Chapter 11 Conic Sections Ex 11.4 Solutions
Question - 11 : - Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13), the conjugate axis is of length 24.
Answer - 11 : -
Foci (0, ±13), theconjugate axis is of length 24.
Here, the foci are on the y-axis.
Therefore, the equation ofthe hyperbola is of the form
.
Since the foci are (0, ±13), c =13.
Since the length of the conjugate axis is24, 2b = 24 ⇒ b = 12.
We know that a2 + b2 = c2.
∴a2 + 122 = 132
⇒ a2 = 169 – 144 = 25
Thus,the equation of the hyperbola is
.
Question - 12 : - Find the equation of the hyperbola satisfying the give conditions: Foci 
, the latus rectum is of length 8.
Answer - 12 : -
Foci, the latus rectum is of length8.
Here, the foci are on the x-axis.
Therefore, the equation ofthe hyperbola is of the form
.
Since the foci are, c =
.
Length of latus rectum = 8
We know that a2 + b2 = c2.
∴a2 + 4a =45
⇒ a2 + 4a –45 = 0
⇒ a2 + 9a – 5a –45 = 0
⇒ (a + 9) (a –5) = 0
⇒ a = –9,5
Since a isnon-negative, a = 5.
∴b2 = 4a = 4× 5 = 20
Thus,the equation of the hyperbola is
.
Question - 13 : - Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12
Answer - 13 : -
Foci (±4, 0), the latusrectum is of length 12.
Here, the foci are on the x-axis.
Therefore, the equation ofthe hyperbola is of the form.
Since the foci are (±4, 0), c =4.
Length of latus rectum =12
We know that a2 + b2 = c2.
∴a2 + 6a =16
⇒ a2 + 6a –16 = 0
⇒ a2 + 8a – 2a –16 = 0
⇒ (a + 8) (a –2) = 0
⇒ a = –8,2
Since a isnon-negative, a = 2.
∴b2 = 6a = 6× 2 = 12
Thus,the equation of the hyperbola is
.
Question - 14 : - Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0), 
Answer - 14 : -
Vertices (±7, 0),
Here, the vertices are on the x-axis.
Therefore, the equation ofthe hyperbola is of the form.
Since the vertices are (±7, 0), a =7.
It is given that
We know that a2 + b2 = c2.
Thus,the equation of the hyperbola is
.
Question - 15 : - Find the equation of the hyperbola satisfying the give conditions: Foci 
, passing through (2, 3)
Answer - 15 : -
Foci, passing through (2, 3)
Here, the foci are on the y-axis.
Therefore, the equation ofthe hyperbola is of the form.
Since the foci are, c =
.
We know that a2 + b2 = c2.
∴ a2 + b2 = 10
⇒ b2 = 10 – a2 … (1)
Since the hyperbola passesthrough point (2, 3),
From equations (1) and(2), we obtain
In hyperbola, c > a,i.e., c2 > a2
∴ a2 = 5
⇒ b2 = 10 – a2 = 10 – 5 = 5
Thus,the equation of the hyperbola is
.