Mathematics. Session Hyperbola Session - 2 Session Objectives.
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Transcript of Mathematics. Session Hyperbola Session - 2 Session Objectives.
Mathematics
Session
HyperbolaSession - 2
Session Objectives
Session Objectives
1. Equation of chord joining two points on the hyperbola
2. Equation of chord whose mid-point is given
3. Equation of pair of tangents from an external point
4. Equation of chord of contact
5. Asymptotes of hyperbola
6. Rectangular hyperbola
7. Equation of rectangular hyperbola referred to its asymptotes as the axes of coordinates
8. Director circle
Equation of the Chord Joining Two
Points on the Hyperbola 12 2
2 2
x y
a b
The equation of the chord joining twopoints
on the hyperbola is
1 1 2 2P asec , b tan and Q asec , b tan
2 2
2 2
x y1
a b
2 1
1 12 1
b tan b tany b tan x asec
asec asecwhich reduces to
1 2 1 2 1 2x ycos sin cos
a 2 b 2 2
Equation of Chord whose Mid-Point is Given
Equation of chord of the hyperbola
whose middle point is
is given by
2 2
2 2
x y1
a b 1 1x , y
2 2
1 1 1 12 2 2 2
xx yy x y1 1
a b a b
1 12 2
xx yyi.e. , where T 1
a b1T = S
2 21 1
1 2 2
x yS 1
a b
Equation of Pair of Tangents Fromand External Point
Equation of pair of tangents from the
point to the hyperbola
is
1 1x , y 2 2
2 2
x y1
a b
22 22 21 1 1 1
2 2 2 2 2 2
x y xx yyx y1 1 1
a b a b a b
i.e. , where 21SS T
2 22 2
1 1 1 112 2 2 2 2 2
xx yy x yx yT 1, S 1, S 1
a b a b a b
Equation Chord of Contact
Equation of chord of contact of point with respect to the hyperbola
is , i.e. T = 0
1 1x , y
2 2
2 2
x y1
a b 1 1
2 2
xx yy1 0
a b
Asymptotes of Hyperbola
An asymptote is a straight line, whichmeets the conic in two points both ofwhich are situated at an infinite distance,but which is itself not altogether(entirely) at infinity.
To Find the Equation of the Asymptotes of the Hyperbola
2 2
2 2
x y= 1
a b
Let the straight line
y = mx + c ... (i)
meet the given hyperbola in points,whose abscissae are given by theequation
22
2 2
mx cx1
a b
or ... (ii) 2 2 2 2 2 2 2 2x b a m 2a mcx a c b 0
To Find the Equation of the Asymptotes of the Hyperbola
2 2
2 2
x y= 1
a b
If the straight line (i) be an asymptote,both roots of equation (ii) must be infinite.
Hence, the coefficients of x2 and x in theequation (ii) must be zero.
We have 2 2 2 2b a m 0 and a mc 0
Hence, and c = 0b
ma
To Find the Equation of the Asymptotes of the Hyperbola
2 2
2 2
x y= 1
a b
Substituting the values of m
and c in
y = mx + c, we get
b
y xa
, i.e.
x y x y
0 and 0a b a b
The combined equation of the
asymptotes is2 2
2 2
x y0
a b
X
Y´
X´
Y
O xa
2
2— – — = 1y
b
2
2
Asymptotes
xa
2
2— – — = – 1y
b
2
2
Asymptotes
Points to Remember
i. A hyperbola and its conjugate hyperbolahave the same asymptotes.
ii. The equation of the pair of asymptotesdiffer the hyperbola and the conjugatehyperbola by the same constant only, i.e
Hyperbola – Asymptotes = Asymptotes – Conjugate Hyperbola
2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
x y x y x y x y1 1
a b a b a b a b
iii. The asymptotes pass through the centre of hyperbola.
iv. The bisectors of the angles between the asymptotes of the
hyperbola are the coordinate axes (or axes of the
hyperbola).
2 2
2 2
x y1
a b
Points to Remember
v. As we know that combined equation of
asymptotes is and equation
of hyperbola is
2 2
2 2
x y0
a b
2 2
2 2
x y1.
a b
Equation of pair of asymptotes and equation of hyperbola differ by a constant only. (Important)
Rectangular Hyperbola or Equilateral Hyperbola
A hyperbola whose asymptotes are atright angles to each other is called arectangular hyperbola.
The equations of asymptotes of the hyperbola
are given by 2 2
2 2
x y1
a b
by x.
a
The angle between two asymptotes is given by
2 2 2
2
b b 2b2aba a atan
b b b a b1 1a a a
Rectangular Hyperbola or Equilateral Hyperbola
If the asymptotes are at right angles,
then 2 2tan a b 0
2
a b 2a 2b
Thus, the transverse and conjugate axesof a rectangular hyperbola are equal.
2 2 2The equation of rectangular hyperbola is x y a .
The equation of the asymptotes of the rectangular
hyperbola are y x.
Cor: Eccentricity of rectangular hyperbola is 2.
Equation of the Rectangular Hyperbola Referred to its Asymptotes as the Axes of Coordinates
Referred to the transverse and conjugateaxes as the axes of coordinates, theequation of the rectangular hyperbola is
2 2 2x y a ... (i)
The equation of asymptotes of the hyperbola (i) is
Each of these two asymptotes is inclined at an angle
of with the transverse axis.
y x.
4
Equation of the Rectangular Hyperbola Referred to its Asymptotes as the Axes of Coordinates
Now rotating the axes through an angle in
clockwise direction, keeping the origin fixed,then the axes coincide with the asymptotes
of the hyperbola and
4
X Yx Xcos Y sin
4 4 2
Y Xand y Xsin Ycos
4 4 2
Putting the values of x and y in (i), we get
2 22X Y Y X
a2 2
Equation of the Rectangular Hyperbola Referred to its Asymptotes as the Axes of Coordinates
2a
XY2
2
2 2 aXY c where c
2
X
Y´
X´
Y
O
This is the transformed equationof rectangular hyperbola (i).
Thus, equation of rectangularhyperbola when its asymptotestaken as coordinate axes is
2xy c .
Cor: If equation of a rectangularhyperbola be thenequation of its conjugatehyperbola will be
2xy c ,
2xy c .
Parametric Form of RectangularHyperbola xy = c2
If ‘t’ is non-zero variable, the coordinates ofany point on the rectangular hyperbola
xy = c2 can be written
cct, .
t
The point is also called point ‘t’.
cct,
t
Equation of Chord Joining Points ‘t1’ and ‘t2’
The equation of the chord joining two
points and of hyperbola
xy = c2 is
11
cct ,
t
22
cct ,
t
2 11
1 2 1
c ct tc
y x ctt ct ct
1 2 1 2x t t y c t t
This is the required equation of chord.
Equation of Tangent in Different Forms
(i) Equation of tangent in point formof the hyperbola xy = c2
21 1xy x y 2c
(ii) Equation of tangent in parametric form
xyt 2c
t
Equation Normal in Different Forms
(i) Equation of normal in point form
2 21 1 1 1xx yy x y
Equation Normal in Parametric Forms
(i) Equation of normal in parametricform
3 4xt yt ct c 0
Note: The equation of normal at is a
fourth degree equation in t. Therefore, in general four normal can be drawn from a point to the hyperbola xy = c2.
cct,
t
Point of Intersection of Tangents at ‘t1’ and ‘t2’ to the Hyperbola xy = c2
The equations to the tangents at thepoints ‘t1’ and ‘t2’ are
1 21 2
x xyt 2c and y t 2c
t t
By solving these equations, we get point ofintersection of tangents.
Coordinates of point of intersection of tangents at ‘t1’ and ‘t2’ is .
1 2
1 2 1 2
2ct t 2c,
t t t t
Director Circle
The locus of intersection of tangentswhich are at right angles is calleddirector circle of Hyperbola.
To find the locus of the point of intersectionof tangents which meet at right angles.
2 2
2 2
x yLet equation of hyperbola is 1 ...(i)
a b
2 2 2Any tangent to the hyperbola is y mx a m b ...(ii)
22 21 1
and its perpendicular tangent is y x a b ... (iii)m m
Director Circle
Let (h, k) be their point of intersection.We have
2 2 2k mh a m b ... (iv)
2 2 2and mk h a b m ... (v)
[By putting the value of (h, k) in equations (iii) and (iv)]
If between (iv) and (v), we eliminate m, we shall have a relation between h and k, i.e. locus of (h, k).
Squaring and adding these equations, we get
2 2 2 2 2 2 2 2k mh mk h a m b a b m
Director Circle
2 2 2 2 2 2k h 1 m a b 1 m
2 2 2 2h k a b
Locus of (h, k) is 2 2 2 2x y a b
This is the equation of director circle, whose centre
is origin and radius is 2 2a b .
Class Test
Class Exercise - 1
If the chord through the points
on the hyperbola passes
through a focus, prove that
asec , b tan and a sec , b tan
2 2
2 2
x y1
a b
1 etan tan .
2 2 1 e
Solution
The equation of the chord joining is
x ycos sin cos
a 2 b 2 2
If it passes through the focus (ae, 0), then
ecos cos2 2
cos12e
cos2
By componendo and dividendo,
cos cos1 e2 21 e
cos cos2 2
1 e
cot cot2 2 1 e
1 e
tan tan2 2 1 e [Proved]
Class Exercise - 2
Chords of the hyperbola touch the parabola Prove thatthe locus of their middle points is thecurve
2 2 2x y a2y 4ax.
2 3y x a x .
Solution
If (h, k) be the mid-point of the chord,then the equation of the chord is T= S1,
2 2 2 2i.e. xh yk a h k a
2 2i.e. xh yk h k
2 2k hhy x = mx + c [Say]
k k
If it touches the parabola y2 = 4ax, then
a
cm
[Condition for tangency for any line y = mx + c to the parabola]
Solution contd..
2 2 2ak h k h
Locus of (h, k) is 2 2 2ay x y x
2 3 i.e. y x a x .
2 2k h akk h
Class Exercise - 3
Find the point of intersection oftangents drawn to the hyperbola
at the points where it
is intersected by the linelx + my + n = 0.
2 2
2 2
x y1
a b
Solution
Let (h, k) be the required point.
Equation of chord of contact drawnfrom (h, k) to the hyperbola is
T = 0
2 2
xh yki.e. 1 0 ... (i)
a b
The given line islx + my + n = 0 ... (ii)
Equations (i) and (ii) represent same line
2 2
h k 1na l b m
Solution contd..
2a l
hn
2b m
kn
Coordinates of the required point
2 2a l b m,
m n
Class Exercise - 4
Prove that the product of theperpendiculars from any point
on the hyperbola to its
asymptotes is equal to
2 2
2 2
x y1
a b
2 2
2 2
a b.
a b
Solution
Let be any point on
the hyperbola
a sec , b tan
2 2
2 2
x y1.
a b
The equation of the asymptotes of the
given hyperbola are x y x y
0 and 0a b a b
1p Length of perpendicular from
x y
a sec , b tan on 0a b
2 2
sec tan
1 1
a b
Solution contd…
2p Length of perpendicular from
a sec , b tan x y
on 0a b
2 2
sec tan
1 1
a b
2 2 2 2
1 2 2 2
2 2
sec tan a bp p Pr oved
1 1 a ba b
Class Exercise - 5
The asymptotes of a hyperbola areparallel to lines 2x + 3y = 0 and3x + 2y = 0. The hyperbola hasits centre at (1, 2) and it passesthrough (5, 3). Find its equation.
Solution
Asymptotes are parallel to lines2x + 3y = 0 and 3x + 2y = 0
Equations of asymptotes are
2x + 3y + k1 = 0 and 3x + 2y + k2 = 0
As we know that asymptotes passes through the centre of the hyperbola.Here centre of hyperbola is (1, 2).
1 22 6 k 0 and 3 4 k 0
1 2k 8, k 7
Solution contd..
The equations of asymptotes are
2x + 3y – 8 = 0 and 3x + 2y – 7 = 0
Equation of hyperbola is
(2x + 3y – 8) (3x + 2y – 7) + c = 0
It passes through (5, 3).
10 9 8 15 6 7 c 0 c 154
Equation of hyperbola is (2x + 3y – 8)(3x + 2y – 7) – 154 = 0
i.e. 6x2 + 13xy + 6y2 – 38x – 37y – 98 = 0
Class Exercise - 6
The chord PP´ of a rectangularhyperbola meets asymptotes inQ and Q´. Show QP = P´Q´.
Solution
Let equation of rectangular hyperbolais xy = c2.
1
1
c cCoordinates of P ct, and P´ ct ,
t t
Equation of chord PP´ is 1 1x y tt c t t 0
It meets asymptotes, i.e. axes at Q and Q´respectively.
1Putting y = 0, Q c t t , 0
1
1 1and putting x = 0, Q´ 0, c
t t
Solution contd..
22
1
cNow PQ ct ct ct
t
21 2
1c t
t
2
2 21
1 1
c c cP´Q´ c t
t t t 21 2
1c t
t
PQ P´Q´ [Proved]
Class Exercise - 7
The normal at the three points P, Q, Ron a rectangular hyperbola, intersectat a point S on the curve. Prove thatcentre of the hyperbola is the centroidof PQR.
Solution
Let equation of the rectangular hyperbolais xy = c2.
Let ‘t’ be the parameter of any of points P, Q, R so that normal is ... (i)
It passes through a point S on the hyperbola.
Let coordinates of point
cS ct́ ,
t́
3 4cFrom i ct́ t .t ct c 0
t́
3 3tt́ t 1 t t́ 1 0
t́
Solution contd..
3 tt́ t 1 1 0
t́
3t t́ 1 0 (Remember this result)
This is a cubic equation in t, and gives us theparameters of the three points P, Q, R, say 1 2 3t , t , t .
2
1 2 3t t t 0 Coeff of t 0
1 2
1 2 3 1 2 3
t t1 1 1and 0 Coeff of t 0
t t t t t t
If (h, k) is the centroid of ,PQR
1 2 3c t t th 0
3
Solution contd..
1 2 3
1 1 1c
t t tk 0
3
Hence, centroid is (0, 0) which is centre of the hyperbola.
Class Exercise - 8
A rectangular hyperbola whose centreis C is cut by any circle of radius r infour points P, Q, R and S. Prove that
2 2 2 2 2CP CQ CR CS 4r .
Solution
Let the equation of rectangular hyperbola is xy = k2 ... (i)
and equation of circle is
2 2x y 2gx 2fy c 0 ... (ii)
where ... (iii) 2 2 2g f c r
Eliminating y between equations (i) and (ii), we get
4 2
22
k kx 2gx 2f c 0
xx
4 3 2 2 4x 2gx cx 2fk x k 0
Solution contd..
This is biquadratic equation in x, whichgives us the abscissae of the four pointsof intersection.
Let x1, x2, x3, x4 are the roots of the equation.
1 2 3 4 1 2x x x x 2g, x x c
22 2 2 21 2 3 4 1 2 3 4 1 2x x x x x x x x 2 x x
24g 2c
Similarly, eliminating x from (i) and (ii), we get
2 2 2 2 21 2 3 4y y y y 4f 2c
Solution contd..
2 2 2 2CP CQ CR CS
2 2 2 2 2 2 2 21 1 2 2 3 3 4 4x y x y x y x y
2 21 1x y
2 24 g f c
24r [Pr oved]
Class Exercise - 9
Prove that the locus of the mid-pointsof the chords of the hyperbola which pass through a fixed
point is a hyperbola whose
centre is
2 2
2 2
x y1
a b
,
, .2 2
Solution
Let (h, k) be the coordinates of mid-pointof the chord.
Equation of chord is T = S1
2 2
2 2 2 2
hx ky h k1 1
a b a b
2 2
2 2 2 2
hx ky h k
a b a b
It passes through a fixed point . ,
2 2
2 2 2 2
h k h k
a b a b
Solution contd..
Locus of (h, k) is
2 2
2 2
x x y y0
a b
2 2
2 22
2 2 4 4
x y12 2
c say4a b a b
This is an equation of hyperbola whose centre is
, [Pr oved]2 2
Class Exercise - 10
From a point, tangents to therectangular hyperbola are drawn and they intersect eachother at an angle of 45o. Prove thatthe locus of the point is the curve
2 2 2x y a
22 2 2 2 2 4x y 4a x y 4a .
Solution
Here equation of rectangular hyperbola is
2 2 2x y a ... (i)
Equation of any tangent to this hyperbola is
2 2 2y mx a m a
If it is passes through (h, k), then 2 2 2 2k mh a m a
2 2 2 2 2m h a 2mkh k a 0
Let m1, m2 be the roots of the above equation, which gives the slopes of two tangents passing through (h, k).
Solution contd..
2 2
1 2 1 22 2 2 2
2kh k am m and m m
h a h a
Given that angle between two tangents are 45o.
1 2
1 2
m mtan45
1 m m 2 2
1 2 1 2m m 1 m m
2 2
1 2 1 2 1 2m m 4m m 1 m m
22 22 2 2 2
2 2 22 22 2
4 k a4h k k a1
h ah ah a
22 2 2 2 2 2 2 24 h k h a k a h k
Locus of (h, k) is 22 2 2 2 2 4x y 4a x y 4a [Pr oved]
Thank you