NCERT Solutions Class 11 Mathematics Conic Sections Exercise 11.4

Question 8:
Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci (0, ±8).

Answer:
Given, Vertices (0, ±5), foci (0, ±8)
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form y²/a² – x²/b² = 1
Since the vertices are (0, ±5), a = 5
Since the foci are (0, ±8), c = 8
We know that a² + b² = c²
=> 5² + b² = 8²
=> b² = 64 – 25
=> b² = 39
Thus, the equation of the hyperbola is y²/25 – x²/39 = 1

Question 9:
Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci (0, ±5).

Answer:
Given, Vertices (0, ±3), foci (0, ±5)
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form y²/a² – x²/b² = 1
Since the vertices are (0, ±3), a = 3
Since the foci are (0, ±5), c = 5
We know that a² + b² = c²
=> 3² + b² = 5²
=> b² = 25 – 9
=> b² = 16
Thus, the equation of the hyperbola is y²/9 – x²/16 = 1

Question 10:
Find the equation of the hyperbola satisfying the give conditions: foci (±5, 0), the transverse axis is of length 8.

Answer:
Given, foci (±5, 0), the transverse axis is of length 8
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form x²/a² – y²/b² = 1
Since the foci are (±5, 0), c = 5
Since the length of transverse axis is 8
So, 2a = 8
=> a = 4
We know that a² + b² = c²
=> 4² + b² = 5²
=> b² = 25 – 16
=> b² = 9
Thus, the equation of the hyperbola is x²/16 – y²/9 = 1

Question 11:
Find the equation of the hyperbola satisfying the give conditions: foci (0, ±13), the conjugate axis is of length 24.

Answer:
Given, foci (0, ±13), the conjugate axis is of length 24.
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form y²/a² – x²/b² = 1
Since the foci are (0, ±13), c = 13
Since the length of conjugate axis is 24
So, 2b = 24
=> b = 12
We know that a² + b² = c²
=> a² + 12² = 13²
=> a² = 169 – 144
=> a² = 25
Thus, the equation of the hyperbola is y²/25 – x²/144 = 1

Question 12:
Find the equation of the hyperbola satisfying the give conditions: Foci (±3√5, 0), the latus rectum is of length 8.

Answer:
Given, Foci = (±3√5, 0) and the latus rectum is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form x²/a² – y²/b² = 1.
Since the foci are (±3√5, 0), c = ±3√5
Length of latus rectum = 8
=> 2b²/a = 8
=> b²/a = 4
=> b² = 4a
We know that a² + b² = c²
=> a² + 4a = (±3√5)²
=> a² + 4a = 45
=> a² + 4a – 45 = 0
=> (a + 9)(a – 5) = 0
=> a = -9, 5
Since a is non-negative, a = 5.
So, b² = 4a = 4 * 5 = 20
Thus, the equation of the hyperbola is x²/25 – y²/20 = 1
 

Question 13:
Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12.

Answer:
Given, Foci = (±4, 0) and the latus rectum is of length 12.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form x²/a² – y²/b² = 1.
Since the foci are (±4, 0), c = 4
Length of latus rectum = 12
=> 2b²/a = 12
=> b²/a = 6
=> b² = 6a
We know that a² + b² = c²
=> a² + 6a = 4²
=> a² + 6a = 16
=> a² + 6a – 16 = 0
=> (a + 8)(a – 2) = 0
=> a = -8, 2
Since a is non-negative, a = 2
So, b² = 6a = 6 * 2 = 12
Thus, the equation of the hyperbola is x²/4 – y²/12 = 1
 
 
 

Question 14:
Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0), e = 4/3.

Answer:
Given, Vertices (±7, 0), e = 4/3
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form x²/a² – y²/b² = 1.
Since the vertices are (±7, 0), a = 7
It is given that e = 4/3
So, c/a = 4/3
=> c/7 = 4/3
=> c = 28/3
We know that a² + b² = c²
=> 7² + b² = (28/3)²
=> b² = 784/9 – 49
=> b² = (789 – 441)/9
=> a² = 343/9
Thus, the equation of the hyperbola is x²/49 – y²/(343/9) = 1
=> x²/49 – 9y²/343 = 1

Question 15:
Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±√10), passing through (2, 3).

Answer:
Given, Foci (0, ±√10), passing through (2, 3).
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form y²/a² – x²/b² = 1.
Since the foci are (0, ±√10), c = √10
We know that a² + b² = c²
=> a² + b² = 10
=> b² = 10 – a²  ……………1
Since the hyperbola passes through point (2, 3),
9/a² – 4/b² = 1    ………..2
From equations 1 and 2, we obtain
      9/a² – 4/(10 – a²)= 1
=> 9(10 – a²) – 4a² = a² (10 – a²)
=> 900 – 9a² – 4a² = 100a² – a⁴
=> a⁴ – 23a² + 90 = 0
=> a⁴ – 18a² – 5a² + 90 = 0
=> a² (a² – 18) – 5(a² – 18) = 0
=> (a² – 18)(a² – 5) = 0
=> a² = 18, 5
In hyperbola, c > a,
=> c²> a²
So, a² = 5
Now, b² = 10 – a²
=> b² = 10 – 5
=> b² = 5
Thus, the equation of the hyperbola is y²/5 – x²/5 = 1.