Conic Section – Exercise 11.4 – Class XI

  1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola (x2)/16 – (y2)/9 = 1

Solution:

The given equation is (x2)/16 – (y2)/9 = 1

or

[(x2)/(42)] – [(y2)/(32)] = 1

On comparing this equation with the standard equation of hyperbola [(x2)/(a2)] – [(y2)/(b2)] = 1 i.e., , we obtain a = 4 and b = 3.

We know that a2 + b2 = c2

c2 = 42 + 32 = 16+ 9 = 25

c = 5

Therefore,

The coordinates of the foci are (±5, 0).

The coordinates of the vertices are (±4, 0).

Eccentricity, e = c/a = 5/4

Length of latus rectum = [2b2]/a = 2×9/4 = 9/2


  1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola (y2)/9 – (x2)/27 = 1

Solution

The given equation is (y2)/9 – (x2)/27 = 1 or (y2)/(32) – (x2)/(√27)2 = 1

On comparing this equation with the standard equation of hyperbola [(x2)/(a2)] – [(y2)/(b2)] = 1  we obtain a = 3 and b = √27

We know that a2 + b2 = c2

c2 = 32 + (√27)2 = 9 + 27 = 36

c = 6

Therefore,

The coordinates of the foci are (0, ±6).

The coordinates of the vertices are (0, ±3).

Eccentricity e = c/a = 6/3 = 2

Length of latus rectum = (2b2)/a = 2×27/3 = 18


  1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y2 – 4x2 = 36

Solution:

The given equation is 9y2 – 4x2 = 36.

It can be written as 9y2 – 4x2 = 36

or

(y2)/4 – (x2)/9 = 1

(y2)/(22) – (x2)/(32) = 1 ——–(1)

On comparing equation (1) with the standard equation of hyperbola [(x2)/(a2)] – [(y2)/(b2)] = 1

we obtain a = 2 and b = 3.

We know that a2 + b2 = c2

c2 = 4 + 9 = 13

c = √13

Therefore,

The coordinates of the foci are (0, ±√13)

The coordinates of the vertices are (0, ±2)

Eccentricity, e = c/a = √13/2

Length of latus rectum = (2b2)/a = 2×9/2 = 9


  1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x2 – 9y2 = 576

Solution:

The given equation is 16x2 – 9y2 = 576.

It can be written as

16x2 – 9y2 = 576

[(x2)/36] – [(y2)/64] = 1

[(x2)/62] – [(y2)/82] = 1 ————(1)

On comparing equation (1) with the standard equation of hyperbola i.e., [(x2)/a2] – [(y2)/b2] = 1 , we obtain a = 6 and b = 8.

We know that a2 + b2 = c2.

c2 = 36 + 64 = 100

c = 10

Therefore,

The coordinates of the foci are (±10, 0).

The coordinates of the vertices are (±6, 0).

Eccentricity, e = c/a = 10/6 = 5/3

Length of latus rectum = (2b2)/a = 2×64/6 = 64/3


  1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y2 – 9x2 = 36

solution:

The given equation is 5y2 – 9x2 = 36.

(y2)/[36/5] – (x2)/4 = 1

(y2)/[6/√5]2  – (x2)/(22) = 1 —————(1)

On comparing equation (1) with the standard equation of hyperbola [(y2)/(a2)] – [(x2)/(b2)] = 1, we obtain a =6/√5 and b = 2.

We know that a2 + b2 = c2

c2 = 36/5 + 4 = 56/5

c = √(56/5) = 2√14/√5

Therefore, the coordinates of the foci are (0, ±2√14/√5)

The coordinates of the vertices are (0, ±6/√5)

Eccentricity, e = c/a = [2√14/√5]/[ 6/√5] = √14/3

Length of latus rectum  = 2b2/a = 2×4/(6/√5) = 4x√5/3


  1. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y2 – 16x2 = 784

solution:

The given equation is 49y2 – 16x2 = 784.

It can be written as

49y2 – 16x2 = 784

(y2)/16 – (x2)/49 = 1

(y2)/(42) – (x2)/(72) = 1 ————(1)

On comparing equation (1) with the standard equation of hyperbola (y2)/(a2) – (x2)/(b2) = 1  i.e., , we obtain a = 4 and b = 7.

We know that a2 + b2 = c2

c2 = 16 + 49 = 65

c = √65

Therefore,

The coordinates of the foci are (0, ±√65)

The coordinates of the vertices are (0, ±4).

Eccentricity, e = c/a = √65/4

Length of latus rectum = (2b2)/a = 2×49/4 = 49/2


  1. Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0)

Solution:

Vertices (±2, 0), foci (±3, 0)

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 (±2, 0), a = 2.

Since the foci are (±3, 0), c = 3.

We know that a2 + b2 = c2

22 + b= 32

b2 = 9 – 4 = 5

Thus, the equation of the hyperbola is (x2)/(4) – (y2)/(5) = 1


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

Solution:

Vertices (0, ±5), foci (0, ±8)

Here, the vertices are on the y-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 = 39

Thus, the equation of the hyperbola is .


 

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

Solution:

Vertices (0, ±3), foci (0, ±5)

Here, the vertices are on the y-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 = 16

Thus, the equation of the hyperbola is (y2)/(9) – (x2)/(16) = 1


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

Solution:

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 the transverse axis is 8, 2a = 8

⇒ a = 4.

We know that a2 + b2 = c2.

∴42 + b2 = 52

⇒ b² = 25 – 16 = 9

b = 3

Thus, the equation of the hyperbola is .(x2)/16 – (y2)/9 = 1


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

Solution:

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 the conjugate axis is 24, 2b = 24 ⇒ b = 12.

We know that a2 + b2 = c2

∴a2 + 122 = 132

⇒ a2 = 169 – 144 = 25

Thus, the equation of the hyperbola is (y2)/25 – (x2)/144 = 1


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

Solution:

Foci (±3√5, 0) 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 = 4a

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 or 5

Since a is non- negative, a = 5

b2 = 4a = 4 x 5 = 20

Thus, the equation of the hyperbola is x2/25 – y2/20 = 1


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

Solution:

Foci (±4, 0), 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 = 6a

We know that a2 + b2 = c2

∴a2 + 6a = 16

a2 + 8a – 6a – 16 = 0

(a + 8)(a – 2) = 0

a = -8 or 2

Since a is non negative, a = 2

b2 = 6a = 6 x 2 = 12

Thus, the equation of the hyperbola is x2/4 – y2/12 = 1


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

Solution:

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 , we know e = c/a

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 = (784 – 441)/9 = 343/9

Thus the equation of the hyperbola x2/49 – (9y2)/343 = 1


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

Solution:

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/a24/b2 = 1 —————-(2)

From equations (1) and (2), we obtain

9/a24/(10-a2) = 1

9(10 – a2) – 4a2 = a2(10 – a2)

90 – 9a2 – 4a2 = a210 – a4

a4– 23a2 + 90 = 0

a4– 18a2– 5a2 + 90 = 0

a2(a2 – 18)-5(a2 – 18) = 0

a2 = 18 or 5

In hyperbola c > a i.e., c2 > a2

a2 = 5

b2 = 10 – a2 = 10 – 5 = 5

Thus the equation of the hyperbola is y2/5 – x2/5 = 1


 

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