Detailed Notes: Hyperbola | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


HYPERBOLA 
Hyperbolas are fascinating conic sections defined by their eccentricity greater than one. This section 
explores the fundamental properties and equations of hyperbolas, including their foci, vertices, axes, 
and directrices. We'll examine the standard forms of hyperbola equations, key relationships between 
their components, and special cases like conjugate and rectangular hyperbolas. Through a mix of 
theoretical explanations and practical problems, this material provides a comprehensive overview of 
hyperbolas, their unique characteristics, and their applications in mathematics and related fields. The 
Hyperbola is a conic whose eccentricity is greater than unity. (e > 1 ). 
1. STANDARD EQUATION & DEFINITION(S): 
Standard equation of the hyperbola is 
?? 2
?? 2
-
?? 2
?? 2
= 1, where ?? 2
= ?? 2
( ?? 2
- 1) 
or ?? 2
?? 2
= ?? 2
+ ?? 2
 i.e. ?? 2
= 1 +
?? 2
?? 2
 
= 1 + (
 ?????????????????? ???????? 
 ???????????????????? ???????? 
)
2
 
 
(a) Foci: 
?? = ( ???? , 0) &?? '
= ( -???? , 0) . 
(b) Equations of directrices: 
?? =
?? ?? &?? = -
?? ?? 
(c) Vertices: 
?? = ( ?? , 0) &?? '
= ( -?? , 0) . 
(d) Latus rectum: 
(i) Equation: ?? = ±???? 
Page 2


HYPERBOLA 
Hyperbolas are fascinating conic sections defined by their eccentricity greater than one. This section 
explores the fundamental properties and equations of hyperbolas, including their foci, vertices, axes, 
and directrices. We'll examine the standard forms of hyperbola equations, key relationships between 
their components, and special cases like conjugate and rectangular hyperbolas. Through a mix of 
theoretical explanations and practical problems, this material provides a comprehensive overview of 
hyperbolas, their unique characteristics, and their applications in mathematics and related fields. The 
Hyperbola is a conic whose eccentricity is greater than unity. (e > 1 ). 
1. STANDARD EQUATION & DEFINITION(S): 
Standard equation of the hyperbola is 
?? 2
?? 2
-
?? 2
?? 2
= 1, where ?? 2
= ?? 2
( ?? 2
- 1) 
or ?? 2
?? 2
= ?? 2
+ ?? 2
 i.e. ?? 2
= 1 +
?? 2
?? 2
 
= 1 + (
 ?????????????????? ???????? 
 ???????????????????? ???????? 
)
2
 
 
(a) Foci: 
?? = ( ???? , 0) &?? '
= ( -???? , 0) . 
(b) Equations of directrices: 
?? =
?? ?? &?? = -
?? ?? 
(c) Vertices: 
?? = ( ?? , 0) &?? '
= ( -?? , 0) . 
(d) Latus rectum: 
(i) Equation: ?? = ±???? 
 
(iii) Ends: ???? ,
?? 2
?? ) , (???? ,
-?? 2
?? ); (-ae, 
?? 2
?? ) , (-???? ,
-?? 2
?? ) 
(e) (i) Transverse Axis: 
The line segment ?? '
?? of length 2a in which the foci ?? '
 & ?? both lie is called the 
Transverse Axis of the Hyperbola. 
(ii) Conjugate Axis: 
The line segment ?? '
?? between the two points ?? '
= ( 0, -?? ) &?? = ( 0, ?? ) is called as the Conjugate Axis of 
the Hyperbola. 
The Transverse Axis & the Conjugate Axis of the hyperbola are together called the 
Principal axes of the hyperbola. 
(f) Focal Property: 
The difference of the focal distances of any point on the hyperbola is constant and equal to transverse 
axis i.e. || ???? | - |?? ?? '
|| = 2?? . The distance ?? ?? '
= focal length. 
(g) Focal distance: 
Distance of any point ?? ( ?? , ?? ) on Hyperbola from foci ???? = ???? - ?? &????
'
= ???? + ?? . 
Problem 1: Find the equation of the hyperbola whose directrix is 2?? + ?? = 1, focus ( 1,2) and 
eccentricity v3. 
Solution: Let ?? ( ?? , ?? ) be any point on the hyperbola and ???? is perpendicular from ?? on the directrix. 
Then by definition  ???? = ?? PM 
  ? ( ???? )
2
= ?? 2
( ???? )
2
? ( ?? - 1)
2
+ ( ?? - 2)
2
= 3 {
2?? + ?? - 1
v4 + 1
}
2
   ? 5( ?? 2
+ ?? 2
- 2?? - 4?? + 5)
= 3( 4?? 2
+ ?? 2
+ 1 + 4???? - 2?? - 4?? )   ? 7?? 2
- 2?? 2
+ 12???? - 2?? + 14?? - 22 = 0  
which is the required hyperbola. 
Problem 2: The eccentricity of the hyperbola 4?? 2
- 9?? 2
- 8?? = 32 is - 
(A) 
v5
3
 
(B) 
v13
3
 
(C) 
v13
2
 
(D) 
3
2
 
Solution:  4?? 2
- 9?? 2
- 8?? = 32 ? 4( ?? - 1)
2
- 9?? 2
= 36 ?
( ?? -1)
2
9
-
?? 2
4
= 1 
Page 3


HYPERBOLA 
Hyperbolas are fascinating conic sections defined by their eccentricity greater than one. This section 
explores the fundamental properties and equations of hyperbolas, including their foci, vertices, axes, 
and directrices. We'll examine the standard forms of hyperbola equations, key relationships between 
their components, and special cases like conjugate and rectangular hyperbolas. Through a mix of 
theoretical explanations and practical problems, this material provides a comprehensive overview of 
hyperbolas, their unique characteristics, and their applications in mathematics and related fields. The 
Hyperbola is a conic whose eccentricity is greater than unity. (e > 1 ). 
1. STANDARD EQUATION & DEFINITION(S): 
Standard equation of the hyperbola is 
?? 2
?? 2
-
?? 2
?? 2
= 1, where ?? 2
= ?? 2
( ?? 2
- 1) 
or ?? 2
?? 2
= ?? 2
+ ?? 2
 i.e. ?? 2
= 1 +
?? 2
?? 2
 
= 1 + (
 ?????????????????? ???????? 
 ???????????????????? ???????? 
)
2
 
 
(a) Foci: 
?? = ( ???? , 0) &?? '
= ( -???? , 0) . 
(b) Equations of directrices: 
?? =
?? ?? &?? = -
?? ?? 
(c) Vertices: 
?? = ( ?? , 0) &?? '
= ( -?? , 0) . 
(d) Latus rectum: 
(i) Equation: ?? = ±???? 
 
(iii) Ends: ???? ,
?? 2
?? ) , (???? ,
-?? 2
?? ); (-ae, 
?? 2
?? ) , (-???? ,
-?? 2
?? ) 
(e) (i) Transverse Axis: 
The line segment ?? '
?? of length 2a in which the foci ?? '
 & ?? both lie is called the 
Transverse Axis of the Hyperbola. 
(ii) Conjugate Axis: 
The line segment ?? '
?? between the two points ?? '
= ( 0, -?? ) &?? = ( 0, ?? ) is called as the Conjugate Axis of 
the Hyperbola. 
The Transverse Axis & the Conjugate Axis of the hyperbola are together called the 
Principal axes of the hyperbola. 
(f) Focal Property: 
The difference of the focal distances of any point on the hyperbola is constant and equal to transverse 
axis i.e. || ???? | - |?? ?? '
|| = 2?? . The distance ?? ?? '
= focal length. 
(g) Focal distance: 
Distance of any point ?? ( ?? , ?? ) on Hyperbola from foci ???? = ???? - ?? &????
'
= ???? + ?? . 
Problem 1: Find the equation of the hyperbola whose directrix is 2?? + ?? = 1, focus ( 1,2) and 
eccentricity v3. 
Solution: Let ?? ( ?? , ?? ) be any point on the hyperbola and ???? is perpendicular from ?? on the directrix. 
Then by definition  ???? = ?? PM 
  ? ( ???? )
2
= ?? 2
( ???? )
2
? ( ?? - 1)
2
+ ( ?? - 2)
2
= 3 {
2?? + ?? - 1
v4 + 1
}
2
   ? 5( ?? 2
+ ?? 2
- 2?? - 4?? + 5)
= 3( 4?? 2
+ ?? 2
+ 1 + 4???? - 2?? - 4?? )   ? 7?? 2
- 2?? 2
+ 12???? - 2?? + 14?? - 22 = 0  
which is the required hyperbola. 
Problem 2: The eccentricity of the hyperbola 4?? 2
- 9?? 2
- 8?? = 32 is - 
(A) 
v5
3
 
(B) 
v13
3
 
(C) 
v13
2
 
(D) 
3
2
 
Solution:  4?? 2
- 9?? 2
- 8?? = 32 ? 4( ?? - 1)
2
- 9?? 2
= 36 ?
( ?? -1)
2
9
-
?? 2
4
= 1 
Here ?? 2
= 9, ?? 2
= 4 
?  eccentricity ?? = v1 +
?? 2
?? 2
= v1 +
4
9
=
v13
3
 
Problem 3: If foci of a hyperbola are foci of the ellipse 
?? 2
25
+
?? 2
9
= 1. If the eccentricity of the hyperbola 
be 2 , then its equation is - 
(A) 
?? 2
4
-
?? 2
12
= 1 
(B) 
?? 2
12
-
?? 2
4
= 1 
(C) 
?? 2
12
+
?? 2
4
= 1 
(D) none of these 
Solution:   For ellipse ?? =
4
5
, so foci = ( ±4,0) 
For hyperbola ?? = 2, so ?? =
????
?? =
4
2
= 2, ?? = 2v4 - 1 = 2v3 
Hence equation of the hyperbola is 
?? 2
4
-
?? 2
12
= 1 
Problem 4: Find the coordinates of foci, the eccentricity and latus-rectum, equations of directrices for 
the hyperbola 9?? 2
- 16?? 2
- 72?? + 96?? - 144 = 0. 
Solution: 
Equation can be rewritten as 
( ?? -4)
2
4
2
-
( ?? -3)
2
3
2
= 1 so ?? = 4, ?? = 3 
?? 2
= ?? 2
( ?? 2
- 1) ?????????? ?? =
5
4
 
Foci: ?? = ± ae, ?? = 0 gives the foci as ( 9,3) , ( -1,3) 
Centre: ?? = 0, ?? = 0 i.e. ( 4,3) 
Directrices: ?? = ±
?? ?? i.e. ?? - 4 = ±
16
5
 ?  directrices are 5?? - 36 = 0; 5?? - 4 = 0 
Latus-rectum =
2 ?? 2
?? = 2 ·
9
4
=
9
2
 
Do yourself - 1: 
(i) Find the eccentricity of the hyperbola 
?? 2
?? 2
-
?? 2
?? 2
= 1 which passes through ( 4,0) &( 3v2, 2) 
(ii) Find the equation to the hyperbola, whose eccentricity is 
5
4
, focus is (a, 0 ) and whose directrix is 
4?? - 3?? = ?? . 
(iii) In the hyperbola 4?? 2
- 9?? 2
= 36, find length of the axes, the co-ordinates of the foci, the 
eccentricity, and the latus rectum. 
Page 4


HYPERBOLA 
Hyperbolas are fascinating conic sections defined by their eccentricity greater than one. This section 
explores the fundamental properties and equations of hyperbolas, including their foci, vertices, axes, 
and directrices. We'll examine the standard forms of hyperbola equations, key relationships between 
their components, and special cases like conjugate and rectangular hyperbolas. Through a mix of 
theoretical explanations and practical problems, this material provides a comprehensive overview of 
hyperbolas, their unique characteristics, and their applications in mathematics and related fields. The 
Hyperbola is a conic whose eccentricity is greater than unity. (e > 1 ). 
1. STANDARD EQUATION & DEFINITION(S): 
Standard equation of the hyperbola is 
?? 2
?? 2
-
?? 2
?? 2
= 1, where ?? 2
= ?? 2
( ?? 2
- 1) 
or ?? 2
?? 2
= ?? 2
+ ?? 2
 i.e. ?? 2
= 1 +
?? 2
?? 2
 
= 1 + (
 ?????????????????? ???????? 
 ???????????????????? ???????? 
)
2
 
 
(a) Foci: 
?? = ( ???? , 0) &?? '
= ( -???? , 0) . 
(b) Equations of directrices: 
?? =
?? ?? &?? = -
?? ?? 
(c) Vertices: 
?? = ( ?? , 0) &?? '
= ( -?? , 0) . 
(d) Latus rectum: 
(i) Equation: ?? = ±???? 
 
(iii) Ends: ???? ,
?? 2
?? ) , (???? ,
-?? 2
?? ); (-ae, 
?? 2
?? ) , (-???? ,
-?? 2
?? ) 
(e) (i) Transverse Axis: 
The line segment ?? '
?? of length 2a in which the foci ?? '
 & ?? both lie is called the 
Transverse Axis of the Hyperbola. 
(ii) Conjugate Axis: 
The line segment ?? '
?? between the two points ?? '
= ( 0, -?? ) &?? = ( 0, ?? ) is called as the Conjugate Axis of 
the Hyperbola. 
The Transverse Axis & the Conjugate Axis of the hyperbola are together called the 
Principal axes of the hyperbola. 
(f) Focal Property: 
The difference of the focal distances of any point on the hyperbola is constant and equal to transverse 
axis i.e. || ???? | - |?? ?? '
|| = 2?? . The distance ?? ?? '
= focal length. 
(g) Focal distance: 
Distance of any point ?? ( ?? , ?? ) on Hyperbola from foci ???? = ???? - ?? &????
'
= ???? + ?? . 
Problem 1: Find the equation of the hyperbola whose directrix is 2?? + ?? = 1, focus ( 1,2) and 
eccentricity v3. 
Solution: Let ?? ( ?? , ?? ) be any point on the hyperbola and ???? is perpendicular from ?? on the directrix. 
Then by definition  ???? = ?? PM 
  ? ( ???? )
2
= ?? 2
( ???? )
2
? ( ?? - 1)
2
+ ( ?? - 2)
2
= 3 {
2?? + ?? - 1
v4 + 1
}
2
   ? 5( ?? 2
+ ?? 2
- 2?? - 4?? + 5)
= 3( 4?? 2
+ ?? 2
+ 1 + 4???? - 2?? - 4?? )   ? 7?? 2
- 2?? 2
+ 12???? - 2?? + 14?? - 22 = 0  
which is the required hyperbola. 
Problem 2: The eccentricity of the hyperbola 4?? 2
- 9?? 2
- 8?? = 32 is - 
(A) 
v5
3
 
(B) 
v13
3
 
(C) 
v13
2
 
(D) 
3
2
 
Solution:  4?? 2
- 9?? 2
- 8?? = 32 ? 4( ?? - 1)
2
- 9?? 2
= 36 ?
( ?? -1)
2
9
-
?? 2
4
= 1 
Here ?? 2
= 9, ?? 2
= 4 
?  eccentricity ?? = v1 +
?? 2
?? 2
= v1 +
4
9
=
v13
3
 
Problem 3: If foci of a hyperbola are foci of the ellipse 
?? 2
25
+
?? 2
9
= 1. If the eccentricity of the hyperbola 
be 2 , then its equation is - 
(A) 
?? 2
4
-
?? 2
12
= 1 
(B) 
?? 2
12
-
?? 2
4
= 1 
(C) 
?? 2
12
+
?? 2
4
= 1 
(D) none of these 
Solution:   For ellipse ?? =
4
5
, so foci = ( ±4,0) 
For hyperbola ?? = 2, so ?? =
????
?? =
4
2
= 2, ?? = 2v4 - 1 = 2v3 
Hence equation of the hyperbola is 
?? 2
4
-
?? 2
12
= 1 
Problem 4: Find the coordinates of foci, the eccentricity and latus-rectum, equations of directrices for 
the hyperbola 9?? 2
- 16?? 2
- 72?? + 96?? - 144 = 0. 
Solution: 
Equation can be rewritten as 
( ?? -4)
2
4
2
-
( ?? -3)
2
3
2
= 1 so ?? = 4, ?? = 3 
?? 2
= ?? 2
( ?? 2
- 1) ?????????? ?? =
5
4
 
Foci: ?? = ± ae, ?? = 0 gives the foci as ( 9,3) , ( -1,3) 
Centre: ?? = 0, ?? = 0 i.e. ( 4,3) 
Directrices: ?? = ±
?? ?? i.e. ?? - 4 = ±
16
5
 ?  directrices are 5?? - 36 = 0; 5?? - 4 = 0 
Latus-rectum =
2 ?? 2
?? = 2 ·
9
4
=
9
2
 
Do yourself - 1: 
(i) Find the eccentricity of the hyperbola 
?? 2
?? 2
-
?? 2
?? 2
= 1 which passes through ( 4,0) &( 3v2, 2) 
(ii) Find the equation to the hyperbola, whose eccentricity is 
5
4
, focus is (a, 0 ) and whose directrix is 
4?? - 3?? = ?? . 
(iii) In the hyperbola 4?? 2
- 9?? 2
= 36, find length of the axes, the co-ordinates of the foci, the 
eccentricity, and the latus rectum. 
(iv) Find the equation to the hyperbola, the distance between whose foci is 16 and whose eccentricity is 
v2. 
2. CONJUGATE HYPERBOLA: 
Two hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate 
& the transverse axes of the other are called Conjugate Hyperbolas of each other. eg. 
?? 2
?? 2
-
?? 2
?? 2
= 1 & -
?? 2
?? 2
+
?? 2
?? 2
= 1 are conjugate hyperbolas of each other . 
Note that: 
(i) If ?? 1
&?? 2
 are the eccentricities of the hyperbola & its conjugate then ?? 1
 
-2
+ ?? 2
 
-2
= 1. 
(ii) The foci of a hyperbola and its conjugate are concyclic and form the vertices of a square. 
(iii) Two hyperbolas are said to be similar if they have the same eccentricity. 
Problem 5: The eccentricity of the conjugate hyperbola to the hyperbola ?? 2
- 3?? 2
= 1 is- 
(A) 2 
(B) 2/v3 
(C) 4 
(D) 4/3 
Solution: Equation of the conjugate hyperbola to the hyperbola ?? 2
- 3?? 2
= 1 is 
-?? 2
+ 3?? 2
= 1 ? -
?? 2
1
+
?? 2
1/3
= 1 
Here ?? 2
= 1, ?? 2
= 1/3 
?  ???????????????????????? ?? = v1 + ?? 2
/?? 2
= v1 + 3 = 2#( ?? )  
 
3. RECTANGULAR OR EQUILATERAL 
HYPERBOLA: 
The particular kind of hyperbola in which the lengths of the transverse & conjugate axis are equal is 
called an Equilateral Hyperbola. Note that the eccentricity of the rectangular hyperbola is v2 and the 
length of it's latus rectum is equal to it's transverse or conjugate axis. 
4. AUXILIARY CIRCLE: 
A circle drawn with centre C & T.A. as a diameter is called the Auxiliary Circle of the hyperbola. 
Equation of the auxiliary circle is ?? 2
 +?? 2
= ?? 2
. 
Note from the figure that ?? &?? are called the "Corresponding Points" 
Page 5


HYPERBOLA 
Hyperbolas are fascinating conic sections defined by their eccentricity greater than one. This section 
explores the fundamental properties and equations of hyperbolas, including their foci, vertices, axes, 
and directrices. We'll examine the standard forms of hyperbola equations, key relationships between 
their components, and special cases like conjugate and rectangular hyperbolas. Through a mix of 
theoretical explanations and practical problems, this material provides a comprehensive overview of 
hyperbolas, their unique characteristics, and their applications in mathematics and related fields. The 
Hyperbola is a conic whose eccentricity is greater than unity. (e > 1 ). 
1. STANDARD EQUATION & DEFINITION(S): 
Standard equation of the hyperbola is 
?? 2
?? 2
-
?? 2
?? 2
= 1, where ?? 2
= ?? 2
( ?? 2
- 1) 
or ?? 2
?? 2
= ?? 2
+ ?? 2
 i.e. ?? 2
= 1 +
?? 2
?? 2
 
= 1 + (
 ?????????????????? ???????? 
 ???????????????????? ???????? 
)
2
 
 
(a) Foci: 
?? = ( ???? , 0) &?? '
= ( -???? , 0) . 
(b) Equations of directrices: 
?? =
?? ?? &?? = -
?? ?? 
(c) Vertices: 
?? = ( ?? , 0) &?? '
= ( -?? , 0) . 
(d) Latus rectum: 
(i) Equation: ?? = ±???? 
 
(iii) Ends: ???? ,
?? 2
?? ) , (???? ,
-?? 2
?? ); (-ae, 
?? 2
?? ) , (-???? ,
-?? 2
?? ) 
(e) (i) Transverse Axis: 
The line segment ?? '
?? of length 2a in which the foci ?? '
 & ?? both lie is called the 
Transverse Axis of the Hyperbola. 
(ii) Conjugate Axis: 
The line segment ?? '
?? between the two points ?? '
= ( 0, -?? ) &?? = ( 0, ?? ) is called as the Conjugate Axis of 
the Hyperbola. 
The Transverse Axis & the Conjugate Axis of the hyperbola are together called the 
Principal axes of the hyperbola. 
(f) Focal Property: 
The difference of the focal distances of any point on the hyperbola is constant and equal to transverse 
axis i.e. || ???? | - |?? ?? '
|| = 2?? . The distance ?? ?? '
= focal length. 
(g) Focal distance: 
Distance of any point ?? ( ?? , ?? ) on Hyperbola from foci ???? = ???? - ?? &????
'
= ???? + ?? . 
Problem 1: Find the equation of the hyperbola whose directrix is 2?? + ?? = 1, focus ( 1,2) and 
eccentricity v3. 
Solution: Let ?? ( ?? , ?? ) be any point on the hyperbola and ???? is perpendicular from ?? on the directrix. 
Then by definition  ???? = ?? PM 
  ? ( ???? )
2
= ?? 2
( ???? )
2
? ( ?? - 1)
2
+ ( ?? - 2)
2
= 3 {
2?? + ?? - 1
v4 + 1
}
2
   ? 5( ?? 2
+ ?? 2
- 2?? - 4?? + 5)
= 3( 4?? 2
+ ?? 2
+ 1 + 4???? - 2?? - 4?? )   ? 7?? 2
- 2?? 2
+ 12???? - 2?? + 14?? - 22 = 0  
which is the required hyperbola. 
Problem 2: The eccentricity of the hyperbola 4?? 2
- 9?? 2
- 8?? = 32 is - 
(A) 
v5
3
 
(B) 
v13
3
 
(C) 
v13
2
 
(D) 
3
2
 
Solution:  4?? 2
- 9?? 2
- 8?? = 32 ? 4( ?? - 1)
2
- 9?? 2
= 36 ?
( ?? -1)
2
9
-
?? 2
4
= 1 
Here ?? 2
= 9, ?? 2
= 4 
?  eccentricity ?? = v1 +
?? 2
?? 2
= v1 +
4
9
=
v13
3
 
Problem 3: If foci of a hyperbola are foci of the ellipse 
?? 2
25
+
?? 2
9
= 1. If the eccentricity of the hyperbola 
be 2 , then its equation is - 
(A) 
?? 2
4
-
?? 2
12
= 1 
(B) 
?? 2
12
-
?? 2
4
= 1 
(C) 
?? 2
12
+
?? 2
4
= 1 
(D) none of these 
Solution:   For ellipse ?? =
4
5
, so foci = ( ±4,0) 
For hyperbola ?? = 2, so ?? =
????
?? =
4
2
= 2, ?? = 2v4 - 1 = 2v3 
Hence equation of the hyperbola is 
?? 2
4
-
?? 2
12
= 1 
Problem 4: Find the coordinates of foci, the eccentricity and latus-rectum, equations of directrices for 
the hyperbola 9?? 2
- 16?? 2
- 72?? + 96?? - 144 = 0. 
Solution: 
Equation can be rewritten as 
( ?? -4)
2
4
2
-
( ?? -3)
2
3
2
= 1 so ?? = 4, ?? = 3 
?? 2
= ?? 2
( ?? 2
- 1) ?????????? ?? =
5
4
 
Foci: ?? = ± ae, ?? = 0 gives the foci as ( 9,3) , ( -1,3) 
Centre: ?? = 0, ?? = 0 i.e. ( 4,3) 
Directrices: ?? = ±
?? ?? i.e. ?? - 4 = ±
16
5
 ?  directrices are 5?? - 36 = 0; 5?? - 4 = 0 
Latus-rectum =
2 ?? 2
?? = 2 ·
9
4
=
9
2
 
Do yourself - 1: 
(i) Find the eccentricity of the hyperbola 
?? 2
?? 2
-
?? 2
?? 2
= 1 which passes through ( 4,0) &( 3v2, 2) 
(ii) Find the equation to the hyperbola, whose eccentricity is 
5
4
, focus is (a, 0 ) and whose directrix is 
4?? - 3?? = ?? . 
(iii) In the hyperbola 4?? 2
- 9?? 2
= 36, find length of the axes, the co-ordinates of the foci, the 
eccentricity, and the latus rectum. 
(iv) Find the equation to the hyperbola, the distance between whose foci is 16 and whose eccentricity is 
v2. 
2. CONJUGATE HYPERBOLA: 
Two hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate 
& the transverse axes of the other are called Conjugate Hyperbolas of each other. eg. 
?? 2
?? 2
-
?? 2
?? 2
= 1 & -
?? 2
?? 2
+
?? 2
?? 2
= 1 are conjugate hyperbolas of each other . 
Note that: 
(i) If ?? 1
&?? 2
 are the eccentricities of the hyperbola & its conjugate then ?? 1
 
-2
+ ?? 2
 
-2
= 1. 
(ii) The foci of a hyperbola and its conjugate are concyclic and form the vertices of a square. 
(iii) Two hyperbolas are said to be similar if they have the same eccentricity. 
Problem 5: The eccentricity of the conjugate hyperbola to the hyperbola ?? 2
- 3?? 2
= 1 is- 
(A) 2 
(B) 2/v3 
(C) 4 
(D) 4/3 
Solution: Equation of the conjugate hyperbola to the hyperbola ?? 2
- 3?? 2
= 1 is 
-?? 2
+ 3?? 2
= 1 ? -
?? 2
1
+
?? 2
1/3
= 1 
Here ?? 2
= 1, ?? 2
= 1/3 
?  ???????????????????????? ?? = v1 + ?? 2
/?? 2
= v1 + 3 = 2#( ?? )  
 
3. RECTANGULAR OR EQUILATERAL 
HYPERBOLA: 
The particular kind of hyperbola in which the lengths of the transverse & conjugate axis are equal is 
called an Equilateral Hyperbola. Note that the eccentricity of the rectangular hyperbola is v2 and the 
length of it's latus rectum is equal to it's transverse or conjugate axis. 
4. AUXILIARY CIRCLE: 
A circle drawn with centre C & T.A. as a diameter is called the Auxiliary Circle of the hyperbola. 
Equation of the auxiliary circle is ?? 2
 +?? 2
= ?? 2
. 
Note from the figure that ?? &?? are called the "Corresponding Points" 
 
 
on the hyperbola & the auxiliary circle. '  ' is called the eccentric angle of the point ' ?? ' on the 
hyperbola. ( 0 = ?? < 2?? ) . 
Parametric Equation: 
The equations ?? = ???????? ?? &?? = ???? ?? ???? together represents the hyperbola 
?? 2
?? 2
-
?? 2
?? 2
= 1 where ?? is a 
parameter. The parametric equations; ?? = ???????? h?? , ?? = ???????? h?? also represents the same hyperbola. 
General Note: 
Since the fundamental equation to the hyperbola only differs from that to the ellipse in having -?? 2
 
instead of ?? 2
 it will be found that many propositions for the hyperbola are derived from those for the 
ellipse by simply changing the sign of ?? 2
. 
5. POSITION OF A POINT 'P' w.r.t. A 
HYPERBOLA: 
The quantity 
?? 1
 
2
?? 2
-
?? 1
 
2
?? 2
= 1 is positive, zero or negative according as the point ( ?? 1
, ?? 1
) lies within , upon 
or outside the curve. 
6. LINE AND A HYPERBOLA: 
The straight line ?? = ???? + ?? is a secant, a tangent or passes outside the hyperbola 
?? 2
?? 2
-
?? 2
?? 2
= 1 
according as: ?? 2
>=< ?? 2
?? 2
- ?? 2
. 
Equation of a chord of the hyperbola 
?? 2
?? 2
-
?? 2
?? 2
= 1 joining its two points ?? ( ?? ) &?? ( ?? ) is 
?? ?? ?????? 
?? -?? 2
-
?? ?? ?????? 
?? +?? 2
= ?????? 
?? +?? 2
 
Problem 6: Show that the line ???????? ?? + ???????? ?? = ?? touches the hyperbola 
?? 2
?? 2
-
?? 2
 ?? 2
= 1 if ?? 2
?????? 2
 ?? -
?? 2
?????? 2
 ?? = ?? 2
. 
Solution:   The given line is ???????? ?? + ???????? ?? = ?? ? ???????? ?? = -???????? ?? + ??