Maths hyperbola notes theory by studysteps .in

Hyperbola

Theory Notes - Hyperbola 1.

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

2. 2.1

VARIOUS FORMS OF HYPERBOLA 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.

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Then, SA = e. AZ, SA = e. ZA  SA + SA  e (AZ  ZA)  2ae  i.e.,

y M’

2SC = 2ae or SC = ae

S’

Similarly by subtraction, SA   SA

Z’

.in

A’

= e(ZA  ZA)  2e.ZC

B’

x (1 – e ) + y = a (1 – e ) i.e.,

x2 y2  1 a 2 a 2 (e2  1)

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 is the 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

... (i)

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

x 2 y2  1. a 2 b2

 b2  2 x 2 y2 e  1  2  . The eccentricity e of the hyperbola 2  2  1 is given by the relation a b  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 same hyperbola 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) (iii) (iv) (v)

a a & x e e Vertices: A = (a, 0) & A = (–a, 0) Transverse Axis: The lines segment AA of length 2a in which the foci S & S both called Transverse axis of the Hyperbola. Conjugate Axis: The line segment BB ( B  (0, b)) and ( B  (0, –b)) is called the Conjugate axis of the hyperbola. The Transverse axis & the Conjugate axis of the hyperbola are together called principal axes of the hyperbola.

Equation of directories: x =

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Hyperbola

(vi)

 b2   2b 2  b2   & L  ae,  , L  ae, Length of latus rectum = a a  a   

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

( x  1) 2

 6



( y  2) 2  1 ( 3)2

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Shifting the origin at (1, 2) without rotating the coordinate axes and denoting the new coordinates with respect to these axes by X and Y, we have X = (x – 1) and Y = (y – 2) .... (i)

x2 Y2  1 2 Using these relations, equation (i) is reduced to ( 6 ) 2 3

 

 6

and b2 =

 3

Illustration 2:

X2 Y2 This equation is of the form 2  2 = –1, where a2 = a b

... (ii)

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

4 . 3

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

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 8 0 33 , centre are   i.e., (4, 3). 2   2

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

( x  4) 2 ( y  3) 2  equation of the hyperbola is =1 a2 b2 Now, distance between the two foci = 2ae



4  (8  0) 2  (3  3) 2 = 2ae  ae = 4  a = 3  e  3   

16   Now, b2 = a2 (e2 – 1)  b2 = 9   1   = 7. 9 

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... (i)

Hyperbola

Thus, the equation of the hyperbola is



( x  4) 2 ( y  3) 2  = 1 [Putting the values of a and b in (i)] 9 7 7x2 – 9y2 – 56x + 54y – 32 = 0 Drill Exercise - 1

Find the coordinates of the vertices, foci, eccentricity and the equations of the directrix of the hyperbola 4x2 – 25y2 = 100.

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

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

Find the eccentricity of the hyperbola

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2 x2 y x 2 y2 1   The foci of the ellipse + 2 = 1 and the hyperbola coincide, then find the value 16 b 144 81 25 of b2.

2.3

x 2 y2 Focal distance: The focal distance of any point (x, y) on the hyperbola 2  2 1 are ex – a and a b ex + a

2.2

x 2 y2  = 1 which passes through (3, 0) and (3 2 , 2) a 2 b2

Another Definition of the Hyperbola The difference of the focal distances of a point

y M’

on the hyperbola is constant. PM and PM  are perpendiculars to the directrices MZ and M Z .

P x

S’

Z’

PS  PS  e (PM   PM) .  eMM   e(2a / e)  2a = constant.

2.4

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Auxiliary Circle A 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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Hyperbola

Note from the figure that P and Q are called the “Corresponding Points” on the hyperbola and the

auxiliary circle. ' ' is called the eccentric angle of the point ‘P’ on the hyperbola 0    2  . 2.5

Parametric Coordinates

x2 y2 The equations x = a sec and y = b tan  together represents the hyperbola 2  2  1 where  a b is a parameter. n other words, (a sec , b tan ) is a point on the hyperbola for all values of

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  .  (2n  1) 2 , n  I The point (a sec , b tan ) is briefly called the point  .

    2   2 x y cos 1 2  sin 1  cos 1 . a 2 b 2 2

Note : Equation of a chord joining 1 &  2 is

General Form

2.6

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 those for the ellipse by simply changing the sign of b2.

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 =

e 2 (lx  my  n ) 2 (l 2  m 2 )

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 on the directrix. Then by definition. SP = ePM

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Hyperbola



(x  1)  (y  2)  3 2

M 2x+y=1

2x  y  1 22  12

z’

(2x  y  1) 5

S(1, 2) Focus



(x – 1)2 + (y – 2)2 = 3



(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 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.

Find the locus of the point which satisfies

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

Find the parametric equation of the hyperbola

The foci of a hyperbola coincide with the foci of the ellipse

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( x  5) 2 ( y  3) 2   1. 36 25

( x  5) 2  y 2  ( x  5) 2  y 2  10 .

x 2 y2 + = 1, then find the equation of 25 9

the hyperbola if its eccentricity is 2. 2.7

Conjugate Hyperbola The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola is called the conjugate hyperbola of the given hyperbola.

x 2 y2 x 2 y2   1 &   1 a 2 b2 a 2 b2 are conjugate hyperbolas of each. e.g.,

Note: If e1 & e2 are the eccentricities of the hyperbola & its conjugate then 3.

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1 1   1. e12 e 22

ASYMPTOTES Definition: If the length of perpendicular drawn from a point on the hyperbola to a straight line tends

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Hyperbola

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

x 2 y2 Let y = mx + c is the asymptote of the hyperbola 2  2 1 . Solving these two we get the quadratic a b 2 2 2 2 2 2 2 2 as (b – a m )x – 2a mcx–a (b + c ) = 0 In 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= 

b & a2mc = 0  c = 0 a

... (1)

A

x y x y   0&  0. a b a b –1 Obviously angle between the asymptotes is 2tan (b/a).



Y BP

equation of asymptote are

A C

R B S

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If we draw lines through B, B parallel to the transverse axis and through A, A parallel to the conjugate axis, then P (a, b), Q (–a, b), R(–a, –b) and S(a, –b) all lie on the asymptotes x2/a2–y2/b2 = 0 so asymptotes are diagonals of the rectangle PQRS. This rectangle is called associated rectangle.

clearly  C + H = 2A { H = hyperbola C = Conjugate hyperbola A = Asymptotes.}

 x 2 y2   x 2 y2   x 2 y2  Note: H  2  2 1 , C  2  2  1 & A  2  2  = 0 b  b b a a  a 

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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 hyperbola by the same constant only.

(v)

The asymptotes pass through the centre of the hyperbola and the bisectors of the angles between the asymptotes are the axes of the hyperbola.

(vi)

The asymptotes of a hyperbola are the diagonals of the rectangle formed by the lines drawn 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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Hyperbola

(viii)

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

f f f f and . Then the point of intersection of = 0 and = 0 gives y y x x the centre of the hyperbola. hyperbola. Find

Illustration 4:

x 2 y2 Show that the acute angle between the asymptotes of the hyperbola 2  2 1 (a2 > b2) is a b 1 2cos–1   , where e is the eccentricity of the hyperbola. e

Solution :

b2x2 – a2y2 = 0



x 2 y2  0 a 2 b2

.in

Equation of the asymptotes of the given hyperbola is

 cos ( / 2)  a 2 /(a 2  b 2 ) = 1/e.

ab so that tan  = 2 a  b2

ab  a 2b2 = 2 2 2 a  b2 b a

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

Drill Exercise - 3

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

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

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

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

Prove that the product of the perpendiculars from any point on the hyperbola

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asymptotes is equal to

a 2b2 . a 2  b2

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x 2 y2   1 to its a 2 b2

Hyperbola

RECTANGULAR OR EQUILATERAL HYPERBOLA A hyperbola is called rectangular if its asymptotes are at right angles. The asymptotes of x2/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) (ii) (iii)

(iv)

Rotating the axes by an angle –  / 4 about the same origin, equation of the rectangular hyperbola x2 – y2 = a2 is reduced to xy = a2/2 or xy = c2, (c2 = a2/2). In xy = c2, asymptotes are coordinate axes. Rectangular hyperbola is also called equilateral hyperbola.

Rectangular Hyperbola referred to its asymptotes as axis of coordinates

(ii)

c , t  R ~ (0}. t Equation of a chord joining the points P (t1) & Q (t2), x + t1 t2y = c (t1 + t2)

(iii)

Equation of the tangent at P(x1, y1) is

(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

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Equation is xy = c2 with parametric representation x = ct, y =

x x y   2 and at P(t) is  ty  2c . t x1 y1

(i)

e

4.1

a2 = a2 (e2 – 1) gives e2 = 2 i.e., e = 2 . Asymptotes are y =  x.

2 c,

and

b g

2 c,  2 c ,the directories are x + y =  2 c and  L. R.  2 2 c = T.A.= C.A. Drill Exercise - 4

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.

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

If a circle cuts a rectangular hyperbola xy = c2 in A, B, C, D and the parameters of these four points be t1, t2, t3 and t4 respectively. Then show that t1 = t2.

If the tangent and normal to a rectangular hyperbola cut off intercepts a1 and a2 on one axis and b1 and

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Hyperbola

b2 on the other show that a1a2 + b1b2 = 0. 5.

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

POSITION OFA POINT P w.r.t HYPERBOLA Let S = 0 be the hyperbola and P (x1, y1) be the point and S1  S(x1, y1). Then S1 < 0  P is in the exterior region S1 > 0  P is in the interior region S1 = 0  P lies on the hyperbola LINE ANDA HYPERBOLA

y Interior region

Exterior region

Interior region

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

.in

according as : c2 > = < a2 m2 – b2.

x 2 y2  1 a 2 b2

Drill Exercise - 5

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

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

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

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

x2 y2 Find the value of m for which y = mx + 6 is a tangent to the hyperbola – = 1. 100 49

7. 7.1

TANGENT AND NORMAL Tangent

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(ii)

xx yy x 2 y2  2  1 at the point (x1 y1) is 21  21  1 . 2 a b a b In general two tangents can be drawn from an external point (x1, y1) to the hyperbola and they are y - y1 = m1(x - x1) and y - y1 = m2(x - x2) , where m1 and m2 are roots of the equation (x12 - a2)m2 - 2 x1y1m + y12 + b2 = 0. If D < 0, then no tangent can be drawn from (x1 y1) to the hyperbola.

(iii)

Equation of the tangent to the hyperbola

(i)

Equation of the tangent to the hyperbola

x 2 y2   1 at the point ( a sec , b tan ) is a 2 b2

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Hyperbola

x sec  y tan    1. a b

(iv)

Point of intersection of the tangents at  1 and  2 is

FG    IJ H 2 K,

cos x= a

cos

1   2 2

FG    IJ H 2 K. yb F    IJ cosG H 2 K sin

    2   2 x y cos 1 2  sin 1  cos 1 . a 2 b 2 2

(v)

Equation of a chord joining 1 &  2 is

(vi)

y = mx  a 2 m 2  b 2 can be taken as the tangent to the hyperbola

x 2 y2  1 . a 2 b2

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Illustration 5: Find the equations of the tangents to the hyperbola 3x2 – y2 = 3, which are perpendicular to the line x + 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  1 m   = – 1  m = 3   3

x 2 y2  1 Now, 3x – y = 3  1 3 2

x 2 y2 This is of the form 2  2 1 , where a2 = 1 and b2 = 3. a b So, the equations of the tangents are y = mx 



y = 3x 

9  3  y = 3x 

a 2m2  b2

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 points whose ordinates are y1 and y2 then prove that 1/y1 + 1/y2 = 2/k.

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

The tangent at any arbitrary point ‘P’ on

x2 a2



y2 b2

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

the locus of mid point of PQ. 4.

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

7.2

Normal (i)

The equation of the normal to the hyperbola

x 2 y2  1 at point P(x1, y1) on the curve a 2 b2

a 2x b2 y  = a2+ b2  a2e2 x1 y1 (ii)

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

(iii)

x 2 y2 ax by  2 1 is  = a2 + b2 = a2 e2. 2 a b sec  tan 

In general, four normals can be drawn to a hyperbola from any point and if , , ,  be the

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concentric angles of these four co-normal points, then        is an odd multiple of  . Drill Exercise - 7

x 2 y2 – = 1 at (–4, 0). 16 9

Find the equation of normal to the hyperbola

x2 y2 The normal to the hyperbola 2  2  1 drawn at an extremity of its latus rectum is parallel to an a b

d1  5i .

2 If the tangent and the normal to the rectangular hyperbola xy = c2, at a point, cuts off intercepts a1 and a2 on the x-axis and b1, b2 on the y-axis, then find the value of a1a2 + b1b2 .

asymptote. Show that the eccentricity is equal to the square root of

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

x 2 y2 If the normal at P to the hyperbola 2  2 =1 meets the transverse axis in G and conjugate axis in a b g and CF be perpendicular to the normal, from the centre then prove that PF.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 Hyperbola Chord of contact of tangents drawn from a point outside the hyperbola is T = 0 i.e., (xx1/a2) – (yy1/b2) = 1. Illustration 6:

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x 2 y2 x 2 y2   1 From any point on the hyperbola 2 tangents are drawn to the hyperbola 2  2  2 . a b2 a b Then show that the area cutoff by the chord of contact on the asymptotes is 4 ab Solution : Let P (x1, y1) be a point on the hyperbola

x 2 y2 x12 y12   1  1 . . Then, a 2 b2 a 2 b2

The chord of contact of tangent from P to the hyperbola

x 2 y2   2 is a 2 b2

xx1 yy1  2 2 a2 b

...(i)

The equations of the asymptotes are

x y x y   0 and   0 a b a b

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

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2a 2b 2a  2b ,y  ,x  ,y  x1 y1 1 x1 y1 2 x1 y1 2 x1 y1     a b a b a b a b     1 1  8ab   4ab Area of the triangle = | x1 y 2  x 2 y1 |  2 2  x12 y12   2 2 b  a

CHORD OF HYPERBOLA WITH SPECIFIED MIDPOINT Chord of hyperbola with specified midpoint (x1, y1) is T = S1 , where S1 and T have usual meanings.



x1 =

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

x 2 y2  1 16 25 Therefore, equation of the chord of this hyperbola in terms of the middle point (5, 3) is (T = S1 ) Equation of the given hyperbola can be written as

5x 3y 52 0  1    1  125x – 48y = 481 16 25 16 25 Illustration 8: Find the locus of the midpoints of the chords of the circle x2 + y2 = 16 which are tangents to the hyperbola. 9x2 – 16y2 = 144 Solution : Let (h, k) be the middle point of a chord of the circle x2 + y2 = 16 Then 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.,

x2 y2  1 16 9

at the point ( ,  ) say, then (i) is identical with Thus

x y  1 16 9

... (iii)

  1  2 = 16 h 9k h  k 2 

16h  9k  2 2 and h k h  k2 2

Since ( ,  ) lies on the hyperbola (ii),

... (ii)

1  16h  1  9k      1 16  h 2  k 2  9  h 2  k 2 

16h2 – 9k2 = (h2 + k2)2  Hence the required locus of (h, k) is (x2 + y2)2 =16x2 – 9y2. PAIR OF TANGENTS

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x 2 y2 Equation of pair of tangents from point (x1, y1) to the hyperbola 2  2 = 1 is a b 2  x 2 y 2   x12 y12   xx yy  1 1       1   1   1  SS1 = T i.e.,  2 = b 2   a 2 b 2   a 2 b2  a

DIRECTOR CIRCLE The locus of the point of intersection of two perpendicular tangents to a hyperbola is called its director circle. Its equation is x2 + y2 = a2 – b2.

10.

(a 2 m 2  b 2 ) ... (i)

Equation of any tangent to x2/a2– y2/ b2 = 1 is y = mx 







1 x  a 2 / m2  b 2 ... (ii) m Locus of point of intersection of these perpendicular tangents i.e., equation of the director circle can be 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.

Tangent perpendicular to (i) is y = –

Drill Exercise - 8 1.

Find the number of point(s) outside the hyperbola

x 2 y2   1 from where two perpendicular tangents 25 36

can be drawn to the hyperbola. 2.

Page 13 of 27

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

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Hyperbola

x2 y2   1 which pass through the If m1 and m2 are the slopes of the tangents to the hyperbola 25 16 point (6, 2) then find the value of m1m2 and m1 + m2.

x 2 y2 If the chord through the points (a sec, b tan) and (a sec, b tan) on the hyperbola 2  2 = 1 a b 1  e 1  e ,   passes through a focus, then prove that tan tan = 1  e 2 2  , 1  e

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x 2 y2 Tangents are drawn from any point on the hyperbola x – y = a + b to the hyperbola 2  2 = 1, a b then prove that they meet the axes in concyclic points. 2

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

Equation of the tangent at P (t) is

(ii)

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

(i)

(iv)

(v)

(iii)

x2 y2 Locus of the feet of the perpendicular drawn from focus of the hyperbola 2  2 = 1 upon a b 2 2 2 any tangent is its auxiliary circle i.e., x + y = a and the product of the feet of these perpendiculars is b2 . The portion of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus.

11.

for focus(ae,0)

for focus(ae,0)

x2 y2   1 upon a 2 b2 any tangent is its auxiliary circle i.e. x 2 + y2 = a2 and the product of the feet of these perpendiculars is b2. (semi C.A.)2. Locus of the feet of the perpendicular drawn from focus of the hyperbola

(vi)

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

(vii)

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

(viii)

The tangent at any point P on a hyperbola

x2 y2   1 with centre C, meets the a 2 b2

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

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Hyperbola

asymptotes and the portion of the tangent intercepted between the asymptote is bisected at 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 standard hyperbola 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 one focus 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 transverse axis, the product of the segments of this line intercepted between the point and the curve is always equal to the square on the semi conjugate axis.

(xiii)

.in

(xii)

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

(xi)

x 2 y2 If the angle between the asymptote of a hyperbola 2  2  1 is 2 , then the eccentricity a b of the hyperbola is sec  . If a circle intersects a rectangular hyperbola at four points, then the mean value of the points of intersection is the midpoint of the line joining the centres of circle and hyperbola.

(b) (c)

If a circle and the rectangular hyperbola xy = c2 meet in the four points t1 , t2, t3 & t4 , then (a) t1 t2 t3 t4 = 1 the centre of the mean position of the four points bisects the distance between the centre of the two curves. the centre of the circle through the points t1, t2 & t3 is

(xiv)

 c   c ,  ct1t 2 t 3  . If  ct i ,  = i = 1, 2, 3 be the angular points P, Q, R then orthocenter is  ti    ti t 2t3 

 c    1  c 1 1 1   t1  t 2  t 3   ,     t1 t 2 t 3   . t1 t 2 t 3  2  t1 t 2 t 3    2 

Drill Exercise - 9 1.

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

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

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

Show that the normal to the rectangular hyperbola xy = c2 at the point ‘t’ meets the curve again at the point ‘t1’ such that t1.t3 = –1. www.StudySteps.in

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Hyperbola

Answer Key Drill Exercise - 1

29 25 , (  5, 0),(  29,0), x   5 29

e

(- 1 , 0) 4.

13 3

4 3

x2  3y2 + 3 = 0

x 2 y2  1 4 12

Drill Exercise - 2 1.

15x2 – y2 = 15

x = 5 + 6 sec, y = –3 + 5 tan

2. pair of rays

Drill Exercise - 3

6x2 + 13xy + 6y2 – 38x – 37y – 98 = 0

3x2 – 5xy – 2y2 + 5x + 11y – 16 = 0

.in

3x + 4y + 5 = 0 and x + 2y + 3 = 0

(LR) = (TA) = (CA)= 12 2 ; e =

2 ; foci  (6 2 , 6 2 ) , ( 6 2 , 6 2 ) ;

Drill Exercise - 4

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

Drill Exercise - 5

point (–7, 3) lies inside ; (2, 7) lies outside

y = 3x + 5, y = 3x – 5

4, 3

x y3 3  0

17 20

Drill Exercise - 6 3.





3 4

y = 3x + 5

m6 - 2m = 0

Drill Exercise - 7 1.

y=0

Drill Exercise - 8 1.

Page 16 of 27

75x – 16y = 418

Drill Exercise - 9

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m1 + m2 = 24/11, m1m2 = 20/11

Hyperbola

1 2 b 2

3. (x2 + y2)2 + 4a2 (x2 – y2) = 4a4

5. 6x2 + 6y2 + 13xy – 38x – 37y – 98 = 0

SOLVED SUBJECTIVE EXAMPLES Example 1 : A tangent to the hyperbola

x 2 y2 x 2 y2   1  1 in points P and Q. Find the locus cuts the ellipse a 2 b2 a 2 b2

of the midpoint of PQ. Solution :

x 2 y2 Let M(x1, y1) be the midpoint of the chord PQ of the ellipse 2  2 1 . a b

2 2 This is tangent to the hyperbola x  y 1 a 2 b2 2

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2 2 2 2  2 2 Equation of PQ is xx21  yy21  x12  y12  y   b 2x1x  b  x12  y12  a b a b a y1 y1  a b 

2 2 2 b4  x 2 y2  b4x 2  2  if 2  12  12   a 2 4 12  b 2   x12  y12   x12  y12 a y1  a b  a y1 b  a b  2

2 2 2 2 Hence locus of (x1, y1) is  x  y   x  y  a 2 b2  a 2 b2  

Example 2 :

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x 2 y2  1 to meet it and the a 2 b2 conjugate hyperbola respectively in the point P and Q. Show that the normals at P and Q to the curves meet on the x-axis. Solution : A straight line is drawn parallel to the conjugate axis of the hyperbola

Let P(a sec  , b tan  ) be a point on the hyperbola, and Q(a tan  , b sec  ) be a point on the conjugate hyperbola. a sec  = atan   sec  = tan   Equation of the normal to the hyperbola a tan  x 2 y2  2 1 at P is y – b tan  =  (x – b sec  ) 2 b sec  a b Equation of the normal to the conjugate hyperbola at Q is

y – b sec  = –

a sec  ( x  a tan ) b tan 

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Hyperbola

Eliminate x and use sec  = tan  We get y (sec  – tan  ) = 0

 y=0 Hence the normals meet on the x-axis. Example 3 : x 2 y2  1 , GL is drawn perpendicular to a 2 b2 one of its asymptotes. Also Gp is a normal to the curve at P. Prove that LP is parallel to the conjugate axis. 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 From a point G on the transverse axis of the hyperbola

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 a 2  b2  2 2 a  b G  sec , 0  x=    sec   a  a The equation of line perpendicular to the asymptote bx – ay = 0 and passing through G, i.e., equation

 a a 2  b2 of GL is y = –  x  sec   b a 

ax + by = (a2 + b2) sec   Its 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 the point 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 at x (4x + c) = 1  4x2 + cx –1  x1 + x2 = –c/4 Also y (y – c) = 4  y2 – cy – 4 = 0  y1 + y2 = c Let the point which divides the line segment in the ratio 1 : 2 be (h, k)



x1  2 x 2 h 3

 x2 = 3h + c/4

 x1 = –c/2 – 3h

y1  2 y 2  k  y2 = 3k – c  y1 = –3k + 2c 3 Now (h, k) lies on the line y = 4x + c  k = 4h + c  c = k – 4h x = –k/2 + 2h – 3h = –h – k/2 and y1 = –3k + 2k – 8h = –k – 8h  1 2 2 16h + k + 10hk = 2. Hence locus of (h, k) is  16x2 + y2 + 10 xy = 2

Also

Example 5 : Prove that if normal to the hyperbola xy = c2 at point t meets the curve again at a point t1 then t3 t1 + 1 = 0.

Page 18 of 27

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Hyperbola

Solution : Equation of normal at point t i.e., (ct, c/t) is c y – xt2 = (1 – t4) t It meets the curve again at t1 then (ct1, c/t1) must satisfy (1)

 

... (1)

c c 1 1  ct1t 2  (1  t 4 )   t1t 2   t 3 t1 t t1 t 1 1 2   t ( t  t1 )  0  ( t  t1 ) (1 + t3t ) = 0 1 t1 t tt 1

Clearly t  t1  t3 t1 + 1 = 0.

... (i)

[ y  2a ( x  ) ]2  ( 2  4a) ( y 2  4ax ) [T2 = SS1] A = coefficient of x2 = 4a2 2H = coefficient of xy = –4 

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



2 H 2  AB 1 = tan45° = AB (A + B)2 = 4 (H2 –AB) 

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

(4a2 + 4 a )2 = 4[a2  2 – (4a)2 (4a  )]

  2  6a  a 2   2  0 or (  3a ) 2   2  8a 2 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 is y = mx + a/m which passes through (  ,  ) if

Page 19 of 27

 = m + a/m or m2  – m  + a = 0 If m1 and m2 are roots of (1). m1 + m2 =  /  and m1m2 = a /  we have m1  m 2 1 = tan45° = 1  m m 1 2 (1 + m1 m2)2 = (m1 – m2)2  (1 + m1m2)2 = (m1 + m2)2 – 4m1 m2 

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... (1)

Hyperbola



(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 of the 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

–

( x  1) 2

 6

 3

 3  6  2

1

.in





( y  2) 2

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 = 2 3 , and conjugate axis = 2 6 . Also b2 = a2 (e2 – 1)



6 = 3 (e2 – 1) i.e., e = 3  In (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 = 

3 / 3   1 or y = 3, y = 1

Example 8 : Find the equation and angle between the asymptotes of the hyperbola x2 + 2xy – 3y2 + x + 7y + 9 = 0 Solution : Let the combined equation of asymptotes x2 + 2xy – 3y2 + x + 7y +  = 0 If it represents pair of straight lines abc + 2fgh – af 2 – bg2 – ch2 = 0  = –23/16  Asymptotes x2 + 2xy – 3y2 + 7y – 23/16 = 0 Required angle =  tan–12. Example 9 : Prove that the locus of a point whose chord of contact touches the circle inscribed on the straight line joining the foci of the hyperbola x2/a2 – y2 / b2 = 1 as diameter is x2/a4 + y2 / b4 = 1/(a2 + b2). Solution :

Page 20 of 27

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Hyperbola

Circle on the join of foci (ae, 0) and (–ae, 0) diameter is (x – ae) (x + ae) + (y – 0) (y – 0) = 0 i.e., x2 + y2 = a2e2 = a2 + b2 ... (i) [ a2e2 = a2 + b2] Let chord of contact of P (x1, y1) touch the cirlce (i) Equation of chord of contact of P is [T = 0] xx1/a2 – yy1 /b2 = 1 i.e., b2x1x–a2 y1y – a2 b2 = 0 ... (ii) 

a 2b2 (b x  a y ) 4

2 1

  (a 2  b 2 )

Hence locus of P (x1, y1) is (b4x2 + a4y2) (a2 + b2) = a4b4 . Example 10 : 1 1  and one focus at the point P , 1 . Its one directrix is common 2 2  2 2 2 2 tangent to the circle x + y = 1 and the hyperbola x – y = 1, nearer to P. The equation of the ellipse in 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

Page 21 of 27

.in

An ellipse has eccentricity

common tangent at these point are x =  1. Since x = 1 is nearer to the focus P 1 ,1 , this is the 2  directrix of the required ellipse.

1 1   = a.e = a × ... (i) 2 2

Then CP =

1 Therefore, the major axis is parallel to the axis passing through the focus P ,1 . Hence the equa2  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) . 1  e   2 

and the distance of the directrix from the centre =



1–  =

a  2a e

a . e

x2 – y 2 = 1

... (ii) x2 + y2 = 1

1 1 and e  . 3 2 If b is the length of the semi minor axis of the ellipse, then b2 = a2(1 – e2)

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



b2 =

1 1 1 1    9  4  12 2

1  x   ( y  1) 2 3 Hence the required equation of the ellipse is   1 2 1 1   12 3

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x=1

Page 22 of 27

Hyperbola 2

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1 2  9  x   + 12(y – 1) = 1 3 

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Hyperbola

SOLVED OBJECTIVE EXAMPLES

Example 1 : The equation of a line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one of its 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 asymptote should 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.

3h  2 = 2 (given) 2k  3 3h = 4k + 4  Required locus is 3x – 4y = 4  Hence (a) is the correct answer.

Its slope is

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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 = S1 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 

Example 3 :

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x 2 y2  =1 meets one of the directrix in F. If PF subtends a 2 b2 an angle  at the corresponding focus, then  equals (a)  / 4 (b)  / 2 (c) 3 / 4 (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. The tangent at a point P on the hyperbola

Equation of tangent at P is

x sec  y tan  = 1. Let F be the intersection point of tangent of directrix,  a b

b(sec   e)    then F =  a / e, e tan   



mSF 

b(sec   e) b tan  , m PS  2 e tan  (a  1) a(sec   e)  mSF .mPS = –1

Hence (b) is the correct answer.

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Hyperbola

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

a 2 b 2 (a 2  b 2 ) 2   l 2 m2 n2

(b)

a 2 b 2 (a 2  b 2 ) 2 (c) 2  2  l m n

a 2 b 2 (a 2  b 2 ) 2   l 2 m2 n2

(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 a cos  b cot  (a 2  b 2 )  = l m n

.in

bl l (a 2  b 2 ) m (a 2  b 2 ) and cot    sin  = am  an  nb

cos  =

Thus

b 2l 2 l 2 (a 2  b 2 ) 2 a 2 b 2 (a 2  b 2 ) 2  1 + or, 2  2  a 2m2 a 2n 2 l m n2



Hence (b) is the correct answer. Example 5 :

If (a sec  , btan  ) and (asec  , btan  ) be the coordinate of the ends of a focal chord of

(a)

e 1 e 1

(c)

1 e 1 c

  x 2 y2  2 = 1, then tan tan equals to 2 2 2 a b

(b)

1 e 1 e

(d)

e 1 e 1

Solution :

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Equation of chord connecting the points (asec  , b tan  ) and (asec  , b tan  ) is x   y     cos    sin    cos   a  2  b  2   2 

    If it passes through (ae, 0); we have, ecos    cos    2   2 

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    cos  1  tan . tan   2   2 2   1 e    tan . tan  e=   1  tan tan 2 2 1 e cos   2 2  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, 2 , is given by s (a) (1, –3) (c) (–2, 4) Solution : x = t2 + 1, y = 2t  x – 1 = y2/2 x = 2s, y = 2/s  xy = 4

(b) (2, 2) (d) (1, 2)

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4 y2  1   y 3  4 y 16  0  y = 2  x = 2 For the point of intersection we have y 4

Hence (b) is the correct answer.

Example 7 :

( x  1) 2 ( y  5) 2  1 (a) 16 9

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

( x  1) 2 ( y  5) 2   1 16 9 Solution : S1  (6, 5); S2  (–4, 5), e = 5/4 S1S2 = 10  2ae = 10  a = 4 and

(c)

5 is 4

x 2 y2  1 (b) 16 9 (d) None of these

 25  b2 = a2 (e2 – 1) = 16   1  9  16 

Centre of the hyperbola is (1, 5)

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Equation of required hyperbola is

( x  1) 2 ( y  5) 2  1 16 9

Example 8 :

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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 :

 lx  my  n   (x –  )2 + (   ) 2 = k (lx + my + n)2 = k (l2 + m2)  2 2   l m  PS  k (l2 + m2)  PM If k(l2 + m2) = 1, P lies on parabola If k(l2 + m2) < 1, P lies on ellipse If k(l2 + m2) > 1, P lies on hyperbola If k = 0, P lies on a point circle Hence (b), (c), (d) are correct.

Example 9 :

(a) (6, 3) (c) (6, –3) Solution :

x 2 y2   1 which is nearest to the line 3x + 2y + 1 = 0 is 24 18 (b) (–6, 3) (d) (–6, –3)

Equation of tangent is ( 24 sec , 18 tan ) is

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The point on the hyperbola

3 2

so its slope = –

x sec  y tan  , then point is nearest to the line 3x + 2y + 1 = 0.  1 24 18

1 sec 18 3    sin  = – 3 2 24 tan  Hence the point is (6, –3) Hence (c) is the correct answer.

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Example 10 : The locus of a point, from where tangents to the rectangular hyperbola x2 – y2 = a2 contain an angle of 45° is (a) (x2 + y2) + a2(x2 – y2) = 4a4 (b) 2(x2 + y2) + 4a2 (x2 – y2) = 4a2 2 2 2 2 2 4 (c) (x + y ) + 4a (x – y ) = 4a (d) (x2 + y2) + a2 (x2 – y2) = a4 Solution :

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Let y = mx  m 2 a 2  a 2 be two tangent and passes through (h, k) then (k – mk)2 = m2a2– a2 m2(h2 – a2) – 2khm + k2 + a2 = 0  2kh k 2  a 2 , using tan45°= m1  m 2 and m m = 1 2 1  m1 m 2 h2  a2 h2  a 2 Hence (c) is the correct answer.

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m1 + m2 =

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