + 4 a )2 = 4[a2 2 – (4a)2 (4a )]

0aa6 222 or 222 a8)a3( The equation of required locus is

(x + 3a)2 – y2 = 8a2

which is a hyperbola.Alternate Solution

Equation of any tangent to hyperbola y2 = 4ax isy = mx + a/m

which passes through ( , ) if

= m + a/m or m2 – m + a = 0 ... (1)If m

1 and m

2 are roots of (1).

m1 + m

2 = / and m

1m

2 = /a we have

1 = tan45° =21

21

mm1

mm

(1 + m1 m

2)2 = (m

1 – m

2)2

(1 + m1m

2)2 = (m

1 + m

2)2 – 4m

1 m

2


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(1 + a/ )2 = 2)/( – 4a/

( + a)2 = 2 – 4a or ( + 3a)2 – 2 = 8a2

The required locus is (x + 3a)2 –y2= 8a2 which is a hyperbola.Example 7 :

Find the centre, eccentricity, foci, directories and the lengths of the transverse and conjugate axes ofthe hyperbola, whose equation is (x – 1)2 –2 (y – 2)2 + 6 = 0

Solution :The equation of the hyperbola can be written as(x – 1)2 – 2(y–2)2 + 6 = 0

or 13

)2y(

6

)1x(– 2

2

2

2

or 2 2

2 2

Y x1

3 6

Where Y = (y–2) and x = (x–1) ... (1) centre: X = 0, Y = 0 i.e., (x – 1) = 0, x = 1 & (y – 2) = 0, y = 2.So a = 3 and b = 6 so transverse axis = 32 , and conjugate axis = 62 .

Also b2 = a2 (e2 – 1) 6 = 3 (e2 – 1) i.e., e = 3In (X, Y) coordinates, foci are (0, ae)i.e., (0, 3). foci are (1, 2 3) i.e., (1, 5) and (1, –1)Equations of directories, Y = a/e.

directrices y – 2 = 13/3 or y = 3, y = 1

Example 8 :Find the equation and angle between the asymptotes of the hyperbolax2 + 2xy – 3y2 + x + 7y + 9 = 0

Solution :Let the combined equation of asymptotes x2 + 2xy – 3y2 + x + 7y + = 0If it represents pair of straight linesabc + 2fgh – af 2 – bg2 – ch2 = 0

= –23/16 Asymptotes x2 + 2xy – 3y2 + 7y – 23/16 = 0Required angle = tan–12.

Example 9 :Prove that the locus of a point whose chord of contact touches the circle inscribed on the straight linejoining the foci of the hyperbola x2/a2 – y2 / b2 = 1 as diameter is x2/a4 + y2 / b4 = 1/(a2 + b2).

Solution :


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Circle on the join of foci (ae, 0) and (–ae, 0) diameter is (x – ae) (x + ae) + (y – 0) (y – 0) = 0i.e., x2 + y2 = a2e2 = a2 + b2 ... (i) [ a2e2 = a2 + b2]Let chord of contact of P (x

1, y

1) touch the cirlce (i)

Equation of chord of contact of P is [T = 0]xx

1/a2 – yy

1 /b2 = 1 i.e., b2x

1x–a2 y

1y – a2 b2 = 0 ... (ii)

)ba()yaxb(

ba 22

21

421

4

22

Hence locus of P (x1, y

1) is (b4x2 + a4y2) (a2 + b2) = a4b4 .

Example 10 :

An ellipse has eccentricity2

1 and one focus at the point

1,

2

1P . Its one directrix is common

tangent to the circle x2 + y2 = 1 and the hyperbola x2 – y2 = 1, nearer to P. The equation of the ellipsein the standard form.

Solution :The circle x2 + y2 = 1 and the hyperbola x2 – y2 = 1 touch each other at the points ( 1, 0) and the

common tangent at these point are x = 1. Since x = 1 is nearer to the focus

1,

2

1P , this is the

directrix of the required ellipse.

Therefore, the major axis is parallel to the axis passing through the focus

1,

2

1P . Hence the equa-

tion of the major axis is y = 1.Let a be the length of the semi major axis of the ellipse and let the coordinates of the centre C of the

ellipse be )1,( .

Then CP = 2

1= a.e = a ×

2

1 ... (i)

2

1e

and the distance of the directrix from the centre =e

a.

1 – = a2e

a ... (ii)

x – y = 12 2

x = 1

x + y = 12 2

From (i) and (ii) we get a =1 1

and e3 2

.

If b is the length of the semi minor axis of the ellipse, then b2 = a2(1 – e2)

b2 =12

1

4

11

9

1

Hence the required equation of the ellipse is 1

121

)1y(

31

31

x 2

2

2


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or2

3

1x9

+ 12(y – 1)2 = 1


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SOLVED OBJECTIVE EXAMPLESExample 1 :

The equation of a line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one ofits asymptotes is 3x – 4y – 6 = 0, the equation of the other asymptote is(a) 4x – 3y + 17 = 0 (b) –4x – 3y + 17 = 0(c) –4x + 3y + 1 = 0 (d) 4x + 3y + 17 = 0

Solution :We know that asymptotes of rectangular hyperbola are mutually perpendicular, thus other asymptoteshould be 4x + 3y + = 0. Intersection point of asymptotes is also the centre of the hyperbola.

Hence intersection point of 4x + 3y + = 0 and 3x – 4y – 6 = 0 should lie on the line x–y–1 = 0,using it can be easily obtained.Hence (d) is the correct answer.

Example 2 :The locus of the middle points of chords of hyperbola 3x2 – 2y2 + 4x – 6y = 0 parallel to y = 2x is(a) 3x – 4y = 4 (b) 3x – 4y + 4 = 0(c) 4x – 4y = 3 (d) 3x – 4y = 2

Solution :Let the mid point be (h, k). Equation of a chord whose mid point is (h, k) would be T = S

1

or 3x h – 2yk + 2(x + h) – 3(y+k) = 3h2 – 2k2 + 4h – 6k x (3h + 2) –y (2k + 3) – (2h + 3k) – 3h2 + 2k2 = 0

Its slope is3k2

2h3

= 2 (given)

3h = 4k + 4 Required locus is 3x – 4y = 4Hence (a) is the correct answer.

Example 3 :

The tangent at a point P on the hyperbola2

2

2

2

b

y

a

x =1 meets one of the directrix in F. If PF subtends

an angle at the corresponding focus, then equals

(a) 4/ (b) 2/(c) 4/3 (d)

Solution :Let directrix be x = a/e and focus be S(ae, 0). Let P (a sec , b tan ) be any point on the curve.

Equation of tangent at P isb

tany

a

secx

= 1. Let F be the intersection point of tangent of directrix,

then F =

tane

)e(secb,e/a

SF PS2

b(sec e) b tanm , m

e tan (a 1) a(sec e)

m

SF .m

PS = –1

Hence (b) is the correct answer.


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Example 4 :The line lx + my + n = 0 will be a normal to the hyperbola b2x2 – a2y2 = a2b2 if

(a) 2

222

2

2

2

2

n

)ba(

m

ba

l(b)

2 2 2 2 2

2 2 2

a b (a b )

m n

l

(c)n

)ba(

m

ba 222

2

2

2

2

l(d) none of these

Solution :Equation of normal at (a sec , btan ) is

ax cos + by cot = a2 + b2

Comparing it with lx + my + n = 0 we get

l

cosa =

2 2bcot (a b )

m n

cos =nb

)ba(mcotand

an

)ba( 2222

l

sin =am

bl

Thus 22

22

ma

b l + 1

na

)ba(22

2222

l

or, 2

222

2

2

2

2

n

)ba(

m

ba

l


Example 5 :If (a sec , btan ) and (asec , btan ) be the coordinate of the ends of a focal chord of

2

2

2

2

b

y

a

x = 1, then tan

2tan

2

equals to

(a)1e

1e

(b)e1

e1

(c)c1

e1

(d)1e

1e

Solution :Equation of chord connecting the points (asec , b tan ) and (asec , b tan ) is

x ycos sin cos

a 2 b 2 2

If it passes through (ae, 0); we have, ecos cos2 2


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e =

cos 1 tan . tan2 2 2

1 tan tancos2 22

tan

e1

e1

2tan.

2

Hence (b) is the correct answer.Example 6 :

The point of intersection of the curves whose parametric equations are x = t3 + 1, y = 2t and x = 2s,

y =2

s , is given by

(a) (1, –3) (b) (2, 2)(c) (–2, 4) (d) (1, 2)

Solution :x = t2 + 1, y = 2t x – 1 = y2/2x = 2s, y = 2/s xy = 4

For the point of intersection we have 016y4y4

y1

y

4 32

y = 2 x = 2


Example 7 :

The equation of the hyperbola whose foci are (6, 5), (–4, 5) and eccentricity4

5 is

(a) 19

)5y(

16

)1x( 22

(b) 19

y

16

x 22

(c) 19

)5y(

16

)1x( 22

(d) None of these

Solution :S

1 (6, 5); S2 (–4, 5), e = 5/4

S1S

2 = 10 2ae = 10 a = 4

and b2 = a2 (e2 – 1) = 16 9116

25

Centre of the hyperbola is (1, 5)

Equation of required hyperbola is 19

)5y(

16

)1x( 22

Example 8 :The equation (x – )2 + (y – )2 = k(lx + my + n)2 represents

(a) a parabola for k < (l2 + m2)–1 (b) an ellipse for 0 < k < (l2 + m2)–1

(c) a hyperbola for k > (l2 + m2)–1 (d) a point circle for k = 0.


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Solution :

(x – )2 + 2)( = k (lx + my + n)2 = k (l2 + m2)

2

22 m

nmyx

l

l

PS

kPM

(l2 + m2)

If k(l2 + m2) = 1, P lies on parabolaIf k(l2 + m2) < 1, P lies on ellipseIf k(l2 + m2) > 1, P lies on hyperbolaIf k = 0, P lies on a point circleHence (b), (c), (d) are correct.

Example 9 :

The point on the hyperbola 118

y

24

x 22

which is nearest to the line 3x + 2y + 1 = 0 is

(a) (6, 3) (b) (–6, 3)(c) (6, –3) (d) (–6, –3)

Solution :Equation of tangent is )tan18,sec24( is

118

tany

24

secx

, then point is nearest to the line 3x + 2y + 1 = 0.

so its slope = –2

3

2

3

tan

18

24

sec

sin = –3

1

Hence the point is (6, –3)Hence (c) is the correct answer.

Example 10 :The locus of a point, from where tangents to the rectangular hyperbola x2 – y2 = a2 contain an angle of45° is(a) (x2 + y2) + a2(x2 – y2) = 4a4 (b) 2(x2 + y2) + 4a2 (x2 – y2) = 4a2

(c) (x2 + y2) + 4a2(x2 – y2) = 4a4 (d) (x2 + y2) + a2 (x2 – y2) = a4

Solution :Let y = mx 222 aam be two tangent and passes through (h, k) then (k – mk)2 = m2a2– a2

m2(h2 – a2) – 2khm + k2 + a2 = 0

m1 + m

2 =

22 ah

kh2

and m

1 m

2 =

22

22

ah

ak

, using tan45°= 1 2

1 2

m m

1 m m

Hence (c) is the correct answer.


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