Conic Section – Exercise 11.4 – Class XI

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

[(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

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

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

(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

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

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

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

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 + b²= 3²

b² = 9 – 4 = 5

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

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

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

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

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

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

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

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/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)

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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